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cricket20 [7]
4 years ago
8

The moon orbits Earth approximately once a month. True False

Physics
1 answer:
Zarrin [17]4 years ago
5 0

Answer:true

Explanation:

It takes the moon approximately a month to complete an orbit round the earth

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What was Great Britain’s policy using African Americans during the war
kifflom [539]

Answer:

they were slaves, so they did practically everything anyone didn't do.

Explanation:

btw, which war? or battle?

6 0
3 years ago
Which statement accurately describes the outer planets?
tiny-mole [99]
The outer planets have a high gravity due to their large size
6 0
3 years ago
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An airplane cabin is pressurized to 570 mmhg. what is the pressure inside the cabin in atmospheres?
abruzzese [7]
1 atm corresponds to 760 mmHg, so we can set up a simple proportion to find how many atmospheres correspond to 570 mmHg:
1 atm: 760 mmHg = x: 570 mmHg
and from this, we find
x= \frac{1 atm \cdot 570 mmHg}{760 mmHg} =0.75 atm
8 0
3 years ago
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Two particles A and B start simultaneously from a Point P with velocities 20 m/s and 30 m/s respectively. A and B move with acce
zysi [14]

Answer:

<u>20 m/s</u>

Explanation:

<u>Given</u>

  • u(A) = 20 m/s
  • u(B) = 30 m/s
  • acceleration equal in magnitude but opposite in direction

<u>Solving</u>

  • Velocity of A at Q = 30 m/s
  • From, P to Q, <u>Δv(A) = 30 - 20 = +10 m/s</u>
  • Therefore, velocity of B at Q will be decreased by 10 as it is equal in magnitude but opposite in direction to A
  • Δv(B) = v(B at Q) - u(B at P)
  • -10 m/s = v(B at Q) - 30 m/s
  • v(B at Q) = 30 - 10 = <u>20 m/s</u>
6 0
3 years ago
A baseball m=.34kg is spun vertically on a massless string of length l=.52m. the string can only support a tension of tmax=9.9n
larisa86 [58]
<span>4.5 m/s This is an exercise in centripetal force. The formula is F = mv^2/r where m = mass v = velocity r = radius Now to add a little extra twist to the fun, we're swinging in a vertical plane so gravity comes into effect. At the bottom of the swing, the force experienced is the F above plus the acceleration due to gravity, and at the top of the swing, the force experienced is the F above minus the acceleration due to gravity. I will assume you're capable of changing the velocity of the ball quickly so you don't break the string at the bottom of the loop. Let's determine the force we get from gravity. 0.34 kg * 9.8 m/s^2 = 3.332 kg m/s^2 = 3.332 N Since we're getting some help from gravity, the force that will break the string is 9.9 N + 3.332 N = 13.232 N Plug known values into formula. F = mv^2/r 13.232 kg m/s^2 = 0.34 kg V^2 / 0.52 m 6.88064 kg m^2/s^2 = 0.34 kg V^2 20.23717647 m^2/s^2 = V^2 4.498574938 m/s = V Rounding to 2 significant figures gives 4.5 m/s The actual obtainable velocity is likely to be much lower. You may handle 13.232 N at the top of the swing where gravity is helping to keep you from breaking the string, but at the bottom of the swing, you can only handle 6.568 N where gravity is working against you, making the string easier to break.</span>
7 0
3 years ago
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