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SOVA2 [1]
3 years ago
7

Gravity provides a centripetal force on the Moon, helping it stay in orbit around Earth.

Physics
2 answers:
Elodia [21]3 years ago
4 0

Answer:

The statement is true.

Both gravity and centrifugal force act on the Moon which causes it get pulled towards Earth (gravity) and get "flung away" so it doesn't hit us (centrifugal force).

Anika [276]3 years ago
3 0

Answer:

true APEX

Explanation:

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U"LL GET 24PTS IF U HELP ME
gavmur [86]

Answer: When your bike stops from brakes.

Explanation: Force and friction affect our daily lives in numerous amounts of ways. For instance, when a football is kicked, it moves faster later after some time its force decreases due to friction. A common example of friciton is when a bike stops. When the brakes are applied the friction on the pads cause the bike to stop.

6 0
3 years ago
A positive test charge q is released from rest at distance r away from a charge of Q and a distance 2r away from a charge of 2Q.
Luba_88 [7]

Answer: Option (b) is the correct answer.

Explanation:

It is given that a positive test charge q is released from rest at a distance r away from a charge of +Q and a distance 2r which is away from a charge of +2Q.

Then test charge to the right immediately after being released.

Therefore, the net force will be as follows.

            F = \frac{kqQ}{r^{2}} - kq\frac{(2Q)}{(2r)^{2}}

               = \frac{4KqQ - 2KqQ}{4r^{2}}

               = \frac{KqQ}{2r^{2}}

           F = \frac{KqQ}{2r^{2}} > 0

Thus, we can conclude that the test charge move to the right immediately after being released.

7 0
3 years ago
What is the spring constant for a supermarket scale that stretches 0.01 m when a force of 4 N is applied
bija089 [108]

Answer:

400N/m

Explanation:

K = F/x

= 4/0.01

= 400 N/m

5 0
3 years ago
The roller-coaster car shown in fig. 6-41 (h1 = 30 m, h2 = 12 m, h3 = 20 m), is dragged up to point 1 where it is released from
maxonik [38]
<span>Since there is no friction, conservation of energy gives change in energy is zero Change in energy = 0 Change in KE + Change in PE = 0 1/2 x m x (vf^2 - vi^2) + m x g x (hf-hi) = 0 1/2 x (vf^2 - vi^2) + g x (hf-hi) = 0 (vf^2 - vi^2) = 2 x g x (hi - hf) Since it starts from rest vi = 0 Vf = squareroot of (2 x g x (hi - hf)) For h1, no hf Vf = squareroot of (2 x g x (hi - hf)) Vf = squareroot of (2 x 9.81 x 30) Vf = squareroot of 588.6 Vf = 24.26 For h2 Vf = squareroot of (2 x 9.81 x (30 – 12)) Vf = squareroot of (9.81 x 36) Vf = squareroot of 353.16 Vf = 18.79 For h3 Vf = squareroot of (2 x 9.81 x (30 – 20)) Vf = squareroot of (20 x 9.81) Vf = 18.79</span>
7 0
3 years ago
Two astronauts on opposite ends of a spaceship are comparing lunches. One has an apple, the other has an orange. They decide to
lbvjy [14]

Answer:

Explanation:

We shall consider direction towards left as positive Let the required velocity be v and let v makes an angle φ

Applying law of conservation of momentum along direction of original motion

m₁ v₁  - m₂ v₂ = m₂v₃ - m₁ v₄

0.132 x 1.25 - .143 x 1.14 = 1.03 cos43 x .143 - v cos θ

v cos θ = .8

Applying law of conservation of momentum along direction perpendicular to direction of original motion

1.03 sin 43 x .143 = .132 x v sinθ

v sinθ = .76

squaring and adding

v² = .76 ² + .8²

v = 1.1 m /s

Tan θ = .76 / .8

θ = 44°

6 0
3 years ago
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