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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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When you're talking about gravity, it's easy to identify the equal
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each direction.

Consider the equal pair of opposite gravitational forces between
you and the Earth.  One force acts on you, and draws you toward
the center of the Earth.  We call that force "your weight". 
The other one acts on the Earth, and draws it toward the center
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7 0
3 years ago
Read 2 more answers
A 878-kg (1940 lb) dragster, starting from rest, attains a speed of 25.9 m/s (57.9 mph) in 0.62 s. (a) Find the average accelera
salantis [7]

Answer:

41.8m/s^2

Explanation:

Since the dragster starts from rest, initial velocity (u) = 0m/s, final velocity (v) = 25.9m/s, time (t) = 0.62s

From the equations of motion, v = u + at

a = (v - u)/t = (25.9 - 0)/0.62 = 25.9/0.62 = 41.8m/s^2

7 0
3 years ago
A car engine produces a force of 2000N while moving a distance of 200m in a time of 10s. What is the power developed by the engi
andreyandreev [35.5K]
Power = work/time
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3 0
3 years ago
How many lines per mm are there in the diffraction grating if the second order principal maximum for a light of wavelength 536 n
grandymaker [24]

To solve this problem it is necessary to apply the concepts related to the principle of superposition and the equations of destructive and constructive interference.

Constructive interference can be defined as

dSin\theta = m\lambda

Where

m= Any integer which represent the number of repetition of spectrum

\lambda= Wavelength

d = Distance between the slits.

\theta= Angle between the difraccion paterns and the source of light

Re-arrange to find the distance between the slits we have,

d = \frac{m\lambda}{sin\theta }

d = \frac{2*536*10^{-9}}{sin(24)}

d = 2.63*10^{-6}m

Therefore the number of lines per millimeter would be given as

\frac{1}{d} = \frac{1}{2.63*10^{-6} }

\frac{1}{d} = 379418.5\frac{lines}{m}(\frac{10^{-3}m}{1 mm})

\frac{1}{d} = 379.4 lines/mm

Therefore the number of the lines from the grating to the center of the diffraction pattern are 380lines per mm

6 0
3 years ago
Need help identifying the rest of the elements!
gregori [183]
I think that number five is lithium
6 0
3 years ago
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