Question 1

Suppose an airliner can travel at 300 km/hr in still air. If the airliner is traveling due east with a tailwind (a wind directed eastward) of 15 km/hr, the ground speed of the craft is

Accepted Answer

The ground speed of the craft is the sum of its speed in still air and the speed of the tailwind, which is 300 km/hr + 15 km/hr = 315 km/hr.

Question 2

Suppose an airliner can travel at 300 km/hr in still air. If the airliner is traveling due east with a headwind (a wind directed westward) of 15 km/hr, the ground speed of the craft is

Accepted Answer

The ground speed of the airliner is the speed of the airliner in still air minus the speed of the headwind, which is 300 km/hr - 15 km/hr = 285 km/hr.

Question 3

Suppose an airliner can travel at 300 km/hr in still air. If the airliner is traveling due east with a crosswind (a wind directed toward the south) of 60 km/hr, the ground speed of the craft is

Accepted Answer

The ground speed of the craft is the vector sum of the airspeed and the wind speed. The airspeed is 300 km/hr due east, and the wind speed is 60 km/hr due south. The vector sum of these two vectors is a vector with magnitude sqrt(300^2 + 60^2) = 294 km/hr and direction arctan(60/300) = 11.3 degrees south of east.

Question 4

The Michelson-Morley experiment demonstrated that

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The answer of The Michelson-Morley experiment demonstrated that
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Question 5

In special relativity time dilation occurs when

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The answer of In special relativity time dilation occurs when
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Question 6

A subatomic particle moving at 0.95 c is measured in the laboratory to live 1.5 $\mu$ s before decaying. The lifetime of the particle in a rest frame is

Accepted Answer

The lifetime of the particle in its rest frame is shorter due to time dilation. Using the time dilation formula $ t = \gamma t_0 $, where $ \gamma = \frac{1}{\sqrt{1-v^2/c^2}} $, and $ v = 0.95c $, we find $ t_0 = \frac{1.5 \mu s}{\gamma} \approx 0.47 \mu s $.

Question 7

For a time dilation to increase the "time" by a factor of 2, the speed of the object is

Accepted Answer

The time dilation factor, often represented by the Greek letter gamma (γ), is calculated using the formula γ = 1 / sqrt(1 - v^2/c^2), where v is the velocity of the object and c is the speed of light. For the time dilation to increase the time by a factor of 2, we set γ = 2, which gives us 1 / sqrt(1 - v^2/c^2) = 2. Solving for v/c gives us v/c = sqrt(1 - (1/4)) = sqrt(3/4) = 0.866. Therefore, the speed of the object must be approximately 0.87 c.

Question 8

A pendulum is noted to swing once every second on Earth. Onboard a spacecraft traveling at 0.89 c the pendulum is noted to swing once every

Accepted Answer

The answer of A pendulum is noted to swing once...

Question 9

The space shuttle orbits the Earth at a speed of 7.5 km/s for one week. The time difference due to time dilation the astronauts experience is

Accepted Answer

Time dilation in special relativity is given by the formula Δt' = Δt / √(1 - v²/c²). For the space shuttle at 7.5 km/s, the effect is very small. Over one week, the time difference is approximately 0.4 milliseconds.

Question 10

A distant source of light is emitting one of the Balmer lines at 656 nm. The light is observed to be at a wavelength of 680 nm due to a motion of the source. The velocity of the source is

Accepted Answer

The observed increase in wavelength (from 656 nm to 680 nm) indicates a redshift, which means the source is moving away from the observer. The velocity can be calculated using the formula for redshift: $z = \frac{\Delta \lambda}{\lambda_{\text{original}}} = \frac{680 - 656}{656} = \frac{24}{656}$. The velocity $v$ is related to the redshift $z$ by $v = zc$, where $c$ is the speed of light. Substituting the values, we get $v \approx 0.036c$, indicating the source is moving away at 3.6% the speed of light.

Question 11

A distant source of light is emitting the Balmer series, and the source of light is moving toward the Earth at a rate of 10 km/s. The wavelength of the 656.28-nm line will be observed at

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Question 12

Pepper, the space cat, travels to the Centauri star system 4.5 light-years away (one light-year is the distance light travels in one year) at a speed of 0.90 c. Relative to the Earth's frame of reference, the time it takes Pepper to reach the star system is

Accepted Answer

The time it takes for Pepper to reach the Centauri star system can be calculated using the formula $t = \frac{d}{v}$, where $t$ is the time, $d$ is the distance (4.5 light-years), and $v$ is the speed (0.90c, where c is the speed of light). Thus, $t = \frac{4.5 \, \text{light-years}}{0.90} = 5.0 \, \text{years}$.

Question 13

Pepper, the space cat, travels to the Centauri star system 4.5 light-years away (one light-year is the distance light travels in one year) at a speed of 0.90 c. Relative to Pepper's frame of reference, the time it takes him to reach the star system is

Accepted Answer

To calculate the time it takes for Pepper to reach the Centauri star system from Pepper's frame of reference, we use the time dilation formula from special relativity: $t' = \frac{t}{\sqrt{1 - v^2/c^2}}$, where $t'$ is the dilated time, $t$ is the time in the stationary frame, $v$ is the velocity of the moving frame, and $c$ is the speed of light. Given that Pepper travels at $0.90c$, the distance is 4.5 light-years, and the speed of light is $c$, the time in Pepper's frame can be calculated as follows: $t = \frac{distance}{speed} = \frac{4.5 \text{ light-years}}{0.90c} = 5 \text{ years}$ in the stationary frame. Applying the time dilation formula, $t' = 5 \text{ years} \times \sqrt{1 - 0.90^2} \approx 2.2 \text{ years}$.

Question 14

Pepper, the space cat, travels toward the Centauri star system 4.5 light-years away at a speed of 0.90 c. He sends back a transmission on his transmitter at a frequency of 31.20 MHz. The frequency received on Earth is

Accepted Answer

Due to the relativistic Doppler effect, the frequency of the transmission received on Earth will be lower than the transmitted frequency because Pepper is moving away from Earth. The formula for the observed frequency is f' = f * sqrt((1 - β) / (1 + β)), where β = v/c. Substituting the given values, the received frequency is approximately 7.158 MHz.

Question 15

Pepper, the space cat, travels back from the Centauri star system 4.5 light-years away at a speed of 0.90 c. He sends back a transmission on his transmitter at a frequency of 31.20 MHz. The frequency received on Earth is

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The answer of Pepper, the space cat, travels back from...

Question 16

A meter stick on a spaceship is measured in the reference frame of the Earth to have a length of 80.0 cm. The velocity of the spaceship is

Accepted Answer

The length contraction formula is $L = L_0\sqrt{1 - v^2/c^2}$, where $L$ is the observed length, $L_0$ is the proper length (the length of the object in its rest frame), $v$ is the velocity of the object relative to the observer, and $c$ is the speed of light. Rearranging the formula to solve for $v$, we get $v = c\sqrt{1 - (L/L_0)^2}$. Substituting $L = 80.0$ cm (or 0.8 m) for the observed length and $L_0 = 1.0$ m for the proper length, we find $v = c\sqrt{1 - (0.8)^2} = c\sqrt{1 - 0.64} = c\sqrt{0.36} = 0.6c$.

Question 17

A meter stick on a spaceship lies at a 45º angle perpendicular to the motion of travel. When the ship is moving at 0.80 c, the length of the meter stick, relative to a stationary observer, is

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Question 18

A meter stick on a spaceship lies at a 45° angle perpendicular to the motion of travel. When the ship is moving at 0.80 c, the angle the meter stick appears to make with respect to a stationary observer is

Accepted Answer

The length contraction in special relativity affects objects in the direction of motion but not perpendicular to it. For a meter stick at a 45° angle, its length component parallel to the motion will appear contracted to a stationary observer, while its perpendicular component remains unchanged. This results in the angle appearing larger than 45°. The correct calculation involves Lorentz transformation, leading to an angle greater than the original 45°, specifically 59°.

Question 19

A simple harmonic oscillator has a period of 2.00 s on Earth. The harmonic oscillator is taken on a spacecraft, and its period is measured on Earth to be 2.50 s. The velocity of the spacecraft is

Accepted Answer

The period of the oscillator increases due to time dilation, which is a relativistic effect. The time dilation formula is T' = T / sqrt(1 - v^2/c^2). Solving for v when T' = 2.50 s and T = 2.00 s gives v = 0.60 c.

Question 20

The speed at which a clock has to travel to run at a rate that is one-third the rate of a stationary clock is

Accepted Answer

The correct answer is derived from the time dilation formula in special relativity, $\Delta t' = \Delta t / \sqrt{1 - v^2/c^2}$, where $\Delta t'$ is the dilated time, $\Delta t$ is the proper time (time measured by a stationary observer), $v$ is the velocity of the moving clock, and $c$ is the speed of light. To make a clock run at one-third the rate of a stationary clock, we set $\Delta t' = 3\Delta t$, leading to $1/\sqrt{1 - v^2/c^2} = 3$. Solving for $v/c$, we get $v/c = \sqrt{1 - (1/3)^2} = \sqrt{8/9}$, which is approximately 0.943.

Quiz 36: The Theory of Special Relativity