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san4es73 [151]
3 years ago
12

Consider a solid sphere and a solid disk with the same radius and the same mass. Explain why the solid disk has a greater moment

of inertia than the solid sphere, even though it has the same overall mass and radius.
Physics
1 answer:
Mandarinka [93]3 years ago
8 0

Answer:

Explanation:

In a Solid sphere; the moment of inertia around its geometrical axis can be expressed by using the formula:

\mathtt{I_s = \dfrac{2}{5} M_s R^2_s}

For the solid disk; the moment of inertia around the central axis is:

\mathtt{I_D= \dfrac{1}{2}M_DR_D^2}

Suppose M_D = M_S; then we can say both to be equal to M

As well as R_D = R_S; then that too can be equal to R

Now;

\mathtt{I_s = \dfrac{2}{5} M R^2} --- (1)

\mathtt{I_D= \dfrac{1}{2}MR^2}---(2)

Multiplying equation (1) by 2, followed by dividing it by 2; we have:

\mathtt{I_s= \dfrac{2}{5}MR^2} \times \dfrac{2}{2}

I_s = \dfrac{4}{5} \times \dfrac{1}{2}MR^2 \\ \\ I_s = \dfrac{4}{5}\times I_D \\ \\ I_s > I_D

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Explanation:

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Answer:

» e. Electrons and protons

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A certain frictionless simple pendulum having a length l and mass m swings with period t. If both l and m are doubled, what is t
iVinArrow [24]

If l and m both are doubled then the period becomes √2*T

what is a simple pendulum?

It is the one which can be considered to be a point mass suspended from a string or rod of negligible mass.

A pendulum is a weight suspended from a pivot so that it can swing freely.

Here,

A certain frictionless simple pendulum having a length l and mass m

mass of pendulum = m

length of the pendulum = l

The period of simple pendulum is:

T = 2\pi \sqrt{\frac{l}{g} }

Where k is the constant.

Now the length and mass are doubled,

m' = 2m

l' = 2l

T' = 2\pi \sqrt{\frac{2l}{g} }

T' = \sqrt{2}* 2\pi \sqrt{\frac{l}{g} }

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Hence,

If l and m both are doubled then the period becomes √2*T

Learn more about Simple Harmonic Motion here:

<u>brainly.com/question/17315536</u>

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A 50-g cube of ice, initially at 0.0°C, is dropped into 200 g of water in an 80-g aluminum container, both initially at 30°C.
MakcuM [25]

Answer:

b. 9.5°C

Explanation:

m_i = Mass of ice = 50 g

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The final equilibrium temperature is 9.50022°C

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