Physics High School

## Answers

**Answer 1**

(a) The ball takes 0.235 seconds to travel the first half of the **distance**.

(b) The ball takes 0.235 seconds to **travel **the second half of the distance.

(c) The ball falls freely 0.055 meters during the first half.

(d) The ball does not fall freely during the second half (0 meters).

To solve this problem, we can use the equations of motion for horizontal and vertical motion separately.

Given:

The initial speed of the baseball (u) = 148 km/h

Distance to the batter (d) = 18.3 m

convert the initial **speed:**

1 km/h = 1000 m/3600 s = 5/18 m/s

(a) Time is taken to travel the first half of the distance:

To find the time taken for the first half of the distance, we can assume that the horizontal **velocity **remains constant throughout the motion. The time taken is given by the formula:

t = d / u

where

t = time taken,

d = distance,

u = initial velocity.

t = (18.3 m) / (148 km/h * 5/18 m/s)

t ≈ 0.235 s

Therefore, the ball takes approximately 0.235 seconds to travel the first half of the distance.

(b) Time is taken to travel the second half of the distance:

Since the horizontal velocity remains constant, the time taken to travel the second half of the distance will also be the same as the first half.

t ≈ 0.235 s

Therefore, the ball also takes approximately 0.235 seconds to travel the second half of the distance.

(c) Distance the ball falls freely during the first half:

Since the initial velocity in the vertical direction is zero, we can use the equation of **motion **for free fall:

s = (1/2) * g * t²

where

s = distance,

g = acceleration due to gravity,

t = time taken.

Substituting the values:

g = 9.8 m/s² (acceleration due to gravity)

s = (1/2) * (9.8 m/s²) * (0.235 s)²

s ≈ 0.055 m

Therefore, the ball falls freely approximately 0.055 meters during the first half.

(d) Distance the ball falls freely during the second half:

The ball does not fall freely during the second half since it has already fallen during the first half. The distance fallen during the second half is zero.

s = 0 m

Therefore, the ball does not fall freely during the second half.

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## Related Questions

A ball thrown vertically upward has an upward velocity of 6.52 m/s at a point 13.2 m above where it was thrown. How long does the ball take to reach that point? X

### Answers

The** time** taken to reach that **point** is 0.6667 seconds.

How to calculate the time taken to reach that point?

In order to calculate the **time** taken to reach that point, we would apply the first equation of motion:

V = U - at

Where:

V represents the final velocity.U represents the initial velocity.a represents the **acceleration** or deceleration.t represents the **time** measured in seconds.

Note: Acceleration due to gravity (g) is equal to 9.8 m/s².

By substituting the given parameters into the first equation of motion, we have;

Time, t = (V - U)/a

Time, t = (6.52 - 0)/9.8

**Time**, t = 0.6667 seconds.

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A particle with a charge of -2.32 μC and a mass of 3.31×10-6 kg is released from rest at point A and accelerates toward point B, arriving there with a speed of 35.8 m/s. The only force acting on the particle is the electric force. What is the potential difference VB - VA between A and B?

### Answers

The **potential difference** VB - VA between points A and B is approximately -76.7 volts. The negative sign indicates that the particle moved from a region of higher potential (A) to a region of lower potential (B).

To determine the potential difference (VB - VA) between points A and B, we can use the equation:

ΔV = qΔV

where ΔV is the potential difference, q is the charge, and ΔV is the change in electric potential.

In this case, the only force acting on the particle is the **electric force**. The work done by the electric force in moving the particle from point A to point B is equal to the change in its electric potential energy. Therefore, the work done by the electric force is given by:

W = ΔPE = q(ΔV)

The work done is also equal to the change in kinetic energy:

W = ΔKE = (1/2)mvB² - (1/2)mvA²

where m is the mass of the particle, vB is the final **velocity **at point B, and vA is the initial velocity at point A (which is 0 since the particle is at rest).

Setting these two expressions for work equal to each other, we have:

q(ΔV) = (1/2)mvB² - (1/2)mvA²

Since vA = 0, the equation simplifies to:

q(ΔV) = (1/2)mvB²

Plugging in the values:

q = -2.32 μC = -2.32 × 10⁻⁶ C

m = 3.31 × 10⁻⁶ kg

vB = 35.8 m/s

We can now solve for ΔV:

ΔV = (1/2)mvB² / q

= (1/2)(3.31 × 10⁻⁶ kg)(35.8 m/s)² / (-2.32 × 10⁻⁶ C)

≈ -76.7 V

Therefore, the **potential difference** VB - VA between points A and B is approximately -76.7 volts. A lower potential region (B) is indicated by the negative sign, which shows that the particle moved from a higher potential region (A).

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A 16 g piece of Styrofoam carries a net charge of −0.5μC and floats above the center of a very large horizontal sheet of plastic that has a uniform charge density on its surface. What is the charge per unit area on the plastic sheet? The acceleration due to gravity is 9.8 m/s

2

and the permittivity of free space is 8.85419×10

−12

C

2

/N/m

2

. Answer in units of μC/m

2

. 01710.0 points Two electrons in an atom are separated by 1.5×10

−10

m, the typical size of an atom. What is the force between them? The Coulomb constant is 9×10

9

N⋅m

2

/C

2

. Answer in units of N.

### Answers

Part 1: The **charge **per unit area on the plastic sheet is 3.5173*[tex]d^{-5}[/tex] μC/m Part 2 : force between the two electrons is 3.6*[tex]10^{-8}[/tex] N.

In order to solve the question, we must consider the forces acting on the Styrofoam. In order for it to float above the plastic sheet, the **electrical force** (FE) must be equal to the gravitational force (FG) acting on it.i.e., FG = FEThe gravitational force acting on the Styrofoam is given by;FG = m*gwhere m is the mass of the Styrofoam and g is the acceleration due to gravity

Using the values given, we get: FG = 16*[tex]10^{3}[/tex]kg * 9.8 m/s2= 0.1568 NThe electrical force acting on the Styrofoam is given by:FE = k*Q*q/[tex]d^{2}[/tex]where k is the Coulomb's constant, Q is the total charge on the sheet of plastic, q is the charge on the Styrofoam, and d is the distance between the Styrofoam and the sheet of plastic.The Styrofoam carries a net charge of -0.5 μC. Hence, q = -0.5 μC = -0.5 * [tex]10^{-6}[/tex] C.

However, since it is very large, we can assume that the Styrofoam is close to its center, and hence that the area is much larger than that of the **Styrofoam**. Let's assume that the area of the sheet is 1 m2. Then, we get:σ = Q/A= 3.5173*[tex]d^{-5}[/tex]C/1 m= 3.5173*[tex]d^{-5}[/tex] μC/m2,Hence, the charge per unit area on the plastic sheet is 3.5173*[tex]d^{-5}[/tex] μC/m

Part 2:According to Coulomb's law, the force (F) between two point charges is given by:F = k*q1*q2/d^2where k is the **Coulomb's constant,** q1 and q2 are the magnitudes of the charges, and d is the distance between them.Substituting the given values into the equation, we get: F = k*q1*q2/d^2= 9*[tex]10^{-98}[/tex] , * / (1.5*10^-10 m)^2= 3.6*[tx]10^{-8}[/tex] N Hence, the **force **between the two electrons is 3.6*[tex]10^{-8}[/tex] N.

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A small sphere of mass m = 6.40 g and charge q1 = 30.4 nC is attached to the end of a string and hangs vertically as in the figure. A second charge of equal mass and charge q2 = −58.0 nC is located below the first charge a distance d = 2.00 cm below the first charge as in the figure.

(a) Find the tension in the string.

(b) If the string can withstand a maximum tension of 0.180 N, what is the smallest value d can have before the string breaks?

### Answers

A small sphere of **mass** m = 6.40 g and charge q1 = 30.4 nC is attached to the end of a string and hangs vertically as in the figure F_electrostatic_max is the maximum tension the string can withstand

(a) To find the tension in the string, we need to consider the forces acting on the hanging sphere. There are two forces acting on the sphere: the gravitational force (weight) and the electrostatic force due to the presence of the second charge.

1. **Gravitational** force:

The gravitational force acting on the sphere can be calculated using the formula:

F_gravity = m * g

where m is the mass of the sphere and g is the acceleration due to gravity.

2. Electrostatic force:

The electrostatic force between the two charges can be calculated using **Coulomb's** law:

F_electrostatic = (k * |q1 * q2|) / d^2

where k is the Coulomb's constant, q1 and q2 are the charges, and d is the distance between the charges.

Since the charges have opposite signs, we take the absolute value in the equation.

Now, we can calculate the **tension** in the string as the vector sum of the gravitational force and the electrostatic force:

Tension = F_gravity + F_electrostatic

(b) To find the smallest value of d before the string breaks, we need to find the maximum electrostatic force that the string can withstand. We can rearrange **Coulomb's** law to solve for the distance d:

d = sqrt((k * |q1 * q2|) / F_electrostatic_max)

Substituting the given values into the equation will give us the smallest value of d before the **string** breaks.

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[] An electric field line points to the direction of higher electric potential. [ ] An equipotential surface is always parallel to electric field lines. [ ] The electrostatic potential generated by multiple charges, is the vector addition of the potentials generated by individual charges. [ ] Inside a conductor, at the electrostatic equilibrium, the electric field and potential are both zero. [ ] The electric fields created by a point charge, a conducting sphere, and an insulator sphere are identical, if they all have the same amount of charge and the field is measured outside the spheres.

### Answers

The statement, "An **electric field** line points to the direction of higher electric potential." is: (False)

The statement, " An equipotential surface is always parallel to electric field lines." is: (False)

The statement, " The electrostatic potential generated by multiple charges is the vector addition of the potentials generated by individual charges." is: (True)

The statement, " Inside a conductor, at the electrostatic equilibrium, the electric field and potential are both zero." is: (True)

The statement, " The electric fields created by a point charge, a conducting sphere, and an insulator sphere are identical if they all have the same amount of charge and the field is measured outside the spheres." is: (False)

- An electric field line points in the direction of the electric field, not necessarily in the direction of higher electric potential. **Electric potential **decreases in the direction of the electric field.

- An equipotential surface is always perpendicular to the electric field lines, not necessarily parallel. This is because the electric field is always perpendicular to the **equipotential surfaces**.

- The electrostatic potential generated by multiple charges is indeed the vector addition of the potentials generated by individual charges. This is due to the principle of superposition.

- Inside a conductor at electrostatic equilibrium, the electric field is zero. The electric potential is also constant throughout the** conductor** and does not vary.

- The electric fields created by a point charge, conducting sphere, and insulator sphere are not identical. The electric field created by a point charge decreases as 1/r², while for conducting and insulator spheres, it depends on the charge distribution and the material properties.

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7) The weights of the two boxes are W1 =100lb and W2

=50lb. The coefficients of friction between the left box and the inclined surface are μs

=0.12 and μk

=0.10. Determine the tension the man must exert on the rope to pull the boxes upward at a constant rate. 8) In Problem 7), for what range of tensions exerted on the rope by the man will the boxes remain stationary?

### Answers

**Answer:**

The** tension** that the man must exert on the rope to pull the boxes upward at a constant rate is 162 lb.

The range of tensions **exerted** on the rope by the man for which the boxes will remain stationary is Tension ≤ 18lb.

7) In order to determine the tension the man must exert on the rope to pull the boxes upward at a constant rate, it is required to calculate the **frictional force**. Using the given values of weight and** friction coefficients,** it is given that;

Fs = μsW1

= 0.12(100)

= 12lb

As the boxes are being pulled at a constant rate, the tension in the rope is equivalent to the sum of frictional force and the total weight of the boxes. Thus;

T = W1 + W2 + Fs

= 100 + 50 + 12

= 162lb

Therefore, the **tension** that the man must exert on the rope to pull the boxes upward at a constant rate is 162 lb.

8) For the boxes to remain stationary, the tension in the rope must be equal to the **maximum static friction. **Therefore;

Fs ≤ μsN

Fs ≤ μs(W1 + W2)

Tension ≤ μs(W1 + W2)

Where;

μs = 0.12

W1 = 100lb

W2 = 50lb

Thus;

Tension ≤ 0.12(100 + 50)

Tension ≤ 18lb

Therefore, the **range of tensions **exerted on the rope by the man for which the boxes will remain stationary is Tension ≤ 18lb.

**Answer:**

The** tension** that the man must exert on the rope to pull the boxes upward at a constant rate is 162 lb.

The range of tensions **exerted** on the rope by the man for which the boxes will remain stationary is Tension ≤ 18lb

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A football is kicked from level ground with a velocity of 5.7 m/s at an angle of 36.9 above the horizontal. How high does it reach (in meters) before falling back to the ground? Enter your answer (only the number, not the unit) rounded to ONE decimal place.

### Answers

A football is kicked from level ground with a velocity of 5.7 m/s at an angle of 36.9 above the **horizontal** the football reaches a maximum height of approximately 1.65 meters before falling back to the ground.

To determine the maximum height reached by the football, we can analyze its vertical motion. We'll consider the initial vertical velocity, the **acceleration** due to gravity, and the time it takes for the football to reach its maximum height.

Given:

Initial velocity (v₀) = 5.7 m/s

Launch angle (θ) = 36.9°

First, we can find the **vertical** component of the initial velocity:

v₀y = v₀ * sin(θ)

Next, we'll calculate the time it takes for the football to reach its maximum height. At the highest point of its trajectory, the vertical **velocity** will be zero (vₓ = 0). Using this information, we can find the time of flight (t) using the equation:

vₓ = v₀x + aₓt

Since there is no horizontal acceleration (aₓ = 0) and the initial horizontal velocity (v₀x) is equal to v₀ * cos(θ), we have:

0 = v₀ * cos(θ) + 0 * t

**Solving** for t, we find:

t = 0

The time of flight is zero, indicating that the football reaches its maximum height instantaneously.

Now, we can determine the maximum **height** (h) using the formula:

h = v₀y² / (2 * g)

Substituting the values we have:

h = (v₀ * sin(θ))² / (2 * g)

Considering the given value of g (acceleration due to **gravity**):

h = (5.7 * sin(36.9°))² / (2 * 9.8)

Calculating the value:

h ≈ 1.65 meters

Therefore, the football reaches a maximum height of approximately 1.65 **meters** before falling back to the ground.

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please help with correct

answer

Calculate the mass, in Msun, of a galaxy cluster where the average galactic velocity is \( 1054.0 \mathrm{~km} / \mathrm{s} \) and the diameter of the cluster is \( 4.5 \mathrm{Mpc} \). Msun

### Answers

The mass, in **Msun**, of a galaxy cluster where the average galactic velocity is 1054.0 km/s and the diameter of the cluster is 4.5 Mpc can be calculated using the formula, \[\mathrm{M} = \frac{\mathrm{vd}^2}{\mathrm{G}}\]Where, M is the mass of the **galaxy cluste**r, v is the average **galactic** velocity and d is the diameter of the cluster. G is the gravitational constant which is equal to \[4.302 \times 10^{-3} \mathrm{~pc} \mathrm{~km}^{-2} \mathrm{~s}^{-2}\]Substituting the given values in the above formula,\[\mathrm{M} = \frac{(1054.0 \mathrm{~km} / \mathrm{s})^2 \times (4.5 \mathrm{Mpc})}{4.302 \times 10^{-3} \mathrm{~pc} \mathrm{~km}^{-2} \mathrm{~s}^{-2}}\]Converting Mpc to km,\[\mathrm{M} = \frac{(1054.0 \mathrm{~km} / \mathrm{s})^2 \times (4.5 \times 10^6 \mathrm{~pc}) \times (3.086 \times 10^{13} \mathrm{~km} / \mathrm{~pc})}{4.302 \times 10^{-3} \mathrm{~pc} \mathrm{~km}^{-2} \mathrm{~s}^{-2}}\]On solving the above equation,\[\mathrm{M} = 9.46 \times 10^{14} \mathrm{~M}_\odot\]Therefore, the mass of the galaxy cluster is 9.46 × 10¹⁴ M_sun.

About G**alactic**

A **Galactic** is a massive, gravitationally bound system consisting of stars, interstellar medium gas and dust, and dark matter – an important but poorly understood component. What does a galaxy consist of? Galactic is a group of stars that form a system. The galaxy consists of more than one large celestial body and is surrounded by other celestial bodies. In astronomy, a galaxy is an extensive system of stars, dust, and gas.

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Given a system which has three identical components, with 2 of those components in cold standby, as depicted in the figure

Assuming the component failure rate is 1 failure per four months, what is the system reliability over 9 months?

### Answers

The** system reliability** over 9 months, given three identical components with 2 in cold standby and a component failure rate of 1 failure per four months, is approximately 0.943.

To calculate the system reliability over 9 months, we need to consider the reliability of each component and their arrangement in the system.

Each **component** has a failure rate of 1 failure per four months, which means its reliability can be calculated as the complement of the failure rate: 1 - (1/4) = 3/4.

Since there are three identical components in the system, and two of them are in cold standby, the system can continue to function as long as at least one component is operational. The reliability of the system in this **configuration** can be calculated using the formula for parallel reliability: R_system = 1 - [tex](1 - R_component)^n[/tex], where R_component is the reliability of a single component and n is the number of components.

Using this formula, the reliability of the system over 9 months is calculated as follows:

R_system = 1 - (1 -[tex](3/4))^3[/tex] = 1 -[tex](1 - 3/4)^3[/tex] = 1 - [tex](1/4)^3[/tex] = 1 - 1/64 = 63/64 ≈ 0.943.

Therefore, the system reliability over 9 months, under the given conditions, is approximately 0.943.

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BULK DEFORMATION PROCESSES Plot the force vs. reduction in height curve in open-die forging of a cylindrical, annealed copper specimen 12 mm high and 30 mm in diameter, up to a reduction of 50%, for the cases of (a) no friction between the flat dies and the specimen, (b) =0.12, and (c)=0.25. Ignore barreling. Use average-pressure formulas.

### Answers

Bulk deformation **processes** are used for the shaping of materials by means of plastic deformation. Open-die forging is one of these processes that are used for the production of simple parts that **require** a decrease in the height of the cylindrical piece of metal.

Open-die forging is a method used for shaping metal where a metal piece is placed on a stationary anvil and then struck repeatedly with a **hammer** until the metal piece is formed into the desired shape. The force vs reduction in height curve can be plotted for a cylindrical, annealed copper specimen 12 mm high and 30 mm in diameter up to a reduction of 50% for the cases of (a) no friction between the flat dies and the specimen, (b) = 0.12, and (c) = 0.25. This can be done with the use of average-pressure formulas.

The curve can be plotted as follows:

(a) No friction between the flat dies and the specimenIn this case, the value of friction is assumed to be zero. In this case, the force required for the **deformation** is equal to the average pressure multiplied by the area of the specimen. The area of the specimen is given by the formula: Area = pi * (Diameter/2)^2 = pi * (30/2)^2 = 706.5 mm^2 The average pressure can be calculated by using the formula: P = (1/2) * F/A, where P is the average pressure, F is the **force**, and A is the area of the specimen. The graph can be plotted by taking the values of average pressure on the y-axis and the values of reduction in height on the x-axis. The curve is then plotted accordingly.

(b) Friction coefficient = 0.12In this case, the value of friction coefficient is assumed to be 0.12. The force required for deformation is given by the formula: F = A * P * (1 + 2 * f), where F is the force, A is the area of the specimen, P is the average pressure, and f is the friction coefficient. The graph can be plotted by taking the values of average pressure on the y-axis and the **values** of reduction in height on the x-axis. The curve is then plotted accordingly.

(c) **Friction** coefficient = 0.25In this case, the value of friction coefficient is assumed to be 0.25. The force required for deformation is given by the formula: F = A * P * (1 + 2 * f), where F is the force, A is the area of the specimen, P is the average pressure, and f is the friction coefficient. The graph can be plotted by taking the values of average pressure on the y-axis and the values of reduction in height on the x-axis. The curve is then plotted accordingly.

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A 2400 V,60 Hz generator is connected to the high-voltage side of a transformer of a 2400/220 V, 100kVA step-down transformer. If the resultant core flux is 0.08 Wb (max), determine a) The number of turns of wire in the secondary coil. b) The new core flux if the driving voltage is increased by 10 percent and the frequency is decreased by 2 percent.

### Answers

The number of turns of wire in the secondary coil is N1 = N2 * 10.909. To determine the new core flux, calculate the new driving voltage and **frequency **using the given **percentages**, and then use the formula to find the new core flux. The number of turns of wire in the secondary coil can be determined using the formula:

V1/V2 = N1/N2

Where V1 and V2 are the **voltages **on the primary and secondary sides respectively, and N1 and N2 are the number of turns on the primary and secondary coils respectively.

In this case, V1 = 2400 V and V2 = 220 V. Substituting these values into the formula, we get:

2400/220 = N1/N2

Simplifying the equation, we find:

N1 = N2 * (2400/220)

Now we can substitute the given **values **to calculate N1:

N1 = N2 * (10.909)

Since the transformer is a step-down **transformer**, the number of turns in the secondary coil is more than the number of turns in the primary coil.

For part b, to determine the new core flux, we need to calculate the new driving voltage and frequency.

The new driving voltage is increased by 10 percent, which means the new voltage is 2400 V + (10/100) * 2400 V.

The new frequency is decreased by 2 percent, which means the new frequency is 60 Hz - (2/100) * 60 Hz.

Substituting these values into the formula, we can calculate the new core flux using the same method as before.

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a) The number of turns of wire in the **secondary** **coil** can be calculated using the turns ratio.

b) The new core flux can be determined by using the formula with the new voltage and frequency values.

To determine the number of turns of wire in the secondary coil, we can use the formula for transformer turns ratio. The turns ratio is given by the ratio of the number of turns in the primary coil to the number of turns in the secondary coil. In this case, the turns ratio is equal to the voltage ratio of the transformer, which is 2400/220.

a) To find the number of turns in the secondary coil, we can rearrange the formula: number of turns in the secondary coil = (turns **ratio**) × (number of turns in the primary coil). Substituting the values, we get:

Number of turns in the secondary coil = (2400/220) × (number of turns in the primary coil)

b) If the driving voltage is increased by 10 percent, the new voltage will be 2400 V + (10/100) × 2400 V. If the frequency is decreased by 2 percent, the new **frequency** will be 60 Hz - (2/100) × 60 Hz.

To find the new core flux, we can use the formula: new core flux = old core flux × (new voltage / old voltage) × (old frequency / new frequency). Substituting the values, we get:

New core flux = 0.08 Wb × [(2400 V + (10/100) × 2400 V) / 2400 V] × [(60 Hz / (60 Hz - (2/100) × 60 Hz)]

This will give us the new **core flux**.

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The calculation of the magnetic moment of a current loop leads to the line integral

I

r × dr.

(a) Integrate around the perimeter of a current loop (in the x y-plane) and show that

the scalar magnitude of this line integral is twice the area of the enclosed surface.

(b) The perimeter of an ellipse is described by r = eˆxa cos θ + eˆyb sin θ. From part (a)

show that the area of the ellipse is πab.

### Answers

The** line integral **is twice the **enclosed surface**. The **area** of the **ellipse **is ab, which comes from the line integral.

How to show that the area of the ellipse is πab.

(a) We must evaluate the **line integral** of the **vector quantity** I(r dr) in order to integrate around the perimeter of a current loop in the** xy plane.**

Assuming the **current loop** has a **radius** of R, the angle can be used to parameterize it as follows:

[tex]r = R(cos \theta, sin \theta, 0).[/tex]

Now, we calculate the **cross-product** [tex]r * dr[/tex]:

[tex]r * dr = (Rcos\theta, Rsin\theta, 0) (-Rsin\thetad\theta, Rcos \thetad\theta, 0)\\ = (0, 0, R^{2} d\theta).[/tex]

The **magnitude** of the cross-product is[tex]R^{2 } d\theta[/tex],

We integrate with regard to[tex]\theta "from 0 to 2$\pi$"[/tex] order to integrate around the loop:

[tex]\int[0 to 2$\pi$] R^{2} d\theta = R^{2}[\theta] from 0 to 2$\pi$ = R^{2}(2$\pi$ - 0) = 2$\pi$R^{2}[/tex]

The **line integral's scalar magnitude** is [tex]2$\pi$R^{2}[/tex], which is twice the **enclosed surface's area.**

(b) We can use the result from part (a) to calculate the** area of an ellipse **described by [tex]r = exa cos \theta+ eyb sin \theta[/tex], where a and b are the semi-major and semi-minor axes, respectively.

The area enclosed by the** loop** is twice the **magnitude** of the line integral.

Therefore, we can see that R2 [tex]R^{2} = ab[/tex] from this equation:[tex]2$\pi$R^{2 = $\pi$ab.[/tex]

As a result, the following gives the ellipse's area:

The **ellipse's area** is therefore ab, as demonstrated by the part (a) result. [tex]Area = R^{2 }= ab[/tex]

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In a single balanced switching mixer using MOSFETs, the drain current through the switching MOSFETs is 50mA and VGS=2V. MOSFET's gm=25mA/V. What value of the drain resistor (in KΩ) is to be used for a voltage conversion gain of 65.6 for the ωRF-ωLO frequency component? Approximate your answers to 2 decimal points

### Answers

The **approximate** value of the drain resistor to be used for a voltage **conversion **gain of 65.6 for the ωRF-ωLO frequency component is 2.62 KΩ.

To calculate the value of the drain resistor in a single balanced switching mixer using MOSFETs, we need to consider the voltage conversion gain and the given **parameters**. The voltage conversion gain (Av) is given as 65.6, and it is defined as the ratio of the output voltage (Vout) to the input voltage (Vin).

In this case, we are interested in the ωRF-ωLO frequency component. We can use the formula for voltage conversion gain: Av = -gm * Rd

where gm is the transconductance of the MOSFET and Rd is the drain resistor. Given that gm is 25mA/V and the drain current through the **switching** MOSFETs is 50mA, we can calculate the value of Rd.

First, let's convert the drain current from mA to A:

Id = 50mA = 0.05A .Since gm = Id/VGS, we can calculate the **gate-to-source** voltage (VGS):

VGS = Id/gm = 0.05A / 25mA/V = 2V

Now, we can substitute the values into the formula for voltage conversion gain: 65.6 = - (25mA/V) * Rd

Solving for Rd, we get: Rd = -65.6 / (25mA/V) Rd ≈ -2.624KΩ

Since we cannot have a **negative** resistance, we take the absolute value: Rd ≈ 2.624KΩ.

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A baggage handler at an airport applies a constant horizontal force with magnitude F1 to push a box, of mass m, across a rough horizontal surface with a very small constant acceleration a.

A. ) The baggage handler now pushes a second box, identical to the first, so that it accelerates at a rate of 2a. How does the magnitude of the force F2 that the handler applies to this box compare to the magnitude of the force F1 applied to the first box?

B. )Now assume that the baggage handler pushes a third box (Figure 2) of mass m/2 so that it accelerates at a rate of 2a. How does the magnitude of the force F3 that the handler applies to this box compare to the magnitude of the force F1 applied to the first box?

### Answers

A. The magnitude of the **force** F₂ applied to the second box is twice the magnitude of the force F₁ applied to the first box ; B. The magnitude of the force F₃ applied to the third box is half the magnitude of the force F₁ applied to the first box.

A. In order to accelerate the second box at a rate of 2a, the magnitude of the force F₂ that the baggage handler applies to this box will be twice the **magnitude** of the force F₁ applied to the first box. This is because Newton's second law states that the force applied to an object is directly proportional to its mass and acceleration.

To understand this concept, let's consider an example. Suppose the first box has a mass of 10 kg and the baggage handler applies a force of 20 N to achieve an acceleration of a m/s². Now, for the second box, which is identical in mass, the baggage handler needs to apply a force of 40 N (twice the magnitude of F₁) in order to achieve an acceleration of 2a m/s².

B. Now, if the baggage handler pushes a third box with a** mass** of m/2 (half the mass of the first box) so that it accelerates at a rate of 2a, the magnitude of the force F₃ that the handler applies to this box can be determined using the same principle as in the previous case. Since the mass of the third box is half that of the first box, the force required to achieve the same acceleration of 2a will also be halved.

Continuing from the previous example, if the mass of the first box is 10 kg, then the mass of the third box will be 10/2 = 5 kg. To achieve an** acceleration** of 2a, the baggage handler will need to apply a force of 10 N (half the magnitude of F₁) to the third box.

In summary:

A. The magnitude of the force F₂ applied to the second box is twice the magnitude of the force F₁ applied to the first box.

B. The magnitude of the force F₃ applied to the third box is half the magnitude of the force F₁ applied to the first box.

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An air-track glider is attached to a spring. The glider is pulled to the right and released from rest at t=0 s. It then oscillates with a period of 1.5 s and a maximum speed of 42 cm/s. Part B What is the glider's position at t=0.28 s ? Express your answer in centimeters.

### Answers

The glider's **position **at t = 0.28 s is approximately 3.355 cm.

To find the glider's **position **at t = 0.28 s, we can use the **equation **for the **simple harmonic motion**:

x = A x cos(ωt + φ)

Where:

x is the position of the glider,

A is the amplitude (maximum displacement),

ω is the angular frequency (2π divided by the period),

t is the time,

φ is the phase constant.

Given:

Period (T) = 1.5 s

Maximum speed [tex](v_{max}) = 42\ \dfrac{cm}{s}[/tex]

We can start by finding the **angular frequency**:

[tex]\omega= \dfrac{2\pi} { T}\\\omega= \dfrac{2\pi} { 1.5 }\\\omega = 4.18879 \ \dfrac{rad}{s}[/tex]

Since the glider is released from rest, the **phase **constant φ is 0.

Now we can find the glider's position at t = 0.28 s:

x = A x cos(ωt)

To find A (**amplitude**), we can use the relationship between the maximum speed and the **angular frequency**:

[tex]v_{max} = A\times\omega[/tex]

42 = A x 4.18879 rad/s

A ≈ 10.053 cm

Now substitute the values into the equation:

x = 10.053 x cos(4.18879 x 0.28 )

x ≈ 10.053 x cos(1.170622)

Using a calculator, we find:

x ≈ 3.355 cm

Therefore, the glider's position at t = 0.28 s is approximately 3.355 cm.

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The peak voltage (V

pk

) of the waveform was given as 50 Volts. By using the analytical method just defined calculate the RMS voltage?

### Answers

The **RMS voltage** of the waveform with a peak voltage of 50 volts is approximately 35.355 V.

The RMS (Root Mean Square) voltage can be calculated using the formula: RMS voltage (Vrms) = **Peak **voltage (Vpk) / √2 In this case, the peak voltage is given as 50 volts.

Substituting this value into the formula, we get: Vrms = 50 V / √2 To simplify the calculation, let's **rationalize **the denominator (√2) by multiplying both the numerator and denominator by √2: Vrms = (50 V / √2) * (√2 / √2) = (50√2 V) / 2 = 25√2 V

Approximating the value of √2 to 1.414, we can calculate the RMS **voltage**: Vrms ≈ 25 × 1.414 V ≈ 35.355 V after using the analytical method, we got the value of peak voltage as 35.355 V

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Each plate of a parallel plate capacitor has a surface area of 2 cm2 and a plate separation of 2 mm. How much charge does the capacitor store when it is connected to a 6 V battery? What is the electric field between the plates? What is the energy density between the plates? Answers, 5.31 x 10-12 C, 3 x 103 V/m, 3.98 x 10-5 J/m3.

### Answers

The charge stored in the **parallel plate capacitor** is 5.31 x 10^-12 C. The electric field between the plates is 3 x 10^3 V/m. The energy density between the plates is 3.98 x 10^-5 J/m³.

The charge stored in the capacitor is calculated by using the formula:

Q = C * V

where Q is the charge, C is the capacitance, and V is the voltage.

The **capacitance **(C) of a parallel plate capacitor is calculated using the formula:

C = (ε₀ * A) / d

where ε₀ is the permittivity of free space (8.85 x 10^-12 F/m), A is the surface area of each plate, and d is the plate separation.

Calculating the charge (Q) first:

A = 2 cm² = 2 x 10^-4 m²

d = 2 mm = 2 x 10^-3 m

C = (8.85 x 10^-12 F/m) * (2 x 10^-4 m²) / (2 x 10^-3 m)

C = 8.85 x 10^-20 F

V = 6 V

Q = (8.85 x 10^-20 F) * (6 V)

Q = 5.31 x 10^-19 C

Therefore, the **charge stored** in the capacitor is 5.31 x 10^-12 C.

The electric field between the plates is calculated using the formula:

E = V / d

where E is the electric field, V is the voltage, and d is the plate separation.

E = (6 V) / (2 x 10^-3 m)

E = 3 x 10^3 V/m

Therefore, the **electric field** between the plates is 3 x 10^3 V/m.

The energy density between the plates is calculated by using the formula:

U = (1/2) * ε₀ * E²

where U is the energy density, ε₀ is the permittivity of free space, and E is the electric field.

U = (1/2) * (8.85 x 10^-12 F/m) * (3 x 10^3 V/m)²

U = 3.98 x 10^-5 J/m³

Therefore, the **energy density** between the plates is 3.98 x 10^-5 J/m³.

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Express the speed of light, 3.00×10

8

m/s in: a) Feet per microseconds (1 point) b) millimeters per picoseconds

### Answers

The **speed** **of light** is approximately 983,571,056.43 ft/μs when expressed in feet per microseconds, and it is 3.00 × 10¹¹ mm/ps when expressed in **millimeters** per picoseconds.

(a) The speed of light, 3.00 × 10⁸ m/s, in feet per **microseconds** is approximately 983,571,056.43 ft/μs.

To convert meters to feet, we use the conversion factor 1 m = 3.28084 ft. Similarly, to convert seconds to **microseconds**, we use the conversion factor 1 s = 1,000,000 μs. Multiplying these conversion factors by the speed of light in meters per second, we get:

3.00 × 10⁸ m/s × (3.28084 ft/1 m) × (1/1,000,000 μs) ≈ 983,571,056.43 ft/μs.

Therefore, the speed of light in feet per microseconds is approximately 983,571,056.43 ft/μs.

(b) The speed of light, 3.00 × 10⁸ m/s, in millimeters per **picoseconds** is 3.00 × 10¹¹ mm/ps.

To convert meters to millimeters, we use the conversion factor 1 m = 1,000 mm. Similarly, to convert seconds to picoseconds, we use the conversion factor 1 s = 1,000,000,000,000 ps. Multiplying these conversion factors by the speed of light in **meters** per second, we get:

3.00 × 10^8 m/s × (1,000 mm/1 m) × (1/1,000,000,000,000 ps) = 3.00 × 10¹¹ mm/ps.

Therefore, the speed of light in millimeters per picoseconds is 3.00 × 10¹¹ **mm/ps**.

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temperature inversions (stable layers) in air layers above the ground can enhance thunderstorm development.

True or False

### Answers

The development of thunderstorms can be aided by** temperature inversions **(stable layers) in the air layers above the ground. This statement is true.

The temperature of the air rises with** height in a layer** of the atmosphere called a temperature inversion. An inversion is present in the lower part of a cap.

The cap is an air mass that is located above the inversion and is relatively warm. As this layer gets cooler than the** surroundings,** air parcels rising into it find it difficult to rise. Air near the ground cools down faster than air in the atmosphere.

Most likely, this will occur when the sky is clear and the wind is little or quiet. In** low areas** (such valleys protected from the wind), cooling will happen more quickly. This frequently occurs in the late afternoon or early evening (before to sunset) and lasts into the next morning (after sunrise) for a number of hours.

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Assuming that the solar system is in uniform dust of density rho, find the force felt by a planet of mass m at a distance r from the center of the sun. Newton's gravitational constant is G and the Sun's mass is M

### Answers

The **force** felt by a planet due to dust in the solar system, assuming uniform dust density ρ, can be calculated using the simplified formula: Force due to dust = (4π/3) × r × ρ × G × M × m.

Assuming that the **solar system** is in uniform dust of density ρ, the force felt by a planet of mass m at a distance r from the center of the Sun can be calculated as follows:

Force due to dust = (4π/3) × (r³) × ρ × G × M × m / r²

Where,

ρ = density of dust

G = Newton's **gravitational constant**

M = mass of the Sun

m = mass of the planet

r = distance from the center of the Sun

The above formula can be simplified to:

Force due to dust = (4π/3) × r × ρ × G × M × m

In conclusion, the force felt by a planet of mass m at a **distance **r from the center of the Sun assuming that the solar system is in uniform dust of density ρ is given by the equation:

Force due to dust = (4π/3) × r × ρ × G × M × m

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A coil of 3 Heny and 5 ohims is connected across a DC supply of 20 volts. Calculate: (a) the time corstant for this coil. econds (b) the time for curent to reach its fuil value. Seconds (c) the final value of current after it has settled down on switch on.

### Answers

a. the **time constant** for this coil is approximately **0.6 seconds.**

b. it takes approximately** 3 seconds** for the current to reach its full value.

c. the final value of the **current** after it has settled down is** 4 Amperes.**

To calculate the **time constant** for the coil, we use the formula:

τ = L / R

where τ is the time constant, L is the inductance of the coil, and R is the resistance of the coil.

Given:

**Inductance** (L) = 3 H

**Resistance** (R) = 5 Ω

(a) Calculating the **time constant**:

τ = 3 H / 5 Ω

τ ≈ 0.6 seconds

Therefore, the time constant for this coil is approximately 0.6 seconds.

(b) The time for the **current** to reach its full value is typically considered to be around 5 time constants. Therefore, the time for the current to reach its full value would be:

Time = 5 * τ

Time = 5 * 0.6 seconds

Time = 3 seconds

So, it takes approximately** 3 seconds** for the current to reach its full value.

(c) The final value of the **current** after it has settled down can be determined using **Ohm's Law**. We know that the voltage across the coil is 20 volts, and the resistance is 5 ohms. Therefore, the final value of the current (I) can be calculated as:

I = V / R

I = 20 V / 5 Ω

I = 4 A

Hence, the final value of the **current** after it has settled down **is 4 Amperes.**

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A clown at a birthday party has brought along a helium cylinder, with which he intends to fill balloons. When full, each balloon contains 0.00450 m3 of helium at an absolute pressure of 1.20 x 105 Pa. The cylinder contains helium at an absolute pressure of 1.50 x 107 Pa and has a volume of 0.00400 m3. The temperature of the helium in the tank and in the balloons is the same and remains constant. What is the maximum number of people who will get a balloon?

### Answers

The **maximum number **of the people that can get **balloons **is 1071 people

What is the Boyle's law?

Boyle's law, named after the physicist Robert Boyle, states that the pressure and volume of a gas are inversely proportional to each other at a constant temperature. One of the fundamental laws of gases, Boyle's law is founded on the fact that, while the temperature is held constant, the volume of a gas has an inverse relationship to the pressure it exerts.

From** Boyle's law **We have that;

Pressure of cylinder * volume of cylinder = n *Pressure of balloon * Volume of balloon

[tex]1.50 * 10^7 Pa * 0.00400 m^3 = n * 1.20 * 10^5 Pa * 0.00450 m^3[/tex]

[tex]n = 1.50 * 10^7 Pa * 0.00400 m^3/1.20 * 10^5 Pa * 0.00450 m^3[/tex]

[tex]n = 6 * 10^4/56\\n = 1071 people[/tex]

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A traveling plane wave (perhaps a transverse wave on a string) is described by either of the following expressions: e

i(kx−ωt)

or e

i(kx+ωt)

. The real part of this function describes the wave. What is the physical difference between the two waves? Explain. b: In a particular situation, ω is replaced by ω−iα. What physical situation(s) do the two waves now describe? Assume that k,ω and α are all positive constants. Explain.

### Answers

The** waves **described by e i(kx−ωt) and e i(kx+ωt) represent right-handed and left-handed waves respectively, while adding damping (ω−iα) results in a damped wave that loses energy over time, characterized by e i(kx−ωt) e^(-αt).

Two traveling waves are described as follows: e i(kx−ωt) and e i(kx+ωt). There is a physical difference between the two waves. A wave that moves in the positive x-direction and a wave that moves in the negative x-direction are described by the two waves. The wave described by e i(kx−ωt) is the wave that moves in the positive x-direction. It is called the "wave moving to the right" and is also known as a** right-handed wave**. The wave described by e i(kx+ωt) is the wave that moves in the negative x-direction. It is called the "wave moving to the left" and is also known as a **left-handed **wave.

In a specific situation, ω is replaced by ω−iα, and the two waves now describe a **damped wave**. This is a wave that loses energy over time. When a wave loses energy, it means that its amplitude decreases. Damping waves are prevalent in natural processes. A damped wave can be described by the equation e i(kx−ωt) e^(-αt), where e^(-αt) is a decreasing function.

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A 5mall object begins a free-fall from a height of 21.0 m. After 1.30 s, a second small object is launched vertically upward fron fie ground with an initial velocity of 33.0 m/s. At what height h above the ground will the two objects first incet? h=

### Answers

After 1.20 seconds, when the second object reaches that height, the two objects will collide at a **height **above the ground. This calculation is based on the assumption of negligible air resistance and the equations of motion for vertical motion.

To determine the height at which the two objects will first meet, we need to find the time it takes for the second object to reach the same height as the first object.

First, let's find the time it takes for the first object to fall from a height of 21.0 m. We can use the equation for vertical **motion**:

h = ut + (1/2)gt²

Where h is the height, u is the initial velocity (which is 0 m/s for free-fall), g is the acceleration due to gravity (approximately -9.8 m/s² for downward motion), and t is the time.

Plugging in the values, we have:

21.0 m = 0 + (1/2)(-9.8 m/s²)t²

Simplifying the equation, we get:

4.9t² = 21.0 m

t² ≈ 4.286

t ≈ 2.07 s

Now, since the second object is launched upward after 1.30 s, we need to find the time it takes for the second object to reach the same **height**. The time for the second object can be found using the equation:

h = ut + (1/2)gt²

Plugging in the values, we have:

h = 33.0 m/s * t + (1/2)(-9.8 m/s²)t²

Since the height h is the same as the height of the first object (21.0 m), we can set up the equation:

21.0 m = 33.0 m/s * t + (1/2)(-9.8 m/s²)t²

Rearranging the equation, we have:

4.9t^2 + 33.0t - 21.0 = 0

Solving this quadratic equation, we find two possible values for t: t ≈ -1.70 s and t ≈ 1.20 s. Since time cannot be negative, we discard the negative solution.

Therefore, the two objects will first meet at a height h above the ground when the second **object **reaches that height, which occurs at approximately 1.20 seconds.

In conclusion, By analyzing the free-fall motion of the first object and the upward motion of the second object, we found that the two objects will meet at a height above the ground when the second object reaches that height after approximately 1.20 seconds.

This calculation is based on the equations of motion for vertical motion and the assumption of negligible air resistance.

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You are planning to test the water quality at a reservoir at

different depths.

Task 1: Explain the planning that you will do

before you leave for Dam. What are the important aspects you must

consider?

### Answers

Be sure to adhere to all safety **precautions** while performing the test, such as wearing protective gear, being cautious of the **surroundings**, and working in teams.

Ensure that you have obtained the necessary permits and permissions from relevant authorities to test the water at the reservoir. Be sure to adhere to all safety precautions while **performing** the test, such as wearing protective gear, being cautious of the surroundings, and working in teams.

Before leaving for the dam, planning is essential. It will allow you to test the water quality in a **systematic** and thorough way. The following are some important aspects that must be considered:

Test **Equipment** Selection - Choose suitable test equipment to test the water quality at various depths. This equipment must be calibrated, cleaned and checked for accuracy prior to testing. Also, take spare parts or additional equipment just in case something happens.Time and Duration - The testing time and duration should be planned to take into account all factors that may affect the test. This **includes** the weather, time of day, depth of the water and other environmental factors.

Depending on the location, the test should be done during the same period to ensure consistency.

Transportation - Make sure that the test equipment is safely transported and can be easily moved around the test site. Consider factors like access points, terrain, and the distance from the test site to the base. Make sure you take enough fuel for transportation.

Test Plan - Develop a test plan that outlines the testing procedure, test objectives, testing schedule, and personnel requirements. This plan will be your guide to ensure that the testing is done properly and within a specific time frame.

Record Keeping - During the test, record the data from all the measurements taken. Record-keeping is crucial for future analysis and comparisons. It is also important to have a detailed map of the site to track the exact location of each **measurement** taken.

Finally, ensure that you have obtained the necessary permits and permissions from relevant authorities to test the water at the reservoir. Be sure to adhere to all safety precautions while performing the test, such as wearing protective gear, being cautious of the surroundings, and working in teams.

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A stepper motor exerts a torque of 500 N⋅m over 30 degrees. a. How much work is done by the stepper motor (provide answer in J)? Recall Work =T⋅θ. 262 J b. How much power is output by the stepper motor if the motion requires 2 seconds (in-lb/s)? 131 W c. How much power is delivered by the stepper motor in horsepower? 0.176hp d. If the efficiency of the stepper motor system (combined electrical and mechanical efficiency) is 70%, then how much electric power in Watts must be supplied to the stepper motor? 187 W

### Answers

a) The **work done** by the stepper motor is 7,854 J.

b) The **power output **of the stepper motor is 3,927 W.

c) The power delivered by the stepper motor is approximately 5.27 hp.

d) The electric power supplied to the stepper motor must be 5,610 W.

How do we calculate?

a) Work = **Torque** × Angle

Work = 500 N⋅m × 30°

Work = 15,000 N⋅m⋅°

We then Convert the degrees to radians:

Work = 15,000 N⋅m × 30° × π/180

Work = 15,000 N⋅m × 0.5236 radians

Work = ** 7,854 J**

b)

the Power = Work / Time

Power = 7,854 J / 2 seconds

Power = **3,927 W**

c) Power (in horsepower) = Power (in watts) / 745.7

Power (in horsepower) = 3,927 W / 745.7

Power = **5.27 hp**

d) electric power = mechanical power / Efficiency

**electric power**= 3,927 W / 0.7

electric power = **5,610 W**

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A turbojet engine with a propulsive efficiency of 0.32 is used to generate thrust for an aircraft cruising at 277 ms-1. If the exhaust gas velocity is 561 ms-1 and the mass flow rate through the engine is 266 kgs-1, calculate the rate at which the engine will need to burn jet fuel if the fuel’s heating value is 47 MJ/kg. Give your answer in kgs-1 to two decimal places.

### Answers

The rate at which the engine will need to burn jet fuel is 16.56 kg/s (approx).

The expression for **propulsive efficiency **is:

ηp=TV/θiG. Where T is the thrust,

V is the forward velocity of the vehicle,

θi is the fuel flow rate's heating value, and

G is the mass flow rate of the vehicle's exhaust gases.

Given:

**Thrust **generated, T = ηpθiG

V = 0.32 x θi x 266/277

Exhaust **gas velocity**, Ve = 561 ms-1

Fuel’s **heating value**, θi = 47 MJ/kg

Mass flow rate through the engine, G = 266 kg s-1

We have to calculate the rate at which the engine will need to burn jet fuel.

In the turbojet engine, the heat supplied to the fuel is Qθi, and the fuel is completely burned, and the heat goes to the exhaust gas.

Now, express the heat in terms of the velocity as:

Qθi = V2-Ve2/2 = (277)2 - (561)2 /2

The rate at which fuel needs to be burned is given by:

mfuel=Qθi/G = (277)2 - (561)2 /2 × θi × 266

Substitute the given values to calculate the value of mfuel.

mfuel = 16.56 kg/s (approx)

Hence, the rate at which the engine will need to burn jet fuel is 16.56 kg/s (approx).

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The uniform dresser has a weight of 120lb and rests on a tile floor for which μs =0.25. If the man pushes on it in the direction Θ=30∘

, determine the smallest magnitude of force F needed to move the dresser. Also, if the man has a weight of 180lb, determine the smallest coefficient of static friction between his shoes and the floor so that he does not slip. ( 25pts)

### Answers

The smallest magnitude of force F needed to move the dresser is 30 lb/sin 30° = 60 lb (approx).The smallest **coefficient** of **static friction** between his shoes and the floor so that he does not slip is 45/180 = 0.25

When a man pushes a uniform dresser on a tile floor of μs = 0.25 in the direction of Θ = 30∘, determine the smallest **magnitude** of force F needed to move the dresser. Also, determine the smallest coefficient of static friction between his shoes and the floor so that he does not slip.

Given,Weight of the uniform dresser, w = 120 lbCoefficient of static friction between the uniform dresser and the tile floor, μs = 0.25The angle between the force applied and the horizontal, Θ = 30°To find the smallest magnitude of force F needed to move the dresser, we need to calculate the **friction force** between the uniform dresser and the tile floor.

Force of friction, f = μs * N, where N is the normal force exerted by the tile floor on the uniform dresser.Since the uniform dresser is at rest on the tile floor, the force exerted on it in the vertical direction will balance the force exerted on it in the horizontal direction.Thus, normal force, N = weight of the uniform dresser, w = 120 lbNow, force of friction, f = μs * N = 0.25 * 120 = 30 lbTo move the uniform dresser, a force greater than 30 lb should be applied at an angle of 30° to the horizontal.

Therefore, the smallest magnitude of force F needed to move the dresser is 30 lb/sin 30° = 60 lb (approx).If the man has a weight of 180 lb, the smallest coefficient of static friction between his shoes and the floor so that he does not slip can be calculated by the following formula:Coefficient of friction, μ = F/N, where F is the force exerted by the man on the tile floor and N is the normal force **exerted** by the tile floor on the man.

To calculate the normal force, we need to balance the force exerted on the man in the vertical direction with the normal force exerted by the tile floor on the man.Since the man is not accelerating in the **vertical** direction, the force exerted on the man in the vertical direction will balance the normal force exerted by the tile floor on the man.Thus, normal force, N = weight of the man, w = 180 lb.

Now, to prevent the man from slipping on the tile floor, the force exerted by the man on the tile floor should be less than or equal to the force of static friction exerted by the tile floor on the man.Force of static friction, Fsf = μs * Nwhere μs is the coefficient of static friction.Now, for the man not to slip, the force exerted by the man on the tile floor should be less than or equal to the force of static friction exerted by the tile floor on the man.

Thus, F ≤ FsfF ≤ μs * N = 0.25 * 180 = 45 lbTherefore, the smallest coefficient of static friction between his shoes and the floor so that he does not slip is 45/180 = 0.25. Hence, the answer is 0.25.

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A battery charger is connected to a dead battery and delivers a current of 7.6 A for 4.9 hours, keeping the voltage across the battery terminals at 12 V in the process. How much energy is delivered to the battery? Number Units

### Answers

The **energy **delivered to the battery is 197,280 Joules.

How do we calculate?

**Energy **is described as the quantitative property that is transferred to a body or to a **physical system,** recognizable in the performance of work and in the form of heat and light.

We have the Energy = Power x Time

where Power = Voltage x Current

**Current **(I) = 7.6 A

Time (t) = 4.9 hours

Voltage (V) = 12 V

**Power** = Voltage x Current

= 12 V x 7.6 A

= 91.2

Energy = Power x Time

Energy = (12 V x 7.6 A) x (4.9 hours x 3600 seconds/hour)

Energy = 12 V x 7.6 A x 4.9 hours x 3600 seconds/hour

Energy = **197280 Joules**

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How much work is required to move a proton and an electron at rest a distance 1×10

−8

m apart to be at rest a distance 3×10

−8

m apart?

### Answers

Work is required to move a proton and an **electron** at rest a distance 1×10m apart to be at rest a distance 3×10m apar Work = (8.99 × 10^9 N m^2/C^2) * (1.6 × 10^(-19) C * (1.6 × 10^(-19) C)) * (1/(3×10^(-8) m) - 1/(1×10^(-8) m))

To calculate the work required to move a proton and an electron from a distance of 1×10^(-8) m apart to a distance of 3×10^(-8) m apart, we can use the formula for the **electrostatic** potential energy.

The electrostatic potential energy between two charged particles is given by the equation:

PE = k * (q1 * q2) / r

Where:

PE is the potential energy,

k is Coulomb's constant (k ≈ 8.99 × 10^9 N m^2/C^2),

q1 and q2 are the charges of the **particles**, and

r is the separation distance between the particles.

In this case, we have a proton and an electron, and their charges are opposite: the charge of a proton is +e (elementary charge) and the charge of an electron is -e. The magnitude of the elementary **charge** is approximately 1.6 × 10^(-19) C.

Using these values, we can calculate the work required to move the particles:

Work = PE_final - PE_initial

Since both the proton and the electron are initially at rest, they have no initial kinetic energy.

PE_initial = k * (q1 * q2) / r_initial

PE_initial = k * (+e * (-e)) / (1×10^(-8) m)

PE_final = k * (q1 * q2) / r_final

PE_final = k * (+e * (-e)) / (3×10^(-8) m)

Now we can calculate the work:

Work = PE_final - PE_initial

Work = k * (+e * (-e)) / (3×10^(-8) m) - k * (+e * (-e)) / (1×10^(-8) m)

Work = k * (+e * (-e)) * (1/(3×10^(-8) m) - 1/(1×10^(-8) m))

**Plugging** in the values and calculating:

Work = (8.99 × 10^9 N m^2/C^2) * (1.6 × 10^(-19) C * (1.6 × 10^(-19) C)) * (1/(3×10^(-8) m) - 1/(1×10^(-8) m))

The final calculation will yield the work required to move the proton and **electron** from their initial separation to the final separation.

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