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VARVARA [1.3K]
3 years ago
13

Question: Why do both the moon and the earth pull on the spaceship the entire time? Why must you be closer to the moon before th

e pull of the earth and moon are equal?
Physics
1 answer:
Ira Lisetskai [31]3 years ago
8 0
-- There's no limit to the distance of gravitational forces. 
There's gravitational force between Pluto and the lint in your
pocket ... not much, but it's there, and it can be calculated.

So there's ALWAYS gravitational force between the Earth and the
spaceship, AND ALSO between the Moon and the spaceship. 
Even before it's ever launched !

-- The Earth has about 80 times as much mass as the Moon has,
so you have to be much closer to the Moon before the gravitational
forces in each direction are equal.
You might be interested in
Did the displacement at this point reach its maximum of 2 mm before or after the interval of time when the displacement was a co
Igoryamba

Answer:

a. before

Explanation:

Did the displacement at this point reach its maximum of 2 mm before or after the interval of time when the displacement was a constant 1 mm?

from the graph given from a source. the vertical axis represents the displacement of the graph motion, whilst the horizontal side is representing the time variable of the motion .

displacement is distance in a specific direction.

before the displacement was maximum at 2mm was instant at time=0.04s.

But later was constant at 0.06s at a displacement point of 1mm

4 0
4 years ago
A drum rotates around its central axis at an angular velocity of 19.8 rad/s. If the drum then slows at a constant rate of 3.02 r
Debora [2.8K]

a) Time taken =6.39 s

b) Angle of rotation while coming to rest = 188.17 rad

<u>Explanation:</u>

Initial angular velocity of the drum \omega_{0} = 19.8 rad/s^{2}

Angular acceleration of the drum \alpha = -3.02 rad/s^{2}

a.) We know that

\omega =\omega_{0} + \alpha  t

0 =  19.8 - 3.02 t

t = \frac{19.8}{3.02}

t = 6.39s

b.) we know that

\theta = \omega_{0} t + \frac{1}{2} \alpha   t^{2}

\theta = 19.8 \times 6.39 + \frac{1}{2} \times 3.02 \times 6.39^{2}

\theta =188.17 rad

3 0
4 years ago
A 600-kg car traveling at 30.0 m/s is going around a curve having a radius of 120 m that is banked at an angle of 25.0°. The coe
ikadub [295]

Answer:

f_{fr}=1590.85 N

Explanation:

Here the total force at the horizontal components will be equal to the centripetal force on the car. So we will have:

f_{fr}cos(25)+Nsin(25)=m\frac{v^{2}}{r} (1)

  • f(fr) is the friction force
  • N is the normal force

Now, the sum of forces at the vertical direction is equal to 0.

Ncos(25)-mg-f_{fr}sin(25)=0 (2)          

Let's combine (1) and (2) to find f(fr)

f_{fr}=\frac{(mv^{2}/r)-mgtan(25)}{cos(25)+tan(25)sin(25)}

f_{fr}=\frac{(600*30^{2}/120)-600*9.81*tan(25)}{cos(25)+tan(25)sin(25)}  

f_{fr}=1590.85 N

I hope it helps you!

5 0
3 years ago
Consider a situation where you are playing air hockey with a friend. The table shoots small streams of air upward to keep the pu
artcher [175]

When we hit the puck from tap the puck will move forward.

This is due to the impulse provided by us at the time of hit. Due to this impulse the puck will move forward and start moving in some direction.

As soon as puck move forward the force on it is zero as the weight of the puck is counterbalanced by the air stream force and there is no other force on it so puck will continue its motion till it will hit at some other point.

So here the motion of the puck will be uniform motion till it will collide with some other points.

So here the correct option will be given as

<em>moves with a constant speed until hitting the other end.</em>

5 0
4 years ago
Which one is a true statement about exothermic reactions?
OLga [1]
<span>D. More heat is given off than is put into the reaction.

Exothermic reactions release heat more than the energy put into the reaction, opposite is an endothermic reaction.</span>
3 0
3 years ago
Read 2 more answers
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