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ioda
3 years ago
10

If the potential energy of the system is zero, the force on the object is zero. When air resistance is taken into account, a bal

l thrown directly upward spends more time on the way down than on the way up. You drop a ball to the ground from rest, and, from the same height, throw a second identical ball down to the ground. Their change in gravitational potential energies are different. The potential energy of a system can be negative. If the force on an object is zero, the potential energy of the system is zero.
Physics
1 answer:
timama [110]3 years ago
3 0

Answer:

Explanation:

Considering it to be True or False Statements

(a)True ,

when Potential Energy of the system is zero then force on the object is zero

F=-\frac{\mathrm{d} U}{\mathrm{d} x}

F=0

(b)True

When ball is thrown upward it faces air resistance of greater magnitude because of the high speed of ball. while on returning the speed of ball gradually increase therefore it  take more time to reach same spot

(c)True

Change in gravitational Potential Energy of two balls will be different as gravitational Potential Energy depends on the distance of object from datum.

Also Potential Energy can also be negative as it depends on reference

(d)False

if Force on object is zero then Potential Energy is constant not zero

F=0

i.e.\frac{\mathrm{d} U}{\mathrm{d} x}=0

U=constant        

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What is the ratio of the object’s relativistic kinetic energy to its rest energy?
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We are given:

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The total energy or relativistic energy of an object is given by the equation:

$E=\frac{m c^{2}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$, where:

- m is the mass of the object.

- $c=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ is the speed of light.

- v is the speed of the object.

According to the special theory of relativity, the rest-mass energy, $E_{0}$, of a mass, m, is given by the equation: $E_{0}=m c^{2}$

Where, $c=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ is the speed of light.

Therefore, the ratio of the two is:

$\begin{aligned} \frac{E}{E_{0}} &=\frac{m c^{2} / \sqrt{1-\frac{v^{2}}{c^{2}}}}{m c^{2}} \\ &=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}} \end{aligned}$

If $v=0.240 c_{\text {r }}$ then the ratio of its total energy to its rest energy is:

$$\begin{aligned}\frac{E}{E_{0}} &=\frac{1}{\sqrt{1-\frac{(0.240 c)^{2}}{c^{2}}}} \\&=\frac{1}{\sqrt{1-(0.240)^{2}}} \\&=\frac{1}{\sqrt{0.9424}} \\& \approx \mathbf{1 . 0 3}\end{aligned}$$

What is Relativistic Energy?

  • The mass-energy equivalence concept states that mass and energy may be converted into one another. Its rest-mass energy is the quantity of energy that corresponds to an object's mass while it is at rest.
  • The entire energy of an object travelling at relativistic speed is referred to as relativistic energy (speed comparable to the speed of light). It is described as the total of an object's kinetic energy and rest mass.

Correct question : Find the ratio of the total energy to the rest energy of a particle of mass m moving with the following speeds.

(a)0.240 c

To learn more about relativistic energy visit:

brainly.com/question/9864983

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