What happens to the motion of an object when the forces acting on it are balanced? A. The motion changes. B. The motion does not
change. C. The motion speeds up. D. The motion slows down.
2 answers:
D. The motion slows down. Because If they are two forces putting force on the opposite sides, then the object would start to balance out and slow down
Question: What happens to the motion of an object when the forces acting on it are balanced?
Choices:
A. The motion changes.
B. The motion does not change.
C. The motion speeds up.
D. The motion slows down.
Answer: <u>D) The motion slows down when a motion of an object forces acting on the balance</u>.
<em>Hope this helps!.</em>
<em>~~~~~~~~~~~</em>
<em>~A.W~ZoomZoom44</em>
You might be interested in
Answer:
- The units are <em>(</em><em>rad</em><em>/</em><em>s</em><em>^</em><em>2</em><em>)</em><em> </em>
Answer:
11.0 kg m/s
Explanation:
The impulse exerted on the cart is equal to its change in momentum:

where
m = 5.0 kg is the mass of the cart
is its change in speed
Substituting numbers into the equation, we find

Answer:
Part a)

Part B)

Part c)

Explanation:
Part a)
As we know that

so we will have





Part B)
Angular speed of the yo-yo

so we have


Part c)
Tangential acceleration is given as



Answer:
a)
b) north edge will rise up
Explanation:
torque on the coil is given as

where N is number of loop = 9 loops
i is current = 7.80 A
-B -earth magnetic field = 
A- area of circular coil


A =0.015 m2
PUTITNG ALL VALUE TO GET TORQUE


b) north edge will rise up
A. gravity is your answer hope this helps