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Kryger [21]
3 years ago
15

Why do both the moon and the earth pull on the spaceship the entire time why must you be closer to the moon for the pull of the

earth and moon are equal
Physics
1 answer:
PilotLPTM [1.2K]3 years ago
5 0

1). Because every speck of matter in the Universe exerts gravitational force on every other speck of matter. The strength of the force depends on the masses of both specks, and the distance between them.

2). Because the Earth has about 80 times as much mass as the moon has.

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Please help me out as soon as you can. Really need help.
olga nikolaevna [1]
The answer would be 6 because 2.0x3= 6

(newton’s 2nd law)

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4 0
3 years ago
As a result of photosynthesis is the production of sugar molecules known as what?
Bezzdna [24]

Answer:

Glucose

Explanation:

4 0
3 years ago
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un movil que parte del reposo alcanza una velocidad de 75 m/s en 13 segundos ¿cual su aceleracion y el espacio que recorrio en l
Dmitriy789 [7]

Answer:

Acceleration = 5.77 m/s²

Distance cover in 13 seconds = 487.56 meter

Explanation:

Given:

Final velocity of mobile device = 75 m/s

initial velocity of mobile device = 0 m/s

Time taken = 13 seconds

Find:

Acceleration

Distance cover in 13 seconds

Computation:

v = u + at

75 = 0 + (a)(13)

13a = 75

a = 5.77

Acceleration = 5.77 m/s²

s = ut + (1/2)(a)(t²)

s = (0)(t) + (1/2)(5.77)(13²)

Distance cover in 13 seconds = 487.56 meter

8 0
3 years ago
The x-axis of a trajectory represents its _____.
sineoko [7]
The x-acis of a trajectory represents its C
3 0
4 years ago
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14. The design for a rotating spacecraft below consists of two rings. The outer ring with a radius of 30 m holds the living quar
Zolol [24]

Answer:

T= 11.0003s

Explanation:

From the question we are told that

The outer ring with a radius of 30 m

inner Gravity Approximately 9.80 m/s'

Outer Gravity Approximately 5.35 m/s.

Generally  the equation for centripetal force is given mathematically as

Centripetal acceleration enables Rotation therefore?

     \omega ^2 r =Angular\ acc

Considering the outer ring,

 \omega ^2 r = 9.8

  \omega ^2= \frac{9.8}{30}

 \omega = \sqrt{\frac{9.8}{30}}

 \omega= 0.571 rad/s

Therefore solving for  Period T

Generally the equation for solving Period T is mathematically given as

 T= \frac{2\pi}{\omega}

 T= \frac{2\pi}{0.571 rad/s}

 T= 11.0003s

5 0
3 years ago
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