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Physics

1st Edition

Walker

ISBN: 9780133256925

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Page 961: Lesson Check

Exercise 21

Step 1

1 of 2

They are connected by mass-energy equivalence relation

$$
E=mc^2.
$$

Result

2 of 2

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Exercise 22

Step 1

1 of 2

Certain mass of matter is annihilated by the same mass of antimatter which produces energy in form of radiation (photons) and the energy produced is determined by mass-energy equivalence relation.

Result

2 of 2

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Exercise 23

Step 1

1 of 2

That is because in the mass-energy equivalence relation the mass is multiplied by $c^2$ and $c$ has really big value, let alone $c^2$.

Result

2 of 2

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Exercise 24

Solution 1

Solution 2

Step 1

1 of 2

$textbf{Given:}$

$m = 1.0$ kg

$textbf{Find:}$ rest energy, $E$

$$
textbf{Formula:}
$$

$$
begin{gather}
E = mc^2
end{gather}
$$

Using Eq (1), we can find $E$ as follows:

$$
begin{align*}
E &= 1.0 text{ kg} times c^2 \
&= 1.0 text{ kg} times (3.0 times 10^8 text{ m/s})^2 \
&= boxed{9 times 10^{16} text{ J}}
end{align*}
$$

Result

2 of 2

$$
E = 9 times 10^{16} text{ J}
$$

Step 1

1 of 2

This is easily calculated

$$
E=1text{ kg}times(3times10^{8}text{m/s})^2=9times 10^{16}text{ J}.
$$

Result

2 of 2

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Exercise 25

Solution 1

Solution 2

Step 1

1 of 2

$textbf{Given:}$

$m = 1.0$ kg

$v = 0.50 c$

$textbf{Find:}$ relativistic energy, $E$

$$
textbf{Formula:}
$$

$$
begin{gather}
E = dfrac{mc^2}{sqrt{1 – dfrac{v^2}{c^2}}}
end{gather}
$$

Using Eq (1), we can find $E$ as follows:

$$
begin{align*}
E &= dfrac{1.0 text{ kg} times c^2}{sqrt{1 – dfrac{(0.50c)^2}{c^2}}} \
&= boxed{1.04 times 10^{17} text{ J}}
end{align*}
$$

Result

2 of 2

$E = 1.04 times 10^{17}$ J

Step 1

1 of 2

Using the formula for relativistic energy we get

$$
E=frac{1text{ kg}times c^2}{sqrt{1-0.5^2}} = 1.04times 10^{17}text{ J}.
$$

Result

2 of 2

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Exercise 26

Solution 1

Solution 2

Step 1

1 of 2

$textbf{Given:}$

$Delta E = 120$ J

a) Due to the heating of the pan, energy was absorbed, since there is an increase in energy in the system, we can say that the mass of the pan increased by virtue of the mass-energy equivalence principle: $E = mc^2$

b) The change in mass can be calculated by using the mass-energy equivalence formula as follows:

$$
begin{align*}
Delta E &= Delta m c^2 \
Delta m &= dfrac{Delta E}{c^2} \
&= dfrac{120 text{ J}}{c^2} \
&= 1.33 times 10^{-15} text{ kg}
end{align*}
$$

The change is mass of the pan is $boxed{1.33 times 10^{-15} text{ kg}}$

Result

2 of 2

$$
1.33 times 10^{-15} text{ kg}
$$

Step 1

1 of 2

a) The mass of the pan is increased since any trapped energy in the system (that has no net momentum as is the case with heat here) contributes to the its’ measured mass.

b) The increase is directly calculated using equivalence relation

$$
Delta m = frac{120text{ J}}{c^2} =1.33times 10^{-15}text{ kg}.
$$

Result

2 of 2

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