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lesya692 [45]
3 years ago
13

How does Newton's third law of motion give a property of process​

Physics
1 answer:
elixir [45]3 years ago
5 0

Answer:

Newton's third law explains the generation of thrust by a rocket engine. In a rocket engine, hot exhaust gas is produced through the combustion of a fuel with an oxidizer. The hot exhaust gas flows through the rocket nozzle and is accelerated to the rear of the rocket. In re-action, a thrusting force is produced on the engine mount.

Explanation:

pa brainliest po :D

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Answer:

It makes it lighter when its closer and heavier when its farther way.

Explanation:

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2 years ago
A 975-kg sports car (including driver) crosses the rounded top of a hill at determine (a) the normal force exerted by the road o
iVinArrow [24]
There are missing data in the text of the problem (found them on internet):
- speed of the car at the top of the hill: v=15 m/s
- radius of the hill: r=100 m

Solution:

(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car W=mg (downwards) and the normal force N exerted by the road (upwards). The resultant of these two forces is equal to the centripetal force, m \frac{v^2}{r}, so we can write:
mg-N=m \frac{v^2}{r} (1)
By rearranging the equation and substituting the numbers, we find N:
N=mg-m \frac{v^2}{r}=(975 kg)(9.81 m/s^2)-(975 kg) \frac{(15 m/s)^2}{100 m}=7371 N

(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:
N=mg-m \frac{v^2}{r}=(62 kg)(9.81 m/s^2)-(62 kg) \frac{(15 m/s)^2}{100 m}=469 N

(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:
mg=m \frac{v^2}{r}
from which we find
v= \sqrt{gr}= \sqrt{(9.81 m/s^2)(100 m)}=31.3 m/s
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3 years ago
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