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pickupchik [31]
4 years ago
8

A pendulum of length =1.0 m is pulled to the side and released on the moon. It's period is measured to be 4.82 seconds. What is

the gravity on the moon?
Physics
2 answers:
Shkiper50 [21]4 years ago
7 0

Answer:

Gravity on the moon, g = 1.69 m/s²

Explanation:

It is given that,

Length of pendulum, l = 1 m

Time period, T = 4.82 seconds

We have to find the gravity of the moon. The time period of the pendulum is given by :

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

g = acceleration due to gravity on moon

g=\dfrac{4\pi^2l}{T^2}

g=\dfrac{4\pi^2\times 1\ m}{(4.82\ s)^2}

g = 1.69 m/s²

Hence, the gravity on the moon is 1.69 m/s².

skelet666 [1.2K]4 years ago
6 0

Answer:

Gravity on the moon, g = 1.69 m/s²

Explanation:

If the pendulum of length is 1.0 m is pulled to the side and released on the moon and It's period is measured to be 4.82 seconds, the gravity on the moon is 1.69 m/s².

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A 1200.0-kg car is traveling at 19m/s. The driver suddenly slams on the brakes and skids to a stop. The coefficient of kinetic f
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<h2>Answer</h2>

option D)

2.4 seconds

<h2>Explanation</h2>

Given in the question,

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speed of car = 19m/s

Force due to direction of travel

F = ma

  = 12000(a)

Force to due frictional force in reverse direction

-F = mg(friction coefficient)

   = -12000(9.81)(0.8)

<h2>-mg(friction coefficient) = ma  </h2>

(cancelling mass from both side of equation)

g(0.8) = a

(9.81)(0.8) = a

a = 7.848 m/s²

<h2>Use Newton Law of motion</h2><h3>vf - vo = a • t</h3>

where vf = final velocity

          vo = initial velocity

          a = acceleration

           t = time

0 - 19 = 7.8(t)

t = 19/7.8

  = 2.436 s

  ≈ 2.4s

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3 years ago
If a green light shines on a red toy, the toy will look:
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How is the energy of the wave affected if the amplitude of the wave increases from 2 meters to 4
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5 0
3 years ago
A 50.0 kg child stands at the rim of a merry-go-round of radius 1.50 m, rotating with an angular speed of 3.00 rad/s. (a) what i
White raven [17]
Weight of the child m = 50 kg 
Radius of the merry -go-around r = 1.50 m
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1.5
 Centripetal Acceleration a = 13.5m/sec^2
 (b)The minimum force between her feet and the floor in circular path
 Circular Path length C = 2 x 3.14 x 1.50 => c = 3 x 3.14 => C = 9.424
 Time taken t = 2 x 3.14 / w => t = 6.28 / 3 => t = 2.09
 Calculating velocity v = distance / time = 9.424 / 2.09 m/s => v = 4.5 m/s
 Calculating force, from equation F x r = mv^2 => F = mv^2 / r => 50 x (4.5)^2

/ 1.5
 F = 50 x 3 x 4.5 => F = 150 x 4.5 => F = 675 N
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 Hence with the force and the friction coefficient she is likely to stay on merry-go-around.
8 0
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