Assuming you exert a constant force on the wagon, how fast is it moving after 5 seconds?

A simple pendulum is used to determine the acceleration due to gravity at the surface of a planet. The pendulum has a length of

SVEN [57.7K]

Answer:

Acceleration due to gravity is 20 $m/sec^2$

So option (E) will be correct answer

Explanation:

We have given length of the pendulum l = 2 m

Time period of the pendulum T = 2 sec

We have to find acceleration due to gravity g

We know that time period of pendulum is given by

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

$2=2\times 3.14 \sqrt{\frac{2}{g}}$

$0.3184= \sqrt{\frac{2}{g}}$

Squaring both side

$0.1014= {\frac{2}{g}}$

$g=19.71=20m/sec^2$

So acceleration due to gravity is 20 $m/sec^2$

So option (E) will be correct answer.

8 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

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

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