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Yuri [45]
3 years ago
14

Why are astronauts weightless in the Space Station? You may find it helpful to watch the video "Newton's Laws of Motion." Why ar

e astronauts weightless in the Space Station? You may find it helpful to watch the video "Newton's Laws of Motion."
A) Because the Space Station is moving at constant velocity.
B) Because the Space Station is constantly in free-fall around the Earth.
C) Because there is no gravity in space.
D) Because the Space Station is traveling so fast.
Physics
1 answer:
Zolol [24]3 years ago
7 0

Answer:

B) Because the Space Station is constantly in free-fall around the Earth.

Explanation:

Anything that is falling experiences an upward force on them. For example when a person is going down in a lift they will experience something that is pushing them upwards. This happens due to the fact that the total acceleration the body is feeling is less than the acceleration due to graviity.

The force on a body which is falling is

F=m(g-a)

Where,

m = Mass of object

g = acceleration due to gravity

a = acceleration the object is experiencing.

a = g. So, the force becomes zero and the object experiences weightlessness.

Hence, the astronauts in the space station experience weightlessness due to fact that the Space Station is constantly in free-fall around the Earth.

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On a touchdown attempt, 95.00 kg running back runs toward the end zone at 3.750 m/s. A 113.0 kg line-backer moving at 5.380 m/s
dsp73

Answer:

(a) 1.21 m/s

(b) 2303.33 J, 152.27 J

Explanation:

m1 = 95 kg, u1 = - 3.750 m/s, m2 = 113 kg, u2 = 5.38 m/s

(a) Let their velocity after striking is v.

By use of conservation of momentum

Momentum before collision = momentum after collision

m1 x u1 + m2 x u2 = (m1 + m2) x v

- 95 x 3.75 + 113 x 5.38 = (95 + 113) x v

v = ( - 356.25 + 607.94) / 208 = 1.21 m /s

(b) Kinetic energy before collision = 1/2 m1 x u1^2 + 1/2 m2 x u2^2

                                               = 0.5 ( 95 x 3.750 x 3.750 + 113 x 5.38 x 5.38)

                                               = 0.5 (1335.94 + 3270.7) = 2303.33 J

Kinetic energy after collision = 1/2 (m1 + m2) v^2                

                                                = 0.5 (95 + 113) x 1.21 x 1.21 = 152.27 J

4 0
3 years ago
How is the Maunder minimum related to climate?
Alexxandr [17]

Maunder minimum is related to climate due to the unusually low sunspot activity correlates to unusually cold climatic events. The answer is letter A. It happened around 1645 and 1715 and also coincided with the phenomena ‘Little Ice Age’ (1500 – 1850) in the Northern Hemisphere.

7 0
3 years ago
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When the air resistance can be ignored the velocity of an object dropped initially from rest is given by the following equation
ad-work [718]

Answer:

I am confused of your question. Do you want final velocity? To get final velocity, use (initial V)+(Gravity*Time)

Explanation:

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3 years ago
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What scientist was the first to propose the heliocentric model of the universe
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This theory was first proposed by Nicolaus Copernicus. Copernicus was a Polish astronomer. He first published the heliocentric system in his book: De revolutionibus <span>orbium </span>coelestium<span> , "On the revolutions of the heavenly bodies," which appeared in 1543.</span>
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3 years ago
A box weighing 12,000 N is parked on a 36° slope. What will be the component of the weight parallel to the plane that balances f
sweet-ann [11.9K]

Given :

A box weighing 12,000 N is parked on a 36° slope.

To Find :

What will be the component of the weight parallel to the plane that balances friction.

Solution :

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