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rjkz [21]
3 years ago
12

In two or more complete sentences, explain how you can prove that the number of degrees that the Moon rotates around the Earth e

very five days is approximately 61⁰.
Physics
2 answers:
ioda3 years ago
7 0
In sidereal month, Moon takes about 13 degrees rotation in one day. To get the precise value, you need to know that Moon takes about 27.32 days to complete its one rotation. And in one rotation there are about 360°. Divide 360 by 27.32. You would get 13.18° per day. Therefore, every five days it would be 13.18° * 5 = 65.9°.

However, in synodic month, it takes about 29.53 days. So for one day, its rotation would be 12.19°. For every five days, it would rotate about 12.19° * 5 = 60.95° or 61°. (I think you are talking about the synodic month)
Finger [1]3 years ago
5 0

Answer:

The number of degrees that the moon rotates around the earth every day is 12.2 degrees, if one day is 12.2 degrees than 5 days is 61 degrees.

Hope this helps! :)

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Hey girls and guys.. I AM TAKING MY FINAL RN IN CLASS!! I NEED HELP AND WUICKLY
nordsb [41]

Same as the speed as continental drift

7 0
3 years ago
A ball is dropped from rest at point O (height unknown). After falling for some time, it passes by a window of height 3 m and it
Usimov [2.4K]

Answer:

Part a)

\Delta v_{down} = 4.8 m/s

Part b)

v_a = 3.71 m/s

Explanation:

Since ball is dropped under uniform gravity

so here we can say that the distance of 3 m moved by the ball under uniform acceleration is given as

d = (\frac{v_f + v_i}{2})t

so we have

3 = (\frac{v_f + v_i}{2})(0.49)

v_f + v_i = 12.24

also we know that

v_f - v_i = at

v_f - v_i = (9.81)(0.49)

v_f - v_i = 4.81

now we will have

v_f = 8.52 m/s

v_i = 3.71 m/s

Part a)

\Delta v_{down} = v_b - v_a

\Delta v_{down} = 8.52 - 3.71

\Delta v_{down} = 4.8 m/s

Part b)

speed at the top of the window is

v_a = 3.71 m/s

3 0
3 years ago
Calculate the time period of simple pendulum whose length is 98.2cm
alekssr [168]

The time period resulting in oscillations will be 1.986 seconds.

<h3>What is the period of oscillation?</h3>

The period is the amount of time it takes for a particle to perform one full oscillation. T is the symbol for it. Taking the reciprocal of the frequency yields the frequency of the oscillation.

The time period of the oscillation is;

\rm T = 2 \pi\sqrt{\frac{L}{g}} \\\\ \rm T = 2 \times 3.14 \times \sqrt{ \frac{98.2 \  \times 10^{-2} \ m}{9.81 \ m/s^2}} \\\\ T= 1.986 \ sec

Hence the time period resulting oscillations will be 1.986 seconds.

To learn more about the time period of oscillation refer to the link;

brainly.com/question/20070798

#SPJ1

5 0
2 years ago
A ball whose mass is 0.3 kg hits the floor with a speed of 5 m/s and rebounds upward with a speed of 2 m/s. If the ball was in c
Sonbull [250]

Answer:

1400 N

Explanation:

Change in momentum equals impulse which is a product of force and time

Change in momentum is given by m(v-u)

Equating this to impulse formula then

m(v-u)=Ft

Making F the subject of the formula then

F=\frac {m(v-u)}{t}

Take upward direction as positive then downwards is negative

Substituting m with 0.3 kg, v with 2 m/s, and u with -5 m/s and t with 0.0015 s then

F=\frac {0.3(2--5)}{0.0015}=1400N

4 0
3 years ago
What does an atomic nucleus give off a particle?
____ [38]
The correct answer is answer choice B.
5 0
3 years ago
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