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hoa [83]
3 years ago
13

The average period of pendulum clock is found to be 1.2s at sea level. The period of the same pendulum on a mountain top is foun

d to be 1.18s. If the acceleration due to gravity at sea level is 9.790m/s2, what is the acceleration due to gravity at the mountain top?
Physics
1 answer:
Kipish [7]3 years ago
5 0

Answer:

g' = 10.12m/s^2

Explanation:

In order to calculate the acceleration due to gravity at the top of the mountain, you first calculate the length of the pendulum, by using the information about the period at the sea level.

You use the following formula:

T=2\pi \sqrt{\frac{l}{g}}         (1)

l: length of the pendulum = ?

g: acceleration due to gravity at sea level = 9.79m/s^2

T: period of the pendulum at sea level = 1.2s

You solve for l in the equation (1):

l=\frac{gT^2}{4\pi^2}\\\\l=\frac{(9.79m/s^2)(1.2s)^2}{4\pi^2}=0.35m

Next, you use the information about the length of the pendulum and the period at the top of the mountain, to calculate the acceleration due to gravity in such a place:

T'=2\pi \sqrt{\frac{l}{g'}}\\\\g'=\frac{4\pi^2l}{T'^2}

g': acceleration due to gravity at the top of the mountain

T': new period of the pendulum

g'=\frac{4\pi^2(0.35m)}{(1.18s)^2}=10.12\frac{m}{s^2}

The acceleration due to gravity at the top of the mountain is 10.12m/s^2

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\\ \sf\longmapsto Work\:Done=Force\times Displacement

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Read 2 more answers
A worker pushed a 25.0 kg block 6.50 m along a level floor at constant speed with a force directed 30.0° below the horizontal. I
Kipish [7]

Answer:

a) 855.41J

b) 855.41J

Explanation:

The force equation is

N - mg - F sinθ = 0

N = mg + F sinθ

The frictonal  force is =

F_x = F cos \theta

U_xN = Fcos\theta

substitute the value of the normal force in the above equation

U_x (mg + Fsin\theta ) = F cos\theta

F = \frac{U_x mg}{cos\theta - U_x sin\theta } \\\\= \frac{0.41 \times 25 \times 9.8 }{cos 30^\circ- 0.41 \times sin30 ^\circ}

= 151.96N

a) work done = F × d

= (151.96) × (6.5) cos30°

= 855.41 J

b) The increase in thermal energy of the block floor system will be equal to the work done by the worker

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= (151.96) × (6.5) cos30°

= 855.41 J

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3 years ago
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