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umka21 [38]
3 years ago
10

A 1000-kg car is driving toward the north along a straight horizontal road at a speed of 20.0 m/s. The driver applies the brakes

and the car comes to a rest uniformly in a distance of 160 m. What are the magnitude and direction of the net force applied to the car to bring it to rest?
Physics
1 answer:
Ludmilka [50]3 years ago
7 0

Answer:

Force applied to stop the car = 1,250 N

Explanation:

Given:

Mass of car (M) = 1,000 kg

Initial velocity (U) = 20 m/s

Final velocity (V) = 0 m/s

Distance (S) = 160 m

Find:

Force applied to stop the car.

Computation:

v^2 = u^2 + 2as\\\\0^2=20^2+2(a)(160)\\\\0=400+320(a)\\\\Acceleration = a = -1.25m/s^2\\\\Force = ma \\\\Force= 1,000(1.25)\\\\Force = 1,250 N

Force applied to stop the car = 1,250 N

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Answer:

F = 4.3671 * 10^{-8}\ Newtons

Explanation:

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F = GMm/d^2

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(G = 6.67408*10^{-11}\ m^3kg^{-1}s^{-2}), M and m are the masses of the corpses and d is the distance between them.

So we have that:

F = 6.67408*10^{-11} * 7.1*10^4 * 73/89^2

F = 4.3671 * 10^{-8}\ Newtons

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the orion nebula (at least the part we can see) is not very old (yet). while several hot, massive stars have had a chance to for
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3 0
1 year ago
A ball rolls horizontally off a table of height 0.6 m with a speed of 9 m/s. How long does it take the ball to reach the ground?
denis23 [38]

Answer: 0.067 s

Explanation:s = Ut + 1/2at^2

0.6 = 9t + 0.5 *10 *t^2

Where a = g =10m/s/s

Solving the quadratic equation

5t^2 + 9t - 0.6=0,

t= 0.067 s and - 1.7 s

Of which 0.067 s is a valid time

4 0
3 years ago
The spring of a spring balance is 6.0 in. long when there is no weight on the balance, and it is 8.4 in. long with 4.0 lb hung f
Sergeeva-Olga [200]

Answer:

W = 55.12 J

Explanation:

Given,

Natural length = 6 in

Force = 4 lb,  stretched length = 8.4 in

We know,

F = k x

k is spring constant

4 = k (8.4-6)

k = 1.67 lb/in

Work done to stretch the spring to 10.1 in.

W =k\int_{6}^{10.1} x

W = \dfrac{k}{2}[x^2]_6^{10.1}

W = \dfrac{1}{2}\times 1.67\times (10.1^2-6.0^2)

W = 55.12 J

Work done in stretching spring from 6 in to 10.1 in is equal to 55.12 J.

3 0
3 years ago
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