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3 years ago

What is the acceleration due to gravity near the surface of earth

2 answers:

4 0

This is simply referring to the fact that the gravitational pull of the earth is greater as you get closer to the surface of the earth, meaning that acceleration increases as well.

beks73 [17]3 years ago

3 0

It's 9.81 meters per second squared // 32.2 feet per second squared. (Both rounded.)

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What is the process of change in a crust due to the motion of hot material underneath?(answer)?

frutty [35]

Im not exactly sure but I think the answer is techtonic plates collide

5 0

3 years ago

10) A soccer player kicks a soccer ball (m = 0.42 kg) accelerating from rest to 32.5m/s in 0.21s. Determine the force that sends

MaRussiya [10]

Explanation:

(10) Mass of a soccer player, m = 0.42 kg

Initial speed, u = 0

Final speed, v = 32.5 m/s

Time, t = 0.21 s

We need to find the force that sends soccer ball towards the goal.

Force, F = ma

$F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{0.42 \times (32.5-0)}{0.21}\\\\F=65\ N$

So, 65 N of force soccer ball sends towards the goal.

(11) Mass of the satellite, m = 72,000 kg

Initial speed, u = 0 m/s

Final speed, v = 0.63 m/s

Time, t = 1296 s

We need to find the force is exerted by the rocket on the satellite.

Force, F = ma

$F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{72,000\times (0.63-0)}{1296}\\\\F=35\ N$

So, 35 N of the force is exerted by the rocket on the satellite.

Hence, this is the required solution.

3 0

2 years ago

As the mass of an object increases, the weight of the object will ______?

Sergio039 [100]

Answer:

also increase

Explanation:

If the mass increases so will the weight.

4 0

2 years ago

Two identical cars A and B are at rest on a loading dock with brakes released. Car C, of a slightly different style but of the s

Nadusha1986 [10]

Answer:

Explanation:

Let the velocity after first collision be v₁ and v₂ of car A and B . car A will bounce back .

velocity of approach = 1.5 - 0 = 1.5

velocity of separation = v₁ + v₂

coefficient of restitution = velocity of separation / velocity of approach

.8 = v₁ + v₂ / 1.5

v₁ + v₂ = 1.2

applying law of conservation of momentum

m x 1.5 + 0 = mv₂ - mv₁

1.5 = v₂ - v₁

adding two equation

2 v ₂= 2.7

v₂ = 1.35 m /s

v₁ = - .15 m / s

During second collision , B will collide with stationary A . Same process will apply in this case also. Let velocity of B and A after collision be v₃ and v₄.

For second collision ,

coefficient of restitution = velocity of separation / velocity of approach

.5 = v₃ + v₄ / 1.35

v₃ + v₄ = .675

applying law of conservation of momentum

m x 1.35 + 0 = mv₄ - mv₃

1.35 = v₄ - v₃

adding two equation

2 v ₄= 2.025

v₄ = 1.0125 m /s

v₃ = - 0 .3375 m / s

3 0

3 years ago

During the Water Wheel simulation you were able to change the flow of water

Law Incorporation [45]

Answer:

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4 0

2 years ago