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Wittaler [7]
3 years ago
15

How would the moon appear from Earth if the moon did not rotate?

Physics
2 answers:
andrey2020 [161]3 years ago
6 0
It would mean that only one side of earth would be light and the other dark all the time also we would only see the sun on one side and on the other we see the moon
n200080 [17]3 years ago
4 0
As the Earth rotates, the Moon rotates the same, hence why we only see one side of the Moon at any time, but if it stopped rotating we would see different parts of it as we rotate...basically it wouldn't look the same all the time (source: my brain, or 7th grade science class: either or)
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The centripetal force acting on a satellite in orbit
krok68 [10]

The centripetal force (of gravity) on a satellite in orbit is an
unbalanced force (because there's no equal force pulling
the satellite away from Earth), changes the direction of the
satellite (into a closed orbit instead of a straight line), and
always acts toward the center of whatever curve the satellite
happens to be on at the moment.

8 0
3 years ago
Read 2 more answers
A boat is rowed perpendicular to the shore of a river that flows at 3.0 m/s as shown in the
vazorg [7]
The magnitude is 5 ms^-1

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2 years ago
Help !!! Pick all the apply
Sergio [31]
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6 0
3 years ago
What is the velocity of an object that has been in free fall for 1.5s?
Crank

Answer:

D. 15 m/s downward

Explanation:

v = at + v₀

v = (-9.8 m/s²) (1.5 s) + (0 m/s)

v = -14.7 m/s

Rounded to two significant figures, the answer is D, 15 m/s downward.

8 0
3 years ago
WILL NAME THE BRAINLIEST! An airplane undergoes the following displacements: It first flies 72 km in a direction of 30° East of
nekit [7.7K]

Answer:

82.1 km

Explanation:

We need to resolve each displacement along two perpendicular directions: the east-west direction (let's label it with x) and the north-south direction (y). Resolving each vector:

A_x = (72) sin 30^{\circ} =36.0 km\\A_y = (72) cos 30^{\circ} = 62.4 km

Vector B is 48 km south, so:

B_x = 0\\B_y = -48

Finally, vector C:

C_x = -(100) cos 30^{\circ} =-86.6 km\\C_y = (100) sin 30^{\circ} = 50.0 km

Now we add the components along each direction:

R_x = A_x + B_x + C_x = 36.0 + 0 +(-86.6)=-50.6 km\\R_y = A_y+B_y+C_y = 62.4+(-48)+50.0=64.6 km

So, the resultant (which is the distance in a straight line between the starting point and the final point of the motion) is

R=\sqrt{R_x^2+R_y^2}=\sqrt{(-50.6)^2+(64.6)^2}=82.1 km

4 0
3 years ago
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