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Pepsi [2]
3 years ago
10

On a distant planet, a rock falls in 48.4 s from the top of a 1.10e+02 m cliff to the planet surface below. What is the accelera

tion due to gravity on this planet?  m/s2 What is the final velocity of the rock just before it hits the planet's surface?  m/s
Physics
1 answer:
pav-90 [236]3 years ago
8 0

this can be solve using the formala of free fall

t = sqrt( 2y/ g)

where t is the time of fall

y is the height

g is the acceleration due to gravity

48.4 s = sqrt (2 (1.10e+02 m)/ g)

G = 0.0930 m/s2

The velocity at impact

V = sqrt(2gy)

= sqrt( 2 ( 0.0930 m/s2)( 1.10e+02 m)

V = 4.523 m/s

<span> </span>

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A disk-shaped merry-go-round of radius 2.13 m and mass 175 kg rotates freely with an angular speed of 0.651 rev/s. A 55.4 kg per
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Answer:

Explanation:

To find out the angular velocity of merry-go-round after person jumps on it , we shall apply law of conservation of ANGULAR momentum

I₁ ω₁ + I₂ ω₂ = ( I₁  + I₂ ) ω

I₁ is moment of inertia of disk , I₂ moment of inertia of running person , I is the moment of inertia of disk -man system , ω₁ and ω₂ are angular velocity of disc and man .

I₁ = 1/2 mr²

= .5 x 175 x 2.13²

= 396.97 kgm²

I₂ = m r²

= 55.4 x 2.13²

= 251.34 mgm²

ω₁ = .651 rev /s

= .651 x 2π rad /s

ω₂ = tangential velocity of man / radius of disc

= 3.51 / 2.13

= 1.65 rad/s

I₁ ω₁ + I₂ ω₂ = ( I₁  + I₂ ) ω

396.97 x  .651 x 2π + 251.34 x 1.65 = ( 396.97 + 251.34 ) ω

ω = 3.14 rad /s

kinetic energy = 1/2 I ω²

= 3196 J

8 0
3 years ago
Ohm’s law describes the relationship between which quantities?
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The cornea is  responsible of refraction light 1/3 in eye.

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3 years ago
Each of 134 identical blocks sitting on a frictionless surface is connected to the next block by a massless string. The first bl
poizon [28]

Answer:

A. T=126N

B. T=63N

Explanation:

To determine the tension in each given blocks, we first determine the acceleration of each block. It obvious that each mass will move with the same acceleration since the string connecting them is massless.

Hence using the equation of force we have

F=ma

Where m=total mass of blocks,

a=acceleration

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For a 134 identical masses with an applied force of 134N, the acceleration of each mass can be computed as

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a. To calculate the tension in the string between the 126 and 127 block, we use the equation below

T=ma

Since the number of blocks before the string is 126, we multiply the mass of each block by 126.

Hence the tension can be computed as

T=126m*a

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B.To calculate the tension in the string between the 63 and 64 block, we use the equation below

T=ma

Since the number of blocks before the string is 63, we multiply the mass of each block by 63.

Hence the tension can be computed as

T=63m*a

Since a=1/m then

T=63m*1/m

T=63N

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3 years ago
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