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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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Samanthawalks along a horizontal path in the direction shown the curved path is a semi circle with a radius of 2 m while the hor
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Answer:

Explanation:

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a semi circle with a radius of 2 m

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A 6000 kg roller coaster goes around a loop of radius 30 m at 6 m/s. What is the centripetal acceleration?
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3 years ago
Jen pushed a box for a distance of 80m with 20 N of force. How much work did she do?
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The answer is 1,600 J.

A work (W) can be expressed as a product of a force (F) and a distance (d):
W = F · d<span>

We have:
W = ?
F = 20 N = 20 kg*m/s</span>²
d = 80 m
_____
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8 0
3 years ago
A 153 g mass is attached to the end of an unstressed vertical spring (of constant 24.7 N/m) and then dropped. The acceleration o
Arte-miy333 [17]

Answer:

The answer to the question is

Its maximum speed is 1.54 m/s

Explanation:

Work done = Kinetic energy

0.5·m·v² = 0.5·k·x²

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m = mass

v = velocity

k =  spring constant

x = extension of the spring

We note that Force F is given by

F = m·a

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= 0.153×9.8 = 1.4994 N

Equating the work done by the force to the work done on the spring gives

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Substituting the value of x into the equation below gives

0.5·m·v² = 0.5·k·x²

0.5×0.153×v² = 12.35×0.121²

v² = 0.182÷0.0765 = 2.379

v = 1.54 m/s

6 0
3 years ago
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