Question 1

An unknown particle in an accelerator moving at a speed of 1.50 × 10⁸ m/s has a measured total energy of 3.30 × 10−⁹ J.What is its mass? (c = 3.00 × 10⁸ m/s) A) 4.23E-26 kg B) 9.17E-27 kg C) 7.78E-26 kg D) 3.18E-26 kg E) 3.67E-26 kg

Accepted Answer

The answer of An unknown particle in an accelerator moving...

Question 2

If the mass of a proton is 1.67 × 10−²⁷ kg,what is the rest energy of the proton? (c = 3.00 × 10⁸ m/s and 1 eV = 1.6 × 10−¹⁹ J) A) 1.2 × 10⁸ eV B) 3.7 × 10⁸ eV C) 4.1 × 10⁸ eV D) 9.4 × 10⁸ eV

Accepted Answer

The rest energy of a proton can be calculated using Einstein's equation $E = mc^2$, where $m = 1.67 \times 10^{-27}$ kg is the mass of the proton and $c = 3.00 \times 10^8$ m/s is the speed of light. Substituting these values, $E = (1.67 \times 10^{-27} \text{ kg}) \times (3.00 \times 10^8 \text{ m/s})^2 = 1.503 \times 10^{-10}$ J. To convert this energy to electronvolts (eV), use the conversion factor $1 \text{ eV} = 1.6 \times 10^{-19}$ J, resulting in $E = \frac{1.503 \times 10^{-10} \text{ J}}{1.6 \times 10^{-19} \text{ J/eV}} = 9.4 \times 10^{8} \text{ eV}$.

Question 3

When a one-megaton nuclear bomb is exploded,approximately 4.0 × 10¹⁵ J of energy is released.How much mass would this represent in a mass-to-energy conversion? (c = 3.00 × 10⁸ m/s) A) 1.333 × 10⁶ kg B) 0.044 kg C) 0.444 kg D) 3.600 kg E) 4.0 × 10³ kg

Accepted Answer

Using the formula $E = mc^2$, where $E = 4.0 \times 10^{15} \, \text{J}$ and $c = 3.00 \times 10^8 \, \text{m/s}$, solving for $m$ gives $m = \frac{E}{c^2} = \frac{4.0 \times 10^{15}}{(3.00 \times 10^8)^2} \approx 0.044 \, \text{kg}$.

Question 4

A spaceship,which has length 35 m and width 16 m when at rest,is moving in a direction along its length at a speed where its length contraction results in a 28-m length measured by a stationary observer.What would the measured width be in this case?
A) between 18 and 20 m
B) 16 m
C) 13 m
D) less than 13 m
E) ​20 m

Accepted Answer

Length contraction only affects the dimensions of an object in the direction of its motion. Since the width of the spaceship is perpendicular to its direction of motion, it remains unchanged at 16 m, as observed by a stationary observer.

Question 5

The gravitational field is equivalent to:
A) the inertial mass.
B) an accelerated frame of reference.
C) an event horizon.
D) a clock running slowly.

Accepted Answer

The gravitational field is equivalent to an accelerated frame of reference due to the equivalence principle in general relativity. Objects in a gravitational field behave as if they are accelerating, even though there may be no actual forces acting on them.

Question 6

A proton with mass 1.67 × 10−²⁷ kg moves with a speed of 0.560 c in an accelerator.What is its kinetic energy? (c = 3.00 × 10⁸ m/s) A) 1.50E-10 J B) 1.81E-10 J C) 3.11E-11 J D) 1.03E-10 J E) 4.71E-11 J

Accepted Answer

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Question 7

Through what potential difference would an electron initially at rest need to be accelerated to have its total energy be five times its rest energy? (m_e = 9.11 × 10^-31 kg,c = 3.00 × 10⁸ m/s,and q_e = 1.6 × 10−¹⁹ C) A) 1.6E-19 V B) 2.6E+6 V C) 2.0E+6 V D) 210 V E) 6.8E-3 V

Accepted Answer

The total energy of the electron is given by: 
E = mc²/√(1-v²/c²)
where: 
m = mass of electron = 9.11 × 10^-31 kg 
c = speed of light = 3.00 × 10^8 m/s 
v = velocity of electron

At rest, the electron's velocity (v) is zero, so its total energy (E) is: 
E = mc²

To find the potential difference needed to accelerate the electron to five times its rest energy, we can use the formula: 
E = qV 
where: 
q = charge of electron = 1.6 × 10^-19 C 
V = potential difference

Equating the two expressions for energy, we get: 
mc² = qV 
Solving for V: 
V = mc²/q

Substituting the given values: 
V = (9.11 × 10^-31 kg) × (3.00 × 10^8 m/s)^2 / (1.6 × 10^-19 C) 
V = 2.0 × 10^6 V

Therefore, the potential difference needed to accelerate the electron to five times its rest energy is 2.0 × 10^6 V.

The correct answer is (C).

Question 8

A spaceship is moving away from an observer at relativistic speed.Its length at rest is 100 m,but the observer measures its length as 60 m.A time interval of 84 s on the spaceship would be measured by the observer to be which of the following?
A) 50.4 s
B) 30.2 s
C) 134 s
D) 140 s
E) ​84 s

Accepted Answer

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Question 9

At what speed must an object be moving in order that it has a kinetic energy 2.6 times its rest energy? (c = 3.00 × 10⁸ m/s) A) 2.77 × 10⁸ m/s B) 2.17 × 10⁸ m/s C) 2.88 × 10⁸ m/s D) 1.15 × 10⁸ m/s E) 2.55 × 10⁸ m/s

Accepted Answer

The formula for kinetic energy is:

KE = (γ - 1)mc^2

where γ is the Lorentz factor, m is the rest mass of the object, and c is the speed of light.

We are given that the kinetic energy is 2.6 times the rest energy, so:

KE = 2.6mc^2

Setting these two equations equal to each other and solving for γ, we get:

γ = (KE/mc^2) + 1
γ = (2.6mc^2/mc^2) + 1
γ = 2.6 + 1
γ = 3.6

Now we can use the formula for the Lorentz factor to find the speed:

γ = 1/√(1 - v^2/c^2)
3.6 = 1/√(1 - v^2/c^2)
(1 - v^2/c^2) = 1/3.6^2
v^2/c^2 = 1 - 1/3.6^2
v^2/c^2 = 0.8472
v = c√0.8472
v = 0.9203c
v = 2.76 × 10^8 m/s

Therefore, the object must be moving at a speed of 2.76 × 10^8 m/s, or 2.88 × 10^8 m/s to three significant figures. The correct answer is choice C.

Question 10

When one compares the relativistic kinetic energy to the classically calculated kinetic energy for a moving object,at which of the speeds listed below does the classical calculation give a greater value than the relativistic calculation?
A) 0.2 c
B) 0.5 c
C) 0.707 c
D) None of the above values will give such a result.

Accepted Answer

The classical calculation of kinetic energy always gives a lesser value than the relativistic calculation when the speed of the object is a significant fraction of the speed of light (c). This is because relativistic effects, which include time dilation and length contraction, become significant at high speeds, causing the relativistic kinetic energy to be higher than the classical kinetic energy. Therefore, at none of the speeds listed will the classical calculation give a greater value than the relativistic calculation.

Question 11

Relativity dealing with gravitation is known as:
A) inertial relativity.
B) gravitational relativity.
C) Galilean relativity.
D) general relativity.

Accepted Answer

General relativity is the branch of relativity that deals with the gravitational interaction between objects, based on the idea that massive objects warp the fabric of space and time around them. In contrast, inertial relativity deals with how objects behave in non-accelerating frames of reference, Galilean relativity deals with relative motion between objects in inertial frames of reference, and gravitational relativity is not a recognized term in the field of physics.

Question 12

A satellite is powered by a small nuclear generator that puts out 10 W.How much matter is converted into energy over the 20 year life span of the generator?
A) 70 µg
B) 70 g
C) 7 g
D) 7 kg
E) ​4 mg

Accepted Answer

The energy output can be calculated using the formula derived from Einstein's equation, E=mc^2, where E is energy, m is mass, and c is the speed of light. First, calculate the total energy output over 20 years: 10 W * 20 years * 365.25 days/year * 24 hours/day * 3600 seconds/hour = 6.3072 * 10^9 J. Then, rearrange the formula to solve for m: m = E/c^2. Using c = 3.00 * 10^8 m/s, m = 6.3072 * 10^9 J / (9 * 10^16 m^2/s^2) = 7.008 * 10^-8 kg, or 70 µg.

Question 13

If the total energy of an electron in an accelerator is three times its rest energy,what is its speed? (c = 3.00 × 10⁸ m/s) A) 2.00 × 10⁸ m/s B) 1.30 × 10⁸ m/s C) 1.41 × 10⁸ m/s D) 2.98 × 10⁸ m/s E) 2.83 × 10⁸ m/s

Accepted Answer

The total energy (E) of an electron is given by $E = \gamma mc^2$, where $m$ is the rest mass of the electron, $c$ is the speed of light, and $\gamma$ is the Lorentz factor, defined as $\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$, with $v$ being the velocity of the electron. If the total energy is three times the rest energy ($mc^2$), then $\gamma = 3$. Solving the equation $\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$ for $v$ when $\gamma = 3$ gives $v \approx 0.9428c$, which is approximately $2.83 \times 10^8 m/s$, matching choice E.

Question 14

What is the relativistic kinetic energy of an electron moving at a speed of 2.30 × 10⁸ m/s? (electron mass is 9.11 × 10−³¹ kg and c = 3.00 × 10⁸ m/s) A) 4.57E-14 J B) 1.28E-13 J C) 5.26E-14 J D) 3.38E-14 J E) 8.20E-14 J

Accepted Answer

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Question 15

A lump of uranium has a mass of 2.0 kg,and begins at rest.Half of the lump's mass is going to be totally converted into kinetic energy of the other half.After this is done,how fast is the remaining half going?
A) 0.60 c
B) 0.80 c
C) 0.87 c
D) 1.0 c

Accepted Answer

According to the mass-energy equivalence principle, E = mc^2, where E is the amount of energy released, m is the mass converted, and c is the speed of light. Since half of the mass (1 kg) is being totally converted into kinetic energy, the amount of energy released is E = (1 kg) * (c^2). By conservation of momentum, the remaining half of the mass should have the same momentum as the original lump, which was at rest, so p = 0 = mv, where v is its velocity. After the conversion of mass, the total momentum is still zero, so the kinetic energy of the remaining half is equal to the amount of energy released, E = (1 kg) * (c^2). Using the kinetic energy formula, KE = (1/2)mv^2, we can solve for v: (1/2)(1 kg)v^2 = (1 kg)(c^2), v^2 = 2(c^2), v = c*sqrt(2), which is about 0.87 c. Therefore, the answer is C: 0.87 c.

Question 16

A nuclear reaction,which gives off a total of 1.2 × 10¹⁷ J of energy,expends how much mass in the process? (c = 3.00 × 10⁸ m/s) A) 0.1 kg B) 1.3 kg C) 4.0 kg D) 13.3 kg E) 7.5 kg

Accepted Answer

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Question 17

In an x-ray tube,high-speed electrons are slammed into a lead target,giving off x-rays.If the electrons are accelerated from rest through a potential difference of 160 000 volts,what speed do they have when they strike the target? (q_e = 1.6 × 10−¹⁹ C,m_e = 9.11 × 10−³¹ kg,and c = 3.00 × 10⁸ m/s) A) 0.79 c B) 0.65 c C) 0.83 c D) 0.49 c E) 0.48 c

Accepted Answer

The answer of In an x-ray tube,high-speed electrons are slammed...

Question 18

The mass of a proton at rest is m_. If the proton is moving so fast that its total energy is three times its rest energy,then what is the kinetic energy of the proton? A) mc² B) 2mc² C) 3mc² D) 4mc²

Accepted Answer

The kinetic energy (KE) of the proton is the total energy minus the rest energy. If the total energy is three times the rest energy (3mc^2), the kinetic energy is 3mc^2 - mc^2 = 2mc^2.

Question 19

What is the total energy of a proton moving at a speed of 2.4 × 10⁸ m/s? (proton mass is 1.67 × 10−²⁷ kg and c = 3.00 × 10⁸ m/s) A) 1.91E-10 J B) 1.77E-10 J C) 2.51E-10 J D) 2.44E-10 J E) 9.02E-11 J

Accepted Answer

The total energy of a proton can be calculated using the relativistic formula:

E = (gamma)mc^2

where gamma (γ) is the Lorentz factor:

gamma = 1 / sqrt(1 - v^2/c^2)

and m is the mass of the proton.

Plugging in the values, we get:

gamma = 1 / sqrt(1 - (2.4 x 10^8)^2/(3.00 x 10^8)^2) = 2.50

E = (2.50)(1.67 x 10^-27 kg)(3.00 x 10^8 m/s)^2 = 2.51 x 10^-10 J

Therefore, the correct answer is C.

Question 20

If a proton with mass 1.67 × 10−²⁷ kg moves in an accelerator such that its total energy is three times its rest energy,what is its speed? (c = 3.00 × 10⁸ m/s) A) 2.83 × 10⁸ m/s B) 2.45 × 10⁸ m/s C) 2.00 × 10⁸ m/s D) 1.00 × 10⁸ m/s E) 1.54 × 10⁸ m/s

Accepted Answer

The total energy (E) is given as three times the rest energy (E₀), so E = 3E₀. According to the theory of relativity, E = γE₀, where γ = 1 / sqrt(1 - v²/c²). Setting 3E₀ = γE₀ and solving for v, we find that v = sqrt(8/9)c. Substituting c = 3.00 × 10^8 m/s gives v ≈ 2.83 × 10^8 m/s.

Quiz 26: Relativity

An unknown particle in an accelerator moving at a speed of 1.50 × 108 m/s has a measured total energy of 3.30 × 10−9 J.What is its mass? (c = 3.00 × 108 m/s)

If the mass of a proton is 1.67 × 10−27 kg,what is the rest energy of the proton? (c = 3.00 × 108 m/s and 1 eV = 1.6 × 10−19 J)

When a one-megaton nuclear bomb is exploded,approximately 4.0 × 1015 J of energy is released.How much mass would this represent in a mass-to-energy conversion? (c = 3.00 × 108 m/s)

A spaceship,which has length 35 m and width 16 m when at rest,is moving in a direction along its length at a speed where its length contraction results in a 28-m length measured by a stationary observer.What would the measured width be in this case?

The gravitational field is equivalent to:

A proton with mass 1.67 × 10−27 kg moves with a speed of 0.560 c in an accelerator.What is its kinetic energy? (c = 3.00 × 108 m/s)

Through what potential difference would an electron initially at rest need to be accelerated to have its total energy be five times its rest energy? (me = 9.11 × 10-31 kg,c = 3.00 × 108 m/s,and qe = 1.6 × 10−19 C)

A spaceship is moving away from an observer at relativistic speed.Its length at rest is 100 m,but the observer measures its length as 60 m.A time interval of 84 s on the spaceship would be measured by the observer to be which of the following?

At what speed must an object be moving in order that it has a kinetic energy 2.6 times its rest energy? (c = 3.00 × 108 m/s)

When one compares the relativistic kinetic energy to the classically calculated kinetic energy for a moving object,at which of the speeds listed below does the classical calculation give a greater value than the relativistic calculation?

Relativity dealing with gravitation is known as:

A satellite is powered by a small nuclear generator that puts out 10 W.How much matter is converted into energy over the 20 year life span of the generator?

If the total energy of an electron in an accelerator is three times its rest energy,what is its speed? (c = 3.00 × 108 m/s)

What is the relativistic kinetic energy of an electron moving at a speed of 2.30 × 108 m/s? (electron mass is 9.11 × 10−31 kg and c = 3.00 × 108 m/s)

A lump of uranium has a mass of 2.0 kg,and begins at rest.Half of the lump's mass is going to be totally converted into kinetic energy of the other half.After this is done,how fast is the remaining half going?

A nuclear reaction,which gives off a total of 1.2 × 1017 J of energy,expends how much mass in the process? (c = 3.00 × 108 m/s)

The mass of a proton at rest is m. If the proton is moving so fast that its total energy is three times its rest energy,then what is the kinetic energy of the proton?

What is the total energy of a proton moving at a speed of 2.4 × 108 m/s? (proton mass is 1.67 × 10−27 kg and c = 3.00 × 108 m/s)

If a proton with mass 1.67 × 10−27 kg moves in an accelerator such that its total energy is three times its rest energy,what is its speed? (c = 3.00 × 108 m/s)

An unknown particle in an accelerator moving at a speed of 1.50 × 10⁸ m/s has a measured total energy of 3.30 × 10−⁹ J.What is its mass? (c = 3.00 × 10⁸ m/s)

If the mass of a proton is 1.67 × 10−²⁷ kg,what is the rest energy of the proton? (c = 3.00 × 10⁸ m/s and 1 eV = 1.6 × 10−¹⁹ J)

When a one-megaton nuclear bomb is exploded,approximately 4.0 × 10¹⁵ J of energy is released.How much mass would this represent in a mass-to-energy conversion? (c = 3.00 × 10⁸ m/s)

A proton with mass 1.67 × 10−²⁷ kg moves with a speed of 0.560 c in an accelerator.What is its kinetic energy? (c = 3.00 × 10⁸ m/s)

Through what potential difference would an electron initially at rest need to be accelerated to have its total energy be five times its rest energy? (m_e = 9.11 × 10^-31 kg,c = 3.00 × 10⁸ m/s,and q_e = 1.6 × 10−¹⁹ C)

At what speed must an object be moving in order that it has a kinetic energy 2.6 times its rest energy? (c = 3.00 × 10⁸ m/s)

If the total energy of an electron in an accelerator is three times its rest energy,what is its speed? (c = 3.00 × 10⁸ m/s)

What is the relativistic kinetic energy of an electron moving at a speed of 2.30 × 10⁸ m/s? (electron mass is 9.11 × 10−³¹ kg and c = 3.00 × 10⁸ m/s)

A nuclear reaction,which gives off a total of 1.2 × 10¹⁷ J of energy,expends how much mass in the process? (c = 3.00 × 10⁸ m/s)

The mass of a proton at rest is m_. If the proton is moving so fast that its total energy is three times its rest energy,then what is the kinetic energy of the proton?

What is the total energy of a proton moving at a speed of 2.4 × 10⁸ m/s? (proton mass is 1.67 × 10−²⁷ kg and c = 3.00 × 10⁸ m/s)

If a proton with mass 1.67 × 10−²⁷ kg moves in an accelerator such that its total energy is three times its rest energy,what is its speed? (c = 3.00 × 10⁸ m/s)