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3 years ago

What is the net force acting on the object above?

Physics

1 answer:

photoshop1234 [79]3 years ago

6 0

The answer would be C (the one you picked)

Hope this helps

Have a great day/night

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The force of gravity on Jupiter is much stronger than the force of gravity on earth what is the answer?

Pachacha [2.7K]

This would be true. On Jupiter you would weigh 234 pounds if you were 100 pounds on Earth.

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3 years ago

Read 2 more answers

Write a hypothesis why the moon has very little liquid water.

Llana [10]

Because the Moon has a very small surface area compared to other spacial geo-bodies, it has cooled down much faster than Earth. Any water on the moon would freeze.

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3 years ago

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

pav-90 [236]

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

8 0

3 years ago

amping equipment weighing 6.0 kN is pulled across a frozen lake by means of ahorizontal rope. The coecient of kinetic friction i

Harman [31]

Answer:

Work done = 422.45 kJ

Explanation:

given,

weight of equipment = 6 kN

coefficient of kinetic friction = 0.05

distance up to which it is pulled = 1000 m

constant acceleration = 0.2 m/s²

Work done by the camper = ?

actual acceleration acting a'

m a = m a' - μ mg

a' = a + μ g

a' = 0.2 + 0.05 x 9.8

a' = 0.69 m/s²

Work done = Force x distance

F = m a'

$F = \dfrac{6000}{9.8} \times 0.69$

F = 422.44897 N

Work done = F x d

Work done = 422.44897 x 1000

Work done = 422449 J

Work done = 422.45 kJ

4 0

3 years ago

You stand on a bridge above a river and drop a rock into the water below from a height of 25 m. (Assume no air resistance)

Ilia_Sergeevich [38]

PART a)

here when stone is dropped there is only gravitational force on it

so its acceleration is only due to gravity

so we will have

$a = g = 9.8 m/s^2$

Part b)

Now from kinematics equation we will have

$y = v_i t + \frac{1}{2} at^2$

now we have

y = 25 m

so from above equation

$25 = 0 + \frac{1}{2}(9.8 )t^2$

$t = 2.26 s$

Part c)

If we throw the rock horizontally by speed 20 m/s

then in this case there is no change in the vertical velocity

so it will take same time to reach the water surface as it took initially

So t = 2.26 s

Part D)

Initial speed = 20 m/s

angle of projection = 65 degree

now we have

$v_x = vcos\theta$

$v_x = 20 cos65 = 8.45 m/s$

$v_y = vsin\theta$

$v_y = 20 sin65 = 18.13 m/s$

PART E)

when stone will reach to maximum height then we know that its final speed in y direction becomes zero

so here we can use kinematics in Y direction

$v_f - v_y = at$

$0 - 18.13 = (-9.8) t$

$t = 1.85 s$

so it will take 1.85 s to reach the top

5 0

3 years ago