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tester [92]
2 years ago
14

The car A has a weight of 4000 lb and is traveling to the right at 3 ft/s. Meanwhile a 2000 lb car B is traveling at 6 ft/s to t

he left. If the cars crash head-on, and at a time instant during the crash the impact force on A is 900 lb to the left, what is the magnitude and direction of the impact force exerted on B at the instant?
(A) 900 lb to the left
(B) 450 lb to the left
(C) 450 lb to the right
(D) 1800 lb to the left
(E) 900 lb to the right
Physics
1 answer:
iVinArrow [24]2 years ago
6 0

To solve this problem we should apply Newton's third law for which it is defined that there must be an equal reaction in the opposite direction.

From this law, if Car A generates a force on car B, that car must have opposed a force exactly the same but in the opposite direction. The car A moved to the left and generates a force of 900lb so the magnitude of the force in the car B is also 900lb but to the right (opposite direction to the first car)

The correct option is 900lb to the right.

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Whats the kinetic energy of an object that has a mass of 12 kilograms and moves with a velocity of 10 m/s
Vesnalui [34]

Answer:

600J

Explanation:

1/2mv^2

1/2(12)(10^2)

6(100)

600

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3 years ago
A car starts from rest and accelerates at 5 m/s/s.
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2 years ago
A boy throws a ball vertically up. It returns to the
NikAS [45]

Answer:

31.25 m

25m/sec

Explanation:

Given :-

Time = 5sec

V = 0 (in going up)

U = 0 (in comming down)

Find :-

H and U by which it is thrown up

Since the total time is 5 sec ,therefore half time will be taken to go up and another half will be taken to go down .

We know that ,

V = U + gt

0 = U - 10*2.5

U = 25 m/sec

Also,

V² = U² +2gs

0 = 625 - 20s

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7 0
3 years ago
A 2500 kg car traveling to the north is slowed down uniformly from an initial velocity of 20 m/s by a 5620 N braking force actin
Dennis_Churaev [7]

Answer:

the distance traveled by the car is 42.98 m.

Explanation:

Given;

mass of the car, m = 2500 kg

initial velocity of the car, u = 20 m/s

the braking force applied to the car, f = 5620 N

time of motion of the car, t = 2.5 s

The decelaration of the car is calculated as follows;

-F = ma

a = -F/m

a = -5620 / 2500

a = -2.248 m/s²

The distance traveled by the car is calculated as follows;

s = ut + ¹/₂at²

s = (20 x 2.5) + 0.5(-2.248)(2.5²)

s = 50 - 7.025

s = 42.98 m

Therefore, the distance traveled by the car is 42.98 m.

5 0
2 years ago
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