The magnetic field strength B around a long current-carrying wire is given byQuestion 15 options:
B=μo I/(2πr).
B=μo I x (2πr)
B=μo I/(2r).

The wavelength of light at an angle of 59° is 0.000897 nm.

Given data:

Separation between the double slits, d = 3 µm

The angle at which the third-order maximum occurs, θ = 59°

We need to calculate the wavelength of light, λ.

Using the formula for the location of the maxima, we can write:

d sinθ = mλ

Here, m is the order of the maximum.

Since we are interested in the third-order maximum, m = 3.

Substituting the given values, we get:

3 × (3 × 10⁻⁶) × sin59° = 3λλ = (3 × (3 × 10⁻⁶) × sin59°)/3= 0.000897 nm

Therefore, the wavelength of light falling on double slits separated by 3 µm if the third-order maximum is at an angle of 59° is 0.000897 nm.

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Carole's hair grows  0.30 m in 1.3 years. The answer is C. 1.3 years.

To calculate the time it takes for Carole's hair to grow 0.30 m, we can use the formula:

Time = Distance / Speed

The speed of Carole's hair growth is given as 3.5 x 10^9 m/s, and the distance is 0.30 m. Plugging these values into the formula:

[tex]Time = 0.30 m / (3.5 x 10^9 m/s)[/tex]

To convert the time from seconds to years, we need to divide by the number of seconds in a year. 1 year is equal to 3.156 x 10^7 seconds:

[tex]Time (in years) = (0.30 m / (3.5 x 10^9 m/s)) / (3.156 x 10^7 s/year)[/tex]

Now, let's calculate the time:

[tex]Time (in years) = (0.30 m / 3.5 x 10^9 m/s) / (3.156 x 10^7 s/year)[/tex]

[tex]= (0.30 / (3.5 x 10^9)) / (3.156 x 10^7)[/tex]

[tex]≈ 0.024 / 0.3156[/tex]

[tex]≈ 0.076[/tex]

Therefore, it takes approximately 0.076 years for Carole's hair to grow 0.30 m.

To find the answer in the given options, we need to convert the decimal into years:

[tex]0.076 years ≈ 0.076 x 3.156 x 10^7 s/year[/tex]

≈ 240,456 seconds

Now, we compare this time with the options:

A. [tex]1.9 years ≈ 1.9 x 3.156 x 10^7 s/year ≈ 59,964,000 seconds[/tex]

B.[tex]2.7 years ≈ 2.7 x 3.156 x 10^7 s/year ≈ 85,212,000 seconds[/tex]

[tex]C. 1.3 years ≈ 1.3 x 3.156 x 10^7 s/year ≈ 40,908,000 seconds[/tex]

[tex]D. 5.4 years ≈ 5.4 x 3.156 x 10^7 s/year ≈ 171,144,000 seconds[/tex]

Since the closest option to 240,456 seconds is option C, the answer is C. 1.3 years.

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"(a) The expression for the velocity of the particle as a function of time is: V = -14.8tj m/s. (b) The acceleration of the particle as a function of time is: a = -14.8j m/s². (c) v = -14.8tj = -14.8(3.00)j = -44.4j m/s."

(a) To find the expression for the velocity of the particle as a function of time, we can differentiate the position vector with respect to time.

From question:

F = 7.20(1 - 7.40t²)j

To differentiate with respect to time, we differentiate each term separately:

dF/dt = d/dt(7.20(1 - 7.40t²)j)

= 0 - 7.40(2t)j

= -14.8tj

Therefore, the expression for the velocity of the particle as a function of time is: V = -14.8tj m/s

(b) The acceleration of the particle is the derivative of velocity with respect to time:

dV/dt = d/dt(-14.8tj)

= -14.8j

Therefore, the acceleration of the particle as a function of time is: a = -14.8j m/s²

(c) To calculate the particle's position and velocity at t = 3.00 s, we substitute t = 3.00 s into the expressions we derived.

Position at t = 3.00 s:

r = ∫V dt = ∫(-14.8tj) dt = -7.4t²j + C

Since we need the specific position, we need the value of the constant C. We can find it by considering the initial position of the particle. If the particle's initial position is given, please provide that information.

Velocity at t = 3.00 s:

v = -14.8tj = -14.8(3.00)j = -44.4j m/s

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(a) The width of the slit is 0.216 μm.

(b) The ratio of the intensity at 3.3 mm from the center of the pattern to the intensity at the center of the pattern is 0.231.

In single-slit diffraction, the central peak refers to the brightest and sharpest peak of light in the diffraction pattern. The given information provides that the width of the central peak is 5.0 mm, wavelength is 600 nm, and the distance of the screen from the slit is 1.8 m. Using the formula of diffraction, we can calculate the width of the slit which comes out to be 0.216 μm.

Secondly, the ratio of intensity at a point of 3.3 mm from the center of the pattern to the intensity at the center of the pattern can be calculated using the formula of intensity. On substituting the given values, the ratio of intensity comes out to be 0.231.

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To find the tension in the rope so that the claw is in equilibrium, we need to analyze the forces acting on the claw and set up equations based on Newton's second law.

Let's break down the forces acting on the claw:

Weight (mg): The weight of the claw acts downward with a magnitude equal to the mass (m) of the claw multiplied by the acceleration due to gravity (g). So, the weight is given by W = mg.

Normal force (N): N is equal to the vertical component of the tension force, which is T * sin(θ), where θ is the angle between the tension force and the vertical direction.

Static friction (f_s): The maximum static friction force can be calculated using the equation f_s = μ_s * N, where μ_s is the coefficient of static friction and N is the normal force.

Tension force (T): The tension force in the rope acts diagonally up and to the left, making an angle of 51 degrees with the vertical direction.

Now let's set up the equations of equilibrium:

In the x-direction:

The net force in the x-direction is zero since the claw is in equilibrium. The horizontal component of the tension force is balanced by the static friction force.

T * cos(θ) = f_s

In the y-direction:

The net force in the y-direction is zero since the claw is in equilibrium. The vertical component of the tension force is balanced by the weight and the normal force.

T * sin(θ) + N = mg

Now, substitute the expressions for f_s and N into the equations:

T * cos(θ) = μ_s * T * sin(θ)

T * sin(θ) + μ_s * T * sin(θ) = mg

Simplify the equations:

cos(θ) = μ_s * sin(θ)

sin(θ) + μ_s * sin(θ) = mg / T

Divide both sides of the second equation by sin(θ):

1 + μ_s = (mg / T) / sin(θ)

Now, solve for T:

T = (mg / sin(θ)) / (1 + μ_s)

Substitute the given values:

m = 2.0 kg

g = 9.8 m/s²

θ = 51 degrees

μ_s = 0.80

T = (2.0 kg * 9.8 m/s²) / sin(51°) / (1 + 0.80)

Calculating this expression will give us the tension in the rope. Let's compute it:

T ≈ 22.58 N

Therefore, the tension in the rope for the claw to be in equilibrium is approximately 22.58 Newtons.

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The value of d, the line spacing in lines/mm for each three scenarios are (m * 500 nm) / sin(30 degrees); (m * 600 nm) / sin(45 degrees) and (m * 600 nm) / sin(45 degrees) respectively.

In the given diffraction equation, dsinθ = mλ, where d represents the line spacing, θ is the angle of diffraction, m is the order of the interference, and λ is the wavelength of light.

To solve for d, we rearrange the equation as follows:

d = (mλ) / sinθ.

Let's consider three different scenarios with corresponding angles and wavelengths to calculate the line spacing in each case.

Scenario 1:

Angle of diffraction (θ) = 30 degrees

Wavelength (λ) = 500 nm

Using the formula:

d = (m * λ) / sinθ

  = (m * 500 nm) / sin(30 degrees)

Scenario 2:

Angle of diffraction (θ) = 45 degrees

Wavelength (λ) = 600 nm

Using the formula:

d = (m * λ) / sinθ

  = (m * 600 nm) / sin(45 degrees)

Scenario 3:

Angle of diffraction (θ) = 60 degrees

Wavelength (λ) = 700 nm

Using the formula:

d = (m * λ) / sinθ

  = (m * 600 nm) / sin(45 degrees)

In each scenario, the line spacing will depend on the order of interference. By substituting the given values into the respective equations, we can calculate the line spacing for each case.

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Holography is the process of creating three-dimensional images called holograms, with applications in security, art, data storage, medicine, engineering, and more.

7. A hologram is a three-dimensional image produced through the process of holography. It is a photographic technique that records the interference pattern of light waves reflected or scattered off an object. When the hologram is illuminated with coherent light, it recreates the original object's appearance, including depth and parallax.

8. Holography has several applications across various fields, including:

- Security: Holograms are used in security features such as holographic labels, ID cards, and banknotes to prevent counterfeiting.

- Art and Entertainment: Holograms are employed in art installations, exhibitions, and performances to create immersive and visually striking experiences.

- Data Storage: Holographic storage technology has the potential for high-capacity data storage with fast access speeds.

- Medical Imaging: Holography finds applications in medical imaging, such as holographic microscopy and holographic tomography, for enhanced visualization and analysis of biological structures.

- Engineering and Testing: Holography is used for non-destructive testing, strain analysis, and deformation measurement in engineering and material science.

- Optical Elements: Holographic optical elements are used as diffractive lenses, beam splitters, filters, and other optical components.

- Virtual Reality (VR) and Augmented Reality (AR): Holography techniques contribute to the development of advanced VR and AR systems, providing realistic 3D visualizations.

These are just a few examples of the wide-ranging applications of holography, which continue to expand as the technology advances.

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Using the concept of the magnetic field generated by current-carrying wire:

(A) The compass needle will point anticlockwise. if you are standing right below it.

(B)The magnets should be directed vertically upward.

(C) The north pole of the bar magnet should point downward.

A straight current-carrying wire generates a circular magnetic field around it as the axis.

A) The compass needle will point anticlockwise if you are standing right underneath the wire. The right-hand rule can be used to figure this out. When viewed from above, the magnetic field produced by the current will move anticlockwise around the wire if the current is exiting the page. The compass needle will point anticlockwise because its north pole lines up with the magnetic field lines.

B) The magnetic field created by the extra magnets should be directed vertically upward to oppose the pull of gravity on the wire and prevent sagging. The upward magnetic force can counterbalance the downward gravitational attraction by positioning the magnetic field in opposition to the gravitational pull.

C) You can place bar magnets in a precise way to provide the necessary upward magnetic field close to the wire. The north poles of the bar magnets should be pointed downward as you position them vertically above the wire. The magnets' south poles should be facing up. By positioning the bar magnets in this way, their magnetic fields will interact to produce an upward magnetic field close to the wire that will work to fight gravity and stop sagging.

Therefore, Using the concept of the magnetic field generated by a current-carrying wire:

(A) The compass needle will point anticlockwise. if you are standing right below it.

(B)The magnets should be directed vertically upward.

(C) The north pole of the bar magnet should point downward.

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The frequency of a sound wave with a wavelength of 0.34 m travelling in room-temperature air (v-340m/s) is 1000

The frequency of a sound wave in room-temperature air can be calculated as follows:f= v/λ where f is the frequency of the sound wave,λ is the wavelength of the sound wave,v is the speed of sound in room-temperature air. We have λ = 0.34 mv = 340 m/s. Substituting these values, we get:

f = 340 m/s / 0.34 mf = 1000 Hz

Hence, the frequency of a sound wave with a wavelength of 0.34 m travelling in room-temperature air (v-340m/s) is 1000 Hz.

Thus, option C is the correct answer.

This question is based on the concept of the relationship between the wavelength, frequency, and velocity of sound waves. The frequency of a sound wave in room-temperature air can be calculated using the formula f = v/λ where f is the frequency of the sound wave, λ is the wavelength of the sound wave, and v is the speed of sound in room-temperature air. The given wavelength of the sound wave is 0.34 m, and the speed of sound in room-temperature air is 340 m/s. We can substitute these values in the formula mentioned above to calculate the frequency of the sound wave as follows:f = v/λf = 340 m/s / 0.34 mf = 1000 Hz

Thus, the frequency of the sound wave with a wavelength of 0.34 m travelling in room-temperature air (v-340m/s) is 1000 Hz.

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The magnitude of the net force applied to the mass is 45N when it is at maximum speed

To find the magnitude of the net force applied to the mass when it is at maximum speed, we need to consider the restoring force exerted by the spring.

In simple harmonic motion, the restoring force exerted by a spring is given by Hooke's law:

F = -kx

where F is the force, k is the spring constant, and x is the displacement from the equilibrium position.

In this case, the mass is attached to the spring and undergoes simple harmonic motion with an amplitude of 20 cm, which corresponds to a maximum displacement from the equilibrium position.

At maximum speed, the mass is at the extreme points of its motion, where the displacement is maximum. Therefore, the force applied by the spring is at its maximum as well.

Substituting the given values into Hooke's law:

F = -(225 N/m)(0.20 m) = -45 N

Since the force is a vector quantity and the question asks for the magnitude of the net force, the answer is:

Magnitude of the net force = |F| = |-45 N| = 45 N

Therefore, the correct option is (a) 45 N.

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The velocity of the center of the rod in vector form is approximately 24.85 m/s. The angular velocity of the rod after the collision is 24844.087 rad/s. The increase in internal energy of the objects is -103.347 J.

(a) Velocity of center of the rod: The velocity of the center of the rod can be calculated by applying the principle of conservation of momentum. Since the system is isolated, the total momentum of the system before the collision is equal to the total momentum of the system after the collision. Using this principle, the velocity of the center of the rod can be calculated as follows:

Let v be the velocity of the center of the rod after the collision.

m1 = 1.7 kg (mass of the rod)

m2 = 0.09 kg (mass of the meteorite)

v1 = 0 m/s (initial velocity of the rod)

u2 = 245 m/s (initial velocity of the meteorite)

i1 = 0° (initial angle of the rod)

i2 = 26° (initial angle of the meteorite)

j1 = 0° (final angle of the rod)

j2 = 82° (final angle of the meteorite)

v2 = 60 m/s (final velocity of the meteorite)

The total momentum of the system before the collision can be calculated as follows: p1 = m1v1 + m2u2p1 = 1.7 kg × 0 m/s + 0.09 kg × 245 m/sp1 = 21.825 kg m/s

The total momentum of the system after the collision can be calculated as follows: p2 = m1v + m2v2p2 = 1.7 kg × v + 0.09 kg × 60 m/sp2 = (1.7 kg)v + 5.4 kg m/s

By applying the principle of conservation of momentum: p1 = p221.825 kg m/s = (1.7 kg)v + 5.4 kg m/sv = (21.825 kg m/s - 5.4 kg m/s)/1.7 kg v = 10.015 m/s

To represent the velocity in vector form, we can use the following equation:

vCM = (m1v1 + m2u2 + m1v + m2v2)/(m1 + m2)

m1 = 1.7 kg (mass of the rod)

m2 = 0.09 kg (mass of the meteorite)

v1 = 0 m/s (initial velocity of the rod)

u2 = 245 m/s (initial velocity of the meteorite)

v = 10.015 m/s (velocity of the rod after the collision)

v2 = 60 m/s (velocity of the meteorite after the collision)

Substituting these values into the equation, we have:

vCM = (1.7 kg * 0 m/s + 0.09 kg * 245 m/s + 1.7 kg * 10.015 m/s + 0.09 kg * 60 m/s) / (1.7 kg + 0.09 kg)

Simplifying the equation:

vCM = (0 + 22.05 + 17.027 + 5.4) / 1.79

vCM = 44.477 / 1.79

vCM ≈ 24.85 m/s

Therefore, the velocity of the center of the rod in vector form is approximately 24.85 m/s.

(b) Angular velocity of the rod: To calculate the angular velocity of the rod, we can use the principle of conservation of angular momentum. Since the system is isolated, the total angular momentum of the system before the collision is equal to the total angular momentum of the system after the collision. Using this principle, the angular velocity of the rod can be calculated as follows:

Let ω be the angular velocity of the rod after the collision.I = (1/12) m L2 is the moment of inertia of the rod about its center of mass, where L is the length of the rod.m = 1.7 kg is the mass of the rod

The angular momentum of the system before the collision can be calculated as follows:

L1 = I ω1 + m1v1r1 + m2u2r2L1 = (1/12) × 1.7 kg × (0.9 m)2 × 0 rad/s + 1.7 kg × 0 m/s × 0.2 m + 0.09 kg × 245 m/s × 0.7 mL1 = 27.8055 kg m2/s

The angular momentum of the system after the collision can be calculated as follows:

L2 = I ω + m1v r + m2v2r2L2 = (1/12) × 1.7 kg × (0.9 m)2 × ω + 1.7 kg × 10.015 m/s × 0.2 m + 0.09 kg × 60 m/s × 0.7 mL2 = (0.01395 kg m2)ω + 2.1945 kg m2/s

By applying the principle of conservation of angular momentum:

L1 = L2ω1 = (0.01395 kg m2)ω + 2.1945 kg m2/sω = (ω1 - 2.1945 kg m2/s)/(0.01395 kg m2)

Here,ω1 is the angular velocity of the meteorite before the collision. ω1 = u2/r2

ω1 = 245 m/s ÷ 0.7 m

ω1 = 350 rad/s

ω = (350 rad/s - 2.1945 kg m2/s)/(0.01395 kg m2)

ω = 24844.087 rad/s

The angular velocity of the rod after the collision is 24844.087 rad/s.

(c) Increase in internal energy of the objects

The increase in internal energy of the objects can be calculated using the following equation:ΔE = 1/2 m1v1² + 1/2 m2u2² - 1/2 m1v² - 1/2 m2v2²

Here,ΔE is the increase in internal energy of the objects.m1v1² is the initial kinetic energy of the rod.m2u2² is the initial kinetic energy of the meteorite.m1v² is the final kinetic energy of the rod. m2v2² is the final kinetic energy of the meteorite.Using the given values, we get:

ΔE = 1/2 × 1.7 kg × 0 m/s² + 1/2 × 0.09 kg × (245 m/s)² - 1/2 × 1.7 kg × (10.015 m/s)² - 1/2 × 0.09 kg × (60 m/s)²ΔE = -103.347 J

Therefore, the increase in internal energy of the objects is -103.347 J.

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Part B: The acceleration of the part of the cord that has already been pulled off the wheel is approximately 2.95 radians per second squared.

Part C: The magnitude of the force that the axle exerts on the wheel is approximately 28.32 N.

Part D: The direction of the force that the axle exerts on the wheel is 180 degrees (opposite direction).

Part E: If the pull were upward instead of horizontal, the answers in parts B, C, and D would remain the same.

Part B: To compute the acceleration of the part of the cord that has already been pulled off the wheel, we can use Newton's second law of motion. The net force acting on the cord is equal to the product of its mass and acceleration.

Radius of the wheel (r) = 0.270 m

Mass of the wheel (m) = 9.60 kg

Pulling force (F) = 36.0 N

The force causing the acceleration is the horizontal component of the tension in the cord.

Tension in the cord (T) = F

The acceleration (a) can be calculated as:

F - Tension due to the wheel's inertia = m * a

F - (m * r * a) = m * a

36.0 N - (9.60 kg * 0.270 m * a) = 9.60 kg * a

36.0 N = 9.60 kg * a + 2.59 kg * m * a

36.0 N = (12.19 kg * a)

a ≈ 2.95 rad/s²

Therefore, the acceleration of the part of the cord that has already been pulled off the wheel is approximately 2.95 radians per second squared.

Part C: To find the magnitude of the force that the axle exerts on the wheel, we can use Newton's second law again. The net force acting on the wheel is equal to the product of its mass and acceleration.

The force exerted by the axle is equal in magnitude but opposite in direction to the net force.

Net force (F_net) = m * a

F_axle = -F_net

F_axle = -9.60 kg * 2.95 rad/s²

F_axle ≈ -28.32 N

The magnitude of the force that the axle exerts on the wheel is approximately 28.32 N.

Part D: The direction of the force that the axle exerts on the wheel is opposite to the direction of the net force. Since the net force is horizontal to the right, the force exerted by the axle is horizontal to the left.

Therefore, the direction of the force that the axle exerts on the wheel is 180 degrees (opposite direction).

Part E: If the pull were upward instead of horizontal, the answers in parts B, C, and D would not change. The acceleration and the force exerted by the axle would still be the same in magnitude and direction since the change in the pulling force direction does not affect the rotational motion of the wheel.

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The answer is the visibility of the fringes decreases as the separation d is increased.

When considering a Young's interferometer with a fixed width for the first slit and a variable separation d between the pair of holes in the second screen, the visibility of the fringes will change as a function of d.

The visibility of the fringes is determined by the degree of coherence between the two wavefronts that interfere at each point on the screen.

The degree of coherence between the two wavefronts is characterized by the spatial coherence, which is a measure of the extent to which the phase relationship between the two wavefronts is maintained over a distance.

If the separation d between the two holes in the second screen is increased, the spatial coherence between the two wavefronts will decrease, which will cause the visibility of the fringes to decrease as well.

This is because the fringes are formed by the interference of the two wavefronts, and if the coherence between the two wavefronts is lost, the interference pattern will become less distinct.

Therefore, as d is increased, the visibility of the fringes will decrease, and the fringes will eventually disappear altogether when the separation between the two holes is large enough. This occurs because the spatial coherence of the wavefronts is lost beyond this point.

The relationship between the visibility of the fringes and the separation d is given by the formula

V = (Imax - Imin)/(Imax + Imin), where Imax is the maximum intensity of the fringes and Imin is the minimum intensity of the fringes. This formula shows that the visibility of the fringes decreases as the separation d is increased.

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Electromagnetic waves carry both electric and magnetic fields and do not have a net charge. The correct option is d. They are transverse waves, with oscillations perpendicular to the direction of propagation. The correct option is a. Light, as the fastest object in the universe, exhibits both wave and particle properties. The correct option is d. Gamma rays have a higher frequency and shorter wavelength compared to radio waves. The correct option is a. For a convex lens, when an object is located beyond its focal point, the resulting image is real, inverted, and diminished in size. The correct option is d.

1.  An electromagnetic wave consists of oscillating electric and magnetic fields that are perpendicular to each other and also perpendicular to the direction of wave propagation.

It does not carry any net charge, but it does have both electric and magnetic fields associated with it.

The correct answer is (d) none of the above.

2.An electromagnetic wave is a transverse wave because the oscillations of the electric and magnetic fields are perpendicular to the direction of wave propagation.

This means that the vibrations of the fields occur in a plane perpendicular to the direction in which the wave is moving.

The correct answer is (a) transverse wave.

3.  Light is indeed the fastest object in the universe as it travels at a constant speed of approximately 299,792,458 meters per second in a vacuum.

It can exhibit both wave-like and particle-like properties. In classical physics, light is described as an electromagnetic wave, while in quantum mechanics, it is considered to have particle-like behavior called photons.

The correct answer is (d) all of the above.

4. Gamma rays have the highest frequency among the electromagnetic spectrum, ranging from about 10^19 to 10^24 Hertz.

This frequency is much higher than the frequency of radio waves, which typically range from about 10^3 to 10^9 Hertz.

The correct answer is (a) greater than.

5. The wavelength of gamma rays is shorter than the wavelength of radio waves.

Gamma rays have very short wavelengths, typically in the range of picometers (10^-12 meters) to femtometers (10^-15 meters), while radio waves have much longer wavelengths, typically ranging from meters to kilometers.

The correct answer is (b) lower.

6. For a convex lens, the image formed depends on the position of the object relative to the focal point.

In this case, since the object (tree) is located beyond the focal point of the convex lens, the image formed will be real, inverted, diminished (smaller in size), and located on the opposite side of the lens compared to the object.

This is a characteristic behavior of convex lenses when the object is located beyond the focal point.

The correct answer is (d) all of the above.

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(a) The work done by a force is given by the equation:

Work = Force * Distance * cos(theta)

In this case, the force applied is 150 N and the distance moved is 5.50 m. Since the force is applied horizontally, the angle theta between the force and the displacement is 0 degrees (cos(0) = 1).

So the work done by the 150 N force is:

Work = 150 N * 5.50 m * cos(0) = 825 J

Therefore, the work done by the 150 N force is 825 Joules (J).

(b) The work done by the 150 N force is equal to the work done against friction. The work done against friction can be calculated using the equation:

Work = Force of friction * Distance

Since the block moves at a constant speed, the net force acting on it is zero. Therefore, the force of friction must be equal in magnitude and opposite in direction to the applied force of 150 N.

So the force of friction is 150 N.

The coefficient of kinetic friction (μk) can be determined using the equation:

Force of friction = μk * Normal force

The normal force (N) is equal to the weight of the block, which is given by:

Normal force = mass * gravity

where gravity is approximately 9.8 m/s².

Substituting the values:

150 N = μk * (47.5 kg * 9.8 m/s²)

Solving for μk:

μk = 150 N / (47.5 kg * 9.8 m/s²) ≈ 0.322

Therefore, the coefficient of kinetic friction between the block and the floor is approximately 0.322.

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The beat frequency observed by the truck passengers is 16 Hz. Thus, correct option is (b).

When two sound waves with slightly different frequencies interfere, they produce a beat frequency equal to the difference between their frequencies. In this scenario, the first police car emits a sound wave with a frequency of 890 Hz, while the second police car emits a sound wave with the same frequency. However, due to the motion of the cars, the frequency observed by the truck passengers is shifted.

The frequency shift, known as the Doppler effect, is given by the formula:

Δf = (v-sound / v-observer) × f-source × (v-source - v-observer)

Where v-sound is the speed of sound, v-observer is the speed of the observer (truck), f-source is the source frequency (890 Hz), and (v-source - v-observer) is the relative velocity between the source and observer.

In this case, the relative velocity between the first police car and the truck is (45 mph - 35 mph) = 10 mph = 4.47 m/s. Plugging the values into the Doppler effect formula, we get:

Δf = (340 m/s / 4.47 m/s) × 890 Hz × 4.47 m/s = 16 Hz.

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The given question is incomplete, complete question is- "Three cars move along a straight highway as follows: in one lane two police cars travel with 45 mph so that they are 300 feet apart with their sirens emitting simultaneously sound at 890 Hz(v sound  =340 m/s). In the other lane a truck travels in the same direction with a speed of 35mph. What beat frequency is observed by the truck passengers while the truck is passed by the first police car but not the second one (see figure).

Select one: a. 7 Hz b. 16 Hz C. 20 Hz d. 23 Hz"

The acceleration of the block, when pulled up the frictionless incline with an angle of 15 degrees and a tension force of 88 N, is approximately 1.23 m/s^2.

To determine the acceleration of the block on the frictionless incline, we can apply Newton's second law of motion. The force component parallel to the incline will be responsible for the acceleration.

The gravitational force acting on the block can be decomposed into two components: one perpendicular to the incline (mg * cos(theta)), and one parallel to the incline (mg * sin(theta)). In this case, theta is the angle of the incline.

The tension force is also acting on the block, in the upward direction parallel to the incline.

Since there is no friction, the net force along the incline is given by:

F_net = T - mg * sin(theta)

Using Newton's second law (F_net = m * a), we can set up the equation:

T - mg * sin(theta) = m * a

mass (m) = 18.8 kg

Tension force (T) = 88 N

angle of the incline (theta) = 15 degrees

acceleration (a) = ?

Plugging in the values, we have:

88 N - (18.8 kg * 9.8 m/s^2 * sin(15 degrees)) = 18.8 kg * a

Solving this equation will give us the acceleration of the block:

a = (88 N - (18.8 kg * 9.8 m/s^2 * sin(15 degrees))) / 18.8 kg

a ≈ 1.23 m/s^2

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The daughter nuclide is Neptunium-237 and its atomic mass is 237.048172 u.

The given Americium isotope, 24Am is unstable and emits a 5.538 MeV alpha particle. The atomic mass of 2Am is 241.0568 u and that of He is 4.0026 u. Identify the daughter nuclide and find its atomic mass.

The daughter nuclide is Neptunium-237 and its atomic mass is 237.048172 u.

How does an isotope decay?

An isotope decays to produce one or more daughter nuclides. The process of isotope decay includes alpha, beta, and gamma decay. Americium 24Am undergoes alpha decay which is a form of radioactive decay that occurs when the nucleus of an atom emits an alpha particle.

The alpha decay equation is 24Am → 4He + 20

Neptunium is a daughter nuclide of Americium. It is denoted by the symbol Np and has an atomic number of 93.  Neptunium-237 is formed when 241Am undergoes alpha decay and emits a 5.538 MeV alpha particle. The mass number of the parent and daughter nuclides must be equal. Therefore,

Atomic mass of 24Am = Atomic mass of 4He + Atomic mass of 237Np

(241.0568 u) = (4.0026 u) + Atomic mass of 237Np

Atomic mass of 237Np = (241.0568 u - 4.0026 u)

Atomic mass of 237Np = 237.048172 u

Hence, the daughter nuclide is Neptunium-237 and its atomic mass is 237.048172 u.

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a. The frequency emitted by truck A will be 200 Hz and the frequency emitted by truck B will be approximately 198.56 Hz

b. The frequency of the beats is 1.44 Hz.

a) Truck A is stationary and truck B is moving away at a constant speed of 30 km/h. Both of the trucks emit a sound of frequency 200 Hz and the speed of sound is 343 m/s, the frequency of sound will be affected by the Doppler effect.

The Doppler effect can be given by:

[tex]f'= \frac {v \pm v_0} {v\pm v_s}f[/tex]

Here, f is the frequency of the sound emitted.

v is the velocity of sound in air ($343 m/s$)

v0 is the velocity of the object emitting the sound and vs is the velocity of the sound wave relative to the stationary object

In this problem, the frequency emitted by the truck A is

[tex]f_{A} = 200[/tex]Hz

v0 = 0m/s

v = 343m/s

The frequency emitted by the truck B is [tex]f_{B} = 200[/tex] Hz

[tex]v0 = - 30km/h \\= - \frac{30 \times 1000}{3600}$ m/s \\= $-\frac{25}{3}$ ms^{-1} \\v= 343m/s[/tex]

On substituting the above values in the Doppler's equation, we get,

For truck A,

[tex]f_{A}' = \frac{v}{v\pm v_{s}}[/tex]

[tex]f_{A}' = \frac{343}{343\pm 0} Hz = 200[/tex] Hz

For truck B,[tex]f_{B}' = \frac{v}{v\pm v_{s}}[/tex]

[tex]f_{B}' = \frac{343} {343 \pm \frac {25}{3}}\text{Hz}[/tex] ≈ 198.56 Hz

Hence the frequency emitted by truck A will be 200 Hz and the frequency emitted by truck B will be approximately 198.56 Hz

b) A beat is produced when two sound waves having slightly different frequencies are superposed.

In this problem, as we see that the frequency of the wave emitted by truck A is 200 Hz and the frequency of the wave emitted by truck B is approximately 198.56 Hz, we can say that a beat will be produced.

To find the frequency of beats, we use the formula for beats:

fbeat = |f1 − f2|

Where,f1 is the frequency of the wave emitted by truck Af2 is the frequency of the wave emitted by truck B

Frequencies of the waves are given by,

f1 = 200 Hz

f2 = 198.56 Hz

fbeat = |200 − 198.56| Hz ≈ 1.44 Hz

Thus, the frequency of the beats is 1.44 Hz.

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a). You will hear a frequency of approximately 195.84 Hz from Truck B.

b). The beat frequency between the two trucks' sounds will be approximately 4.16 Hz.

a) To determine the frequency you will hear from each truck, we need to consider the Doppler effect. The Doppler effect describes how the perceived frequency of a sound wave changes when the source of the sound or the listener is in motion relative to each other.

For the stationary Truck A, there is no relative motion between you and the truck. Therefore, the frequency you hear from Truck A will be the same as its emitted frequency, which is 200 Hz.

For the moving Truck B, which is moving away from you at a constant speed of 30 km/h, the frequency you hear will be lower than its emitted frequency due to the Doppler effect. The formula for the Doppler effect when a source is moving away is given by:

f' = f * (v_sound + v_observer) / (v_sound + v_source)

where f is the emitted frequency, v_sound is the speed of sound (approximately 343 m/s), v_observer is the speed of the observer (you, assumed to be stationary), and v_source is the speed of the source (Truck B).

Converting the speed of Truck B from km/h to m/s:

v_source = 30 km/h * (1000 m/km) / (3600 s/h) = 8.33 m/s

Plugging in the values:

f' = 200 Hz * (343 m/s + 0 m/s) / (343 m/s + 8.33 m/s)

Simplifying the equation:

f' ≈ 195.84 Hz

Therefore, you will hear a frequency of approximately 195.84 Hz from Truck B.

b) Yes, there will be a beat if the frequencies of the two trucks are slightly different. The beat frequency is equal to the absolute difference between the frequencies of the two sounds.

Beat frequency = |f_A - f_B|

Substituting the values:

Beat frequency = |200 Hz - 195.84 Hz|

Simplifying:

Beat frequency ≈ 4.16 Hz

So, the beat frequency between the two trucks' sounds will be approximately 4.16 Hz.

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(a) The final temperature of the mixture is 47.0°C.

(b) The minimum mass of the ice cube that will result in a final mixture at exactly 0°C is 194.36 kg, or 194,360 g.

(a) To determine the final temperature of the mixture, we can use the principle of conservation of energy. The energy gained by the ice melting must be equal to the energy lost by the milk.

First, let's calculate the energy gained by the ice melting:

Energy gained = mass of ice * heat of fusion of ice

The heat of fusion of ice is the amount of energy required to melt one gram of ice without changing its temperature, which is 334,000 J/kg.

Energy gained = (86 g) * (334,000 J/kg) = 28,804,000 J

Now, let's calculate the energy lost by the milk:

Energy lost = mass of milk * specific heat of milk * change in temperature

The specific heat of milk is 3,930 J/(kg·°C).

The change in temperature is the difference between the final temperature of the mixture and the initial temperature of the milk, which is (final temperature - 39.0°C).

Energy lost = (933 g) * (3,930 J/(kg·°C)) * (final temperature - 39.0°C)

Since the energy gained and energy lost are equal, we can set up an equation:

28,804,000 J = (933 g) * (3,930 J/(kg·°C)) * (final temperature - 39.0°C)

Simplifying the equation, we can solve for the final temperature:

final temperature - 39.0°C = 28,804,000 J / (933 g * 3,930 J/(kg·°C))

final temperature - 39.0°C = 8.00°C

Adding 39.0°C to both sides of the equation, we find:

final temperature = 8.00°C + 39.0°C

final temperature = 47.0°C

Therefore, the final temperature of the mixture is 47.0°C.

(b) To determine the minimum mass of the ice cube that will result in a final mixture at exactly 0°C, we can use the same approach as in part (a) but set the final temperature to 0°C.

Setting the final temperature to 0°C in the equation:

0°C - 39.0°C = 28,804,000 J / (mass of milk * 3,930 J/(kg·°C))

Simplifying the equation, we can solve for the minimum mass of the milk:

mass of milk = 28,804,000 J / (3,930 J/(kg·°C) * (39.0°C - 0°C))

mass of milk = 194.36 kg

Therefore, the minimum mass of the ice cube that will result in a final mixture at exactly 0°C is 194.36 kg, or 194,360 g.

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The work done by the force of gravity as the box slides down the ramp is approximately 75.54 J.

The minimum force required, acting at an angle of 40° to the horizontal, to slide the box back up the ramp is approximately 18.94 N.

(a) To determine the work done by the force of gravity as the box slides down the ramp, we first calculate the vertical height (h) using the formula

h = l * sin(θ), where

l is the length of the ramp and

θ is the angle of inclination.

In this case, the vertical height is h = 6 m * sin(18°) ≈ 1.928 m.

Next, we can calculate the work done by gravity using the formula

W = mgh, where

m is the mass of the box,

g is the acceleration due to gravity (approximately 9.8 m/s²), and

h is the vertical height.

Plugging in the values, we have

W = 4 kg * 9.8 m/s² * 1.928 m

≈ 75.5416 J.

Therefore, the work done by the force of gravity as the box slides down the ramp is approximately 75.54 J.

(b) To determine the minimum force required to slide the box back up the ramp, we use the formula

F = mg / sin(θ), where

m is the mass of the box,

g is the acceleration due to gravity, and

θ is the angle of inclination.

Plugging in the values, we have

F = 4 kg * 9.8 m/s² / sin(18°)

≈ 24.851 N.

However, in this scenario, the force is applied at an angle of 40° to the horizontal. To find the component of force along the ramp, we use the formula

F_ramp = F_total * cos(40°).

Plugging in the value of the total force (F = 24.851 N), we have

F_ramp = 24.851 N * cos(40°)

≈ 18.935 N.

Therefore, the minimum force required, acting at an angle of 40° to the horizontal, to slide the box back up the ramp is approximately 18.94 N.

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[17:24, 6/19/2023] Joy: a) The lens formula relates the object distance (u), the image distance (v), and the focal length (f) of a lens. It is given by:

1/f = 1/v - 1/u

In this case, the object distance (u) is 45.0 cm, and the focal length (f) is 20.0 cm. We need to find the image distance (v).

the values into the lens formula:

1/20 cm = 1/v - 1/45 cm

Rearranging the equation:

1/v = 1/20 cm + 1/45 cm

To add the fractions, we need a common denominator:

1/v = (45 + 20) / (45 * 20) cm

1/v = 65 / 900 cm

Now we can find v by taking the reciprocal of both sides:

v = 900 cm / 65

v ≈ 13.85 cm

Therefore, the distance between the image and the lens is approximately 13.85 cm.

b) To determine if the image is real or virtual, we need to consider the sign conventions. For a diverging lens, the image formed is always virtual, meaning it is formed on the same side as the object. So, the image is virtual.

c) To find the height of the image, we can use the magnification formula:

m = -v/u

where m is the magnification, v is the image distance, and u is the object distancES.

Substituting the given values:

m = -13.85 cm / 45.0 cm

m ≈ -0.307

The negative sign indicates an inverted image.

The height of the image can be calculated using the magnification formula:

m = h'/h

where h' is the height of the image and h is the height of the object.

Rearranging the equation:

h' = m * h

h' = -0.307 * 3.0 cm

h' ≈ -0.921 cm

The height of the image is approximately -0.921 cm. The negative sign indicates that the image is inverted.

To summarize:

a) The distance between the image and the lens is approximately 13.85 cm.

b) The image is virtual.

c) The height of the image is approximately -0.921 cm.

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Possible equivalent resistances are as follows:

Using one resistor: 20Ω, 30Ω

Using two resistors: 40Ω, 50Ω, 60Ω, 10Ω, 13.33Ω, 20Ω

Using all three resistors: 70Ω

To find all possible equivalent resistances using the given resistors, we can consider different combinations of resistors in series and parallel arrangements. Here are the possible arrangements and their equivalent resistances:

Using one resistor:

20Ω resistor

30Ω resistor

Using two resistors:

a) Series arrangement:

20Ω + 20Ω = 40Ω (20Ω + 20Ω in series)

20Ω + 30Ω = 50Ω (20Ω + 30Ω in series)

30Ω + 20Ω = 50Ω (30Ω + 20Ω in series)

30Ω + 30Ω = 60Ω (30Ω + 30Ω in series)

b) Parallel arrangement:

10Ω (1 / (1/20Ω + 1/20Ω) in parallel)

13.33Ω (1 / (1/20Ω + 1/30Ω) in parallel)

13.33Ω (1 / (1/30Ω + 1/20Ω) in parallel)

20Ω (1 / (1/30Ω + 1/30Ω) in parallel)

Using all three resistors:

20Ω + 20Ω + 30Ω = 70Ω (20Ω + 20Ω + 30Ω in series)

Sketching the resistor arrangements for each case:

Using one resistor:

Single resistor: R = 20Ω

Single resistor: R = 30Ω

Using two resistors:

a) Series arrangement:

Two resistors in series: R = 40Ω

Resistor and series combination: R = 50Ω

Resistor and series combination: R = 50Ω

Two resistors in series: R = 60Ω

b) Parallel arrangement:

Two resistors in parallel: R = 10Ω

Resistor and parallel combination: R = 13.33Ω

Resistor and parallel combination: R = 13.33Ω

Two resistors in parallel: R = 20Ω

Using all three resistors:

Three resistors in series: R = 70Ω

Note: The resistor arrangements can be represented using circuit diagrams, where the resistors in series are shown in a straight line, and resistors in parallel are shown with parallel lines connecting them.

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An object oscillates with an angular frequency [tex]ω = 5 rad/s. At t = 0[/tex], the object is at [tex]x0 = 6.5 cm.[/tex]It is moving with velocity vx0 = 14 cm/s in the positive x-direction.

The position of the object can be described through the equation x(t) = A cos(ωt + φ).The phase constant φ of the oscillation in radiansThe formula used for the displacement equation is,[tex]x(t) = A cos(ωt + φ)[/tex]Given that, ω = 5 rad/s, x0 = 6.5 cm, and vx0 = 14 cm/sSince the velocity is given.

Therefore it is assumed that the particle is moving with simple harmonic motion starting from x0. Hence the phase constant φ can be obtained from the displacement equation by substituting the initial values,[tex]x0 = A cos (φ)6.5 = A cos (φ)On solving,φ = cos-1 (x0 / A)[/tex]The equation for the amplitude .

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In order to calculate the absolute error (AL) for the given three measurements L1, L2, and L3 which are 9.3 cm, 9.7 cm, and 9.5 cm respectively,

we need to first calculate the average length and then find the difference of each measurement from the average length.

Then, the absolute error (AL) for each measurement is calculated by taking the absolute value of the difference between the measurement and the average length.

Finally, the average of these absolute errors is taken as the absolute error (AL).

Thus, the absolute error (AL) is given as:

AL = (|9.3 - 9.5| + |9.7 - 9.5| + |9.5 - 9.5|)/3

     = (0.2 + 0.2 + 0)/3

     = 0.13 cm

Therefore, the correct option is

5.0.13.

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The formulas provided, the optical line of sight (d = 3.57h) and the effective line of sight (d = 3.57Kh), can be proven using the concept of refraction and basic trigonometry.

The optical line of sight formula, d = 3.57h, is derived based on the assumption that light travels in straight lines. When an antenna is at height h, the distance d to the horizon is the line of sight along a straight line. This formula is valid for situations where the effects of atmospheric refraction are negligible.

On the other hand, the effective line of sight formula, d = 3.57Kh, takes into account the adjustment factor K, which accounts for the effects of atmospheric refraction. Refraction occurs when light bends as it passes through different media with varying refractive indices. In the atmosphere, the refractive index varies with factors such as temperature, pressure, and humidity.

By introducing the adjustment factor K, which is commonly approximated as 4/3, the effective line of sight formula compensates for the bending of light due to atmospheric refraction. This allows for more accurate calculations of the distance d between the antenna and the horizon.

Both formulas are derived using basic trigonometry and the concept of similar triangles. By considering the height of the antenna and the line of sight to the horizon, the ratios of the sides of the triangles can be established, leading to the formulas d = 3.57h and d = 3.57Kh.

It's important to note that while these formulas provide useful approximations, they are not exact and may vary depending on atmospheric conditions.

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The width of the well is approximately [tex]\(4.351 \times 10^{-10}\)[/tex] meters.

The energy difference between two energy levels of an electron trapped in an infinitely deep square well is given by the formula:

[tex]\[\Delta E = \frac{{\pi^2 \hbar^2}}{{2mL^2}} \left( n_f^2 - n_i^2 \right)\][/tex]

where [tex]\(\Delta E\)[/tex] is the energy difference, [tex]\(\hbar\)[/tex] is the reduced Planck's constant, [tex]\(m\)[/tex] is the mass of the electron, [tex]\(L\)[/tex] is the width of the well, and [tex]\(n_f\)[/tex] and [tex]\(n_i\)[/tex] are the final and initial quantum numbers, respectively.

We can rearrange the formula to solve for [tex]\(L\)[/tex]:

[tex]\[L = \sqrt{\frac{{\pi^2 \hbar^2}}{{2m \Delta E}}} \cdot \frac{{n_f \cdot n_i}}{{\sqrt{n_f^2 - n_i^2}}}\][/tex]

Given that [tex]\(n_i = 6\), \(n_f = 2\)[/tex], and the wavelength of the emitted photon is [tex]\(\lambda = 532 \, \text{nm}\)[/tex], we can calculate the energy difference [tex]\(\Delta E\)[/tex] using the relation:

[tex]\[\Delta E = \frac{{hc}}{{\lambda}}\][/tex]

where [tex]\(h\)[/tex] is the Planck's constant and [tex]\(c\)[/tex] is the speed of light.

Substituting the given values:

[tex]\[\Delta E = \frac{{(6.626 \times 10^{-34} \, \text{J} \cdot \text{s}) \cdot (2.998 \times 10^8 \, \text{m/s})}}{{(532 \times 10^{-9} \, \text{m})}}\][/tex]

Calculating the result:

[tex]\[\Delta E = 3.753 \times 10^{-19} \, \text{J}\][/tex]

Now we can substitute the known values into the equation for [tex]\(L\)[/tex]:

[tex]\[L = \sqrt{\frac{{\pi^2 \cdot (6.626 \times 10^{-34} \, \text{J} \cdot \text{s})^2}}{{2 \cdot (9.109 \times 10^{-31} \, \text{kg}) \cdot (3.753 \times 10^{-19} \, \text{J})}}} \cdot \frac{{2 \cdot 6}}{{\sqrt{2^2 - 6^2}}}\][/tex]

Calculating the result:

[tex]\[L \approx 4.351 \times 10^{-10} \, \text{m}\][/tex]

Therefore, the width of the well is approximately [tex]\(4.351 \times 10^{-10}\)[/tex] meters.

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An electromagnetic wave, often abbreviated as EM wave, is a transverse wave consisting of mutually perpendicular electric and magnetic fields that fluctuate simultaneously and propagate through space.

The electric and magnetic field components of an electromagnetic wave (EM wave) are inextricably linked, with each of them being perpendicular to the other and in phase with one another. As a result, one cannot claim that one field component carries more energy than the other. The electric and magnetic fields both carry the same amount of energy and are equal to each other.

In an electromagnetic wave, the electric and magnetic field components are inextricably linked, with each of them being perpendicular to the other and in phase with one another. Therefore, one cannot claim that one field component carries more energy than the other. The electric and magnetic fields both carry the same amount of energy and are equal to each other. Thus, both the electric and magnetic field components have the same energy density.

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The image distance from the spherical mirror is -60 cm.

SPHERICAL MIRROR

Calculation of image distance:Given,Radius of the convex mirror,

r = 30 cm

Object distance, u = -20 cm (Negative sign indicates the object is in front of the mirror)

f = -r/2 = -15 cm

Using mirror formula,

1/f = 1/v + 1/u Where,

f = focal length of the mirror

v = image distance from the mirror1/-15 = 1/v + 1/-20V

= -60 cm

So, the image distance from the mirror is -60 cm.

Calculation of image height:magnification formula is given by,magnification,

m = v/u

Image height = m × object height Where,object height,

h = 0.3 m And,

v = -60 cm,

u = -20 cm

So, the magnification of the spherical convex mirror is -0.6.

Image height is calculated as -0.18 m.c.

Calculation of magnification:

We have,magnification, m = v/u

We have already calculated the image distance and object distance from the mirror in

m = -60 / -20 = -3

So, the magnification of the spherical convex mirror is -3.

Summary of the properties of the image formed:Location: The image is formed 60 cm behind the mirror.Orientation: The image is inverted.

Size: The size of the image is smaller than that of the object.

Type: Real, inverted, and diminished.

Set up using graphical methods (ray diagramming):The following ray diagram shows the graphical method to determine the properties of the image formed by the spherical convex mirror:
THIN LENSES

Calculation of image distance:

Given,Object distance,

u = -8 cm (negative sign indicates that the object is in front of the lens)

Focal length of the converging lens,

f = 6 cmUsing lens formula,1/f = 1/v - 1/u

Where,

v = image distance from the lens

1/6 = 1/v - 1/-8v

= 24/7 cm

So, the image distance from the converging lens is 24/7 cm.b. Calculation of image height:magnification formula is given by,magnification,

m = v/uObject height, h = 4 cm

Given, v = 24/7 cm,

u = -8 cmm = 24/7 / -8m

= -3/7

Thus, the magnification of the converging lens is -3/7.Image height is calculated as -12/7 cm.c. Calculation of magnification:magnification,

m = v/u

= 24/7 / -8

= -3/7

Thus, the magnification of the converging lens is -3/7.

Summary of the properties of the image formed:

Location: The image is formed at a distance of 24/7 cm on the other side of the lens.

Orientation: The image is inverted.

Size: The size of the image is smaller than that of the object.

Type: Real, inverted, and diminished.

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The efficiency of the skier is 86%

(a) The efficiency of the skier can be calculated by finding the ratio of the mechanical energy at the top of the hill to the mechanical energy at the bottom of the hill.

The mechanical energy of an object can be defined as the sum of its kinetic energy and its potential energy.

In this case, the skier starts from rest, so her initial kinetic energy is zero.

Her initial potential energy can be calculated using the formula:

mgh = (75 m)(9.8 m/s²)(63 kg)

       = 45,765 J

where

m = the mass of the skier,

g = the acceleration due to gravity,

h = the height of the hill.

Using the principle of conservation of energy, we know that the skier's mechanical energy at the bottom of the hill must be equal to her mechanical energy at the top of the hill, so her final kinetic energy is given by:

K = (1/2)mv²

  = (1/2)(63 kg)(25 m/s)²

  = 39,375 J

Her final potential energy is zero, since she is at ground level, so her mechanical energy at the bottom of the hill is equal to her final kinetic energy:

K = 39,375 J

Therefore, the efficiency of the skier is given by the ratio of her mechanical energy at the bottom of the hill to her mechanical energy at the top of the hill:

Efficiency = K/mgh

                = 39,375 J/45,765 J

                = 0.86 or 86%

(b)  Here's an energy flow diagram ,

  [Skier at Rest] ------(1)------> [Gravitational Potential Energy]

                                        |

                                        |

                                        |

                                        |

                                        V

   [Gravitational Potential Energy] ----(2)-----> [Kinetic Energy]

                                        |

                                        |

                                        |

                                        |

                                        V

    [Kinetic Energy] --------(3)-------> [Air Resistance/ Frictional Heat]

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