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3 years ago

Select all the correct answers.

Physics

2 answers:

EleoNora [17]3 years ago

8 0

Correct me if I’m wrong, but I believe it’s A and D, though I’m not fully sure.

Mila [183]3 years ago

7 0

A and d i’m sure
good luck

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Astronomers observe two separate solar systems each consisting of a planet orbiting a sun. The two orbits are circular and have

Oxana [17]

Answer:

(D) 3

Explanation:

The angular momentum is given by:

$\vec{L}=\vec{r}\ X \ \vec{p}$

Thus, the magnitude of the angular momenta of both solar systems are given by:

$L_1=Rm_1v_1=Rm_1(\omega R)=R^2m_1(\frac{2\pi}{T_1})=2\pi R^2\frac{m_1}{T_1}\\\\L_2=Rm_2v_2=2\pi R^2\frac{m_2}{T_2}$

where we have taken that both systems has the same radius.

By taking into account that T1=3T2, we have

$L_1=2\pi R^2\frac{m_1}{3T_2}=\frac{1}{3}2\pi R^2\frac{1}{T_2}m_1=\frac{1}{3}\frac{L_2}{m_2}m_1$

but L1=L2=L:

$L=\frac{1}{3}L\frac{m_1}{m_2}\\\\\frac{m_1}{m_2}=3$

Hence, the answer is (D) 3

HOPE THIS HELPS!!

3 0

3 years ago

A building that is 105.50 ft. feet tall is casting a shadow that is 37.50 ft. feet. If the building next to it, is casting a sha

Arlecino [84]

115.35 ft

Set the proportion up 37.50/105.50 = 41/x and solve for x

7 0

3 years ago

Would the frequency of the angular simple harmonic motion (SHM) of the balance wheel increase or decrease if the dimensions of t

storchak [24]

Answer:

Yes the frequency of the angular simple harmonic motion (SHM) of the balance wheel increases three times if the dimensions of the balance wheel reduced to one-third of original dimensions.

Explanation:

Considering the complete question attached in figure below.

Time period for balance wheel is:

$T=2\pi\sqrt{\frac{I}{K}}$

$I=mR^{2}$

m = mass of balance wheel

R = radius of balance wheel.

Angular frequency is related to Time period as:

$\omega=\frac{2\pi}{T}\\\omega=\sqrt{\frac{K}{I}} \\\omega=\sqrt{\frac{K}{mR^{2}}$

As dimensions of new balance wheel are one-third of their original values

$R_{new}=\frac{R}{3}$

$\omega_{new}=\sqrt{\frac{K}{mR_{new}^{2}}}\\\\\omega_{new}=\sqrt{\frac{K}{m(\frac{R}{3})^{2}}}\\\\\omega_{new}={3}\sqrt{\frac{K}{mR^{2}}}\\\\\omega_{new}={3}\omega$