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SSSSS [86.1K]
3 years ago
13

A satellite orbiting earth once per day. From the reference frame of the moon, which orbit Earth once about every 27 days, what

is its motion?
A.The satellite never moves relative to the moon
B.The satellite travels continuously away from the moon.
C. The satellite remains at a constant distance from the moon.
D.The satellite continuously move closer and farther away from the mood.
Physics
2 answers:
sineoko [7]3 years ago
7 0
If each day the satelite orbits the earth its going faster than the moon and is passing it, so the satellite continuously moves closer and further away from the moon. like if you were standing watching a race then the racers would be moving away then towards you then away again
egoroff_w [7]3 years ago
6 0

Answer: D.The satellite continuously move closer and farther away from the moon.

The moon rotates on its axis and revolves around Earth in 27 days. The satellite is orbiting Earth once per day.

In moon 's frame of reference, the satellite would be in motion. Sometime it would be away and sometimes closer. From moon, we would observe satellite to rotate about the Earth. The satellite would get eclipsed by the Earth sometimes when it would go behind it as viewed from moon.

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A rocket in deep space has an empty mass of 220 kg and exhausts the hot gases of burned fuel at 2500 m/s. What mass of fuel is n
Scilla [17]

Answer:

Explanation:

Let fuel is released at the rate of dm / dt where m is mass of the fuel

thrust created on rocket

= d ( mv ) / dt

= v dm / dt

this is equal to force created on the rocket

= 220 dv / dt

so applying newton's law

v dm / dt = 220 dv / dt

v dm = 220 dv

dv / v = dm / 220

integrating on both sides

∫ dv / v    = ∫ dm / 220

lnv =  ( m₂ - m₁ ) / 220

ln4000 - ln 2500 = ( m₂ - m₁ ) / 220

( m₂ - m₁ ) = 220 x ( ln4000 - ln 2500 )

( m₂ - m₁ ) = 220 x ( 8.29  - 7.82 )

= 103.4 kg .

8 0
4 years ago
A 2kg hockey puck is sliding across the ice skating rink at 2 m/s. A player hits the puck so it's velocity increases to 10 m/s.
konstantin123 [22]

The work done on the puck is 96 J

Explanation:

According to the work-energy theorem, the work done on the hockey puck is equal to the change in kinetic energy of the puck.

Mathematically:

W=K_f -K_i= \frac{1}{2}mv^2-\frac{1}{2}mu^2

where

K_f = \frac{1}{2}mv^2 is the final kinetic energy of the puck, with

m = 2 kg being the mass of the puck

v = 10 m/s is the final speed

K_i = \frac{1}{2}mu^2 is the initial kinetic energy of the puck, with

u = 2 m/s being the initial speed of the puck

Substituting numbers into the equation, we find the work done by the player on the puck:

W=\frac{1}{2}(2)(10)^2 - \frac{1}{2}(2)(2)^2=96 J

Learn more about work and kinetic energy:

brainly.com/question/6763771  

brainly.com/question/6443626  

brainly.com/question/6536722

#LearnwithBrainly

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3 years ago
Why doesn't an object thrown in an upward direction fall the same distance in each time interval as it descends toward Earth? (A
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It's mostly used in CHEMICAL PROCESSES.
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Read 2 more answers
The potential energy stored in the compressed spring of a dart gun, with a spring constant of 32.50 N/m, is 0.640 J. Find by how
liraira [26]

Answer:

A

   x = 0.198456 \ m

B

    h  =  1.3061 \  m

C

 v =  5.06 \  m/s

D

  d = 4.0273 \  m

Explanation:

Considering the first question

From the question we are told that

   The spring constant is  k  =  32.50 N/m

    The potential energy is  PE  =  0.640 \ J

Generally the potential  energy stored in spring  is mathematically represented as   PE  =  \frac{1}{2}  *  k  *  x^2

=>    0.640=  \frac{1}{2}  * 32.50  *  x^2  

=>    x = \sqrt{0.03938}  

=>    x = 0.198456 \ m  

Considering the second question

 From the question we are told that

   The mass of the dart is  m =  0.050 kg

Generally from the law of energy conservation

         PE =  mgh

=>       0.640   =  0.050 *  9.8  *  h

=>      h  =  1.3061 \  m

Considering the third  question

   The height at which the dart was fired horizontally is  H  =   3.90\  m

Generally  from the law of energy conservation

         PE = KE

Here  KE is kinetic energy of the dart which is mathematical represented as

     KE  =  \frac{1}{2}  *  mv^2

=>      0.640 =  \frac{1}{2}  * 0.050 *  v^2

=>       v^2 = 25.6

=>       v =  5.06 \  m/s

Considering the fourth question

Generally the total time of flight of the dart is mathematically represented as

       t  =  \frac{ 2 *  H }{g}

=>     t  =  \frac{ 2 * 3.90 }{9.8 }

=>     t  =  0.7959 \ s

Generally the  horizontal distance from the equilibrium position to the ground is  mathematically represented as

       d =  v  *   t

=>     d = 5.06  *   0.7959

=>     d = 4.0273 \  m

5 0
3 years ago
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