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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

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Answer:

Part a)

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Explanation:

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1)

As we know that

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now we have

\vec v \times \vec B = 1.44\hat k - 5.76 \hat i

so we have

E = 1.44 (0.50) = 0.72 V

2)

If the length vector is along Y direction then we have

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\vec v \times \vec B = 1.44\hat k - 5.76 \hat i

so we have

EMF = 0

6 0
3 years ago
What type of friction present when you wrench on a car?
dlinn [17]

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Explanation:

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3 years ago
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8 0
3 years ago
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Answer:

Explanation:

Given

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Answer:

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