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3 years ago

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Analogies exist between rotational and translational physical quantities. Identify the rotational term analogous to each of the

following: acceleration, force, mass, work, translational kinetic energy, linear momentum, impulse.

1 answer:

damaskus [11]3 years ago

4 0

Explanation:

Acceleration. Angular acceleration: Is the rate of change of the angular velocity of a body with respect to time.

Force. Torque: Is also called rotational force, since an applied torque will change the rotational motion of a body.

Mass. Moment of inertia: It is the resistance that opposes a body to rotates.

Work. Work: In a rotational motion, the work is done by the torque.

Translational kinetic energy. Rotational kinetic energy: is the kinetic energy due to the rotational motion of a body.

Linear momentum. Angular momentum: Represents the quantity of rotational motion of a body.

Impulse. Angular impulse: Is the change in angular momentum of a body.

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Image will form at distance 22.22 cm behind the mirror

Explanation:

As we know that the mirror formula is given as

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now we have

$\frac{1}{-50} + \frac{1}{d_i} = \frac{1}{40}$

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Why are some substances dissolve both polar and nonpolar substances?

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3 years ago

An unstable particle at rest breaks up into two fragments of unequal mass. The mass of the lighter fragment is equal to 2.90 ✕ 1

motikmotik

Answer:

The speed of the heavier fragment is 0.335c.

Explanation:

Given that,

Mass of the lighter fragment $M_{l}=2.90\times10^{-28}\ kg$

Mass of the heavier fragment $M_{h}=1.62\times10^{-27}\ Kg$

Speed of lighter fragment = 0.893c

We need to calculate the speed of the heavier fragment

Let v is the speed of the second fragment after decay

Using conservation of relativistic momentum

$0=\drac{m_{1}v_{1}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}-\drac{m_{2}v_{2}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}$

$\drac{m_{1}v_{1}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}=\drac{m_{2}v_{2}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}$

$\dfrac{2.90\times10^{-28}\times0.893c}{\sqrt{1-(0.893)^2}}=\dfrac{1.62\times10^{-27}v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}$

$\dfrac{v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}=\dfrac{2.90\times10^{-28}\times0.893c}{1.62\times10^{-27}\times0.45}$

$\dfrac{v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}=0.355c$

$\dfrac{v_{2}}{1-\dfrac{v_{2}^{2}}{c^2}}=(0.355c)^2$

$\dfrac{1-\dfrac{v_{2}^2}{c^2}}{v_{2}^2}=\dfrac{1}{(0.355c)}$

$\dfrac{1}{v_{2}^2}-\dfrac{1}{c^2}=\dfrac{1}{(0.355c)^2}$

$\dfrac{1}{v_{2}^2}=\dfrac{1}{c^2}+\dfrac{1}{0.126c^2}$

$\dfrac{1}{v_{2}^2}=\dfrac{1}{c^2}(1+\dfrac{1}{0.126})$

$\dfrac{1}{v_{2}^2}=\dfrac{8.93}{c^2}$

$v_{2}^2=\dfrac{c^2}{8.93}$

$v_{2}=0.335c$

Hence, The speed of the heavier fragment is 0.335c.

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3 years ago