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Serhud [2]
3 years ago
7

By what factor should the length of the pendulum of a clock be changed if you want it to run 2 times faster?

Physics
2 answers:
Anarel [89]3 years ago
8 0

The period of a simple pendulum is given by T=2\pi \sqrt{ \frac{l}{g} }.

Here T is the period, l length of the pendulum and g acceleration due to gravity. When the pendulum runs 2 times faster, the period gets halved. That is

\frac{T}{2}=2\pi \sqrt{ \frac{\frac{l}{4}}{g} }. The new length of the pendulum is \frac{l}{4}.

That is the length of the pendulum decreases by a factor of 4.

Semenov [28]3 years ago
7 0
You have to decrease the length of the pendulum by 4 times in order to make the clock go 2 times faster
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s2008m [1.1K]

E = <u>kQ</u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u>

(r + h)²

where,

k = 9 × 10^9Nm²C^-2

Q = total charge, 300uC = 300 × 10^ -6C

r = 8 × 10^ -2m

h = 16 × 10^ -2m

then,

E = <u>9</u><u>e</u><u>9</u><u> </u><u>*</u><u> </u><u>3</u><u>0</u><u>0</u><u>e</u><u>^</u><u>-</u><u>6</u><u> </u><u> </u><u> </u><u> </u>

(8e^-2 + 16e^-2)²

E = 4687500N/C

6 0
2 years ago
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Answer:

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5 0
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5 0
3 years ago
A 51.0 kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 10.0 m th
shepuryov [24]

Answer:

The final speed of the crate is 12.07 m/s.

Explanation:

For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:

F = ma

a = \frac{F}{m} = \frac{225 N}{51.0 kg} = 4.41 m/s^{2}

Now, we can calculate the final speed of the crate at the end of 10.0 m:

v_{f}^{2} = v_{0}^{2} + 2ad_{1}                  

v_{f} = \sqrt{0 + 2*4.41 m/s^{2}*10.0 m} = 9.39 m/s    

For the next 10.5 meters we have frictional force:

F - F_{\mu} = ma

F - \mu mg = ma

So, the acceleration is:

a = \frac{F - \mu mg}{m} = \frac{225 N - 0.17*51.0 kg*9.81 m/s^{2}}{51.0 kg} = 2.74 m/s^{2}

The final speed of the crate at the end of 10.0 m will be the initial speed of the following 10.5 meters, so:

v_{f}^{2} = v_{0}^{2} + 2ad_{2}  

v_{f} = \sqrt{(9.39 m/s)^{2} + 2*2.74 m/s^{2}*10.5 m} = 12.07 m/s  

Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.  

I hope it helps you!                              

7 0
3 years ago
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