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3 years ago

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A ball of mass m and radius R is both sliding and spinning on a horizontal surface so that its rotational kinetic energy equals

its translational kinetic energy. What is the ratio of the ball’s center-ofmass speed to the speed due to rotation only of a point on the ball’s surface? The moment of inertia of the ball is 0.63 m R2 .

1 answer:

spin [16.1K]3 years ago

4 0

Answer:

$\frac{v_{cm}}{\omega} = 1.122\cdot R$

Explanation:

According to the statement of the problems, the following identity exists:

$K_{t} = K_{r}$

$\frac{1}{2}\cdot m \cdot v_{cm}^{2} = 0.63\cdot m \cdot R^{2} \cdot \omega^{2}$

After some algebraic handling, the ratio is obtained:

$\frac{v_{cm}^{2}}{\omega^{2}}=1.26\cdot R^{2}$

$\frac{v_{cm}}{\omega} = 1.122\cdot R$

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Ostrovityanka [42]

Answer:

The change in gravitational potential energy of the climber-Earth system is $\Delta PE = 396900 \ J$

Explanation:

From the question we are told that

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The change in gravitational potential is mathematically represented as

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substituting values

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frutty [35]

radio waves,X-rays,

Explanation:

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A runner taking part in the 200-m dash must run around the end of a track that has a circular arc with a radius of curvature of

Leto [7]

Answer: $2.27\ m/s^2$

Explanation:

Given

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the radius of curvature of the track $r=29.5\ m$

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romanna [79]

Answer:

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Explanation:

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Before the collision we have:

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After the collision we have:

$p_f=(m_1+m_2)v_f$

since they clong together, acting as one body.

This means we have:

$m_1v_1+m_2v_2=(m_1+m_2)v_f$

Or:

$v_f=\frac{m_1v_1+m_2v_2}{m_1+m_2}$

Which for our values is:

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3 years ago

A point charge Q is placed at the center of a cube of side l. what is the flux through one face of the cube?

dimaraw [331]

Answer:

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Explanation:

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So, the flux through one face of the cube is $\dfrac{q}{6\epsilon_o}$ .

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3 years ago