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SSSSS [86.1K]
3 years ago
13

A and B, move toward one another. Object A has twice the mass and half the speed of object B. Which of the following describes t

he forces the objects exert on each other when the collide and provides the best explanation? (A) The force exerted by A on B will be twice as great as the force exerted by B on A, as great as the force exerted by B on A other are the same, because the product of because A has twice the mass of B (B) The force exerted by A on B will be half because A has half the speed of B (C) The forces exerted by each object on the mass and speed is the same for both objects. (D) The forces exerted by each object on the other are the same, because inter objects cannot exert forces of difi magnitude on each other.
Physics
1 answer:
BARSIC [14]3 years ago
4 0

Answer:D

Explanation:

Given

mass of A is twice the mass of B half the velocity of B

Suppose F_a and F_b be the average force exerted on A and B respectively

and According to Newton third law of motion Force on the body A is equal to Force on body B but opposite in direction as they are action and reaction force.

Thus F_a=-F_b  and option d is correct

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A. A land speed car can decelerate at 9.8m/s. How long does it take the car to come to a complete stop from a run of 885 km/hr (
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Answer:

A. 25.08 s

B. 3082.53 m

C. 3×10⁵ m/s²

Explanation:

A. Determination of the time.

This can be obtained as illustrated below:

Acceleration (a) = –9.8 m/s²

Initial velocity (u) = 245.8 m/s

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Time (t) =.?

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0 = 245.8 – 9.8t

Collect like terms

0 – 245.8 = – 9.8t

– 245.8 = – 9.8t

Divide both side by –9.8

t = –245.8 / –9.8

t = 25.08 s

Therefore, it will take 25.08 s for the car to come to a complete stop.

B. Determination of the distance travelled by the car.

Acceleration (a) = –9.8 m/s²

Initial velocity (u) = 245.8 m/s

Final velocity (v) = 0 m/s

Distance (s) =?

v² = u² + 2as

0² = 245.8² + (2 × –9.8 × s)

0 = 60417.64 – 19.6s

Collect like terms

0 – 60417.64 = – 19.6s

– 60417.64 = – 19.6s

Divide both side by –19.6

s = –60417.64 / –19.6

s = 3082.53 m

Thus, the car travelled a distance of 3082.53 m before stopping completely.

C. Determination of the acceleration of the object.

Initial velocity (u) = 0 m/s

Final velocity (v) = 600 m/s

Distance (s) = 0.6 m

Acceleration (a) =?

v² = u² + 2as

600² = 0² + (2 × a × 0.6)

360000 = 0 + 1.2a

360000 = 1.2a

Divide both side by 1.2

a = 360000 / 1.2

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