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Arte-miy333 [17]
3 years ago
13

Simone is walking her dog on a leash. The dog is pulling with a force of 32 N to the right and Simone is pulling backward with a

force of 16 N. What is the net force on them?
Physics
2 answers:
sammy [17]3 years ago
6 0

Answer:

Net force acting on them is 16 N and it is acting to the right side.

Explanation:

It is given that,

Force acting by the dog, F_1 = 32\ N (right side)

Force acting by Simone , F_2 = -16\ N (backward)

Let backward direction is taken to be negative while right side is taken to be positive.

The net force will act in the direction where the magnitude of force is maximum. Net force is given by :

F=F_1+F_2

F=32+(-16)    

F = 16 N

So, the net force is 16 N and it is acting to the right side.

belka [17]3 years ago
5 0

Answer:

16 N (towards right)

Explanation:

Pulling force towards right, F1 = 32 N

Pulling force towards left, F2 = 16 N

Net force = F1 - F2

F = 32 - 16 = 16 N (towards right)

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A ball has a mass of 1.5kg and is thrown straight up with a speed of 60m/s, what is the ball’s momentum:
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Answer:

Assumption: the air resistance on this ball is negligible. Take g = 10\; \rm m \cdot s^{-2}.

a. The momentum of the ball would be approximately 60\;\rm kg \cdot m \cdot s^{-1} two seconds after it is tossed into the air.

b. The momentum of the ball would be approximately \rm \left(-45\; \rm kg \cdot m \cdot s^{-1}\right) three seconds after it reaches the highest point (assuming that it didn't hit the ground.) This momentum is smaller than zero because it points downwards.

Explanation:

The momentum p of an object is equal its mass m times its velocity v. That is: \vec{p} = m \cdot \vec{v}.

Assume that the air resistance on this ball is negligible. If that's the case, then the ball would accelerate downwards towards the ground at a constant g \approx -10\; \rm m \cdot s^{-2}. In other words, its velocity would become approximately 10\; \rm m \cdot s^{-1} more negative every second.

The initial velocity of the ball is 60\; \rm m \cdot s^{-1}. After two seconds, its velocity would have become 60\;\rm m \cdot s^{-1} + 2\; \rm s \times \left(-10\;\rm m \cdot s^{-1}\right) = 40\; \rm m \cdot s^{-1}. The momentum of the ball at that time would be around p = m \cdot v \approx 60\; \rm kg \cdot m \cdot s^{-1}.

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