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photoshop1234 [79]
2 years ago
8

What is the speed of a car traveling 31,450 meters in 68450 seconds?

Physics
1 answer:
atroni [7]2 years ago
3 0

Answer:

2,152,752,500 meters per second

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A mass of 2 kg traveling at 3 m/s undergoes a one-dimensional elastic collision with a group of four 1kg masses that are at rest
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To pop a balloon you stab it with a pencil. If the area of the pencil tip is .01 cm² and the pressure applied by the pencil to t
Romashka-Z-Leto [24]

Answer:

Push with force of 1N

Explanation:

I have explained in the paper.

Goodluck

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3 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
7 0
4 years ago
2. A net force is _____.
Art [367]

The answer would be "the vector sum of forces acting on a particle or body."

Hope that helped ^^

8 0
3 years ago
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