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RSB [31]
3 years ago
15

A 2500-N net force acting on a 880-kg car accelerates it at a rate of ______ m/s/s

Physics
1 answer:
Bond [772]3 years ago
8 0

Answer:

a = 2.84 m/s²

Explanation:

Given that,

Net force, F = 2500 N

Mass of the car, m = 880 kg

We need to find the acceleration of the car. Net force is given by :

F = ma

a=\dfrac{F}{m}\\\\a=\dfrac{2500\ N}{880\ m/s^2}\\\\a=2.84\ m/s^2

So, the acceleration of the car is 2.84 m/s².

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A<br> B <br> C <br> D<br><br> Plz help me.
Montano1993 [528]

Answer:

The correct option is;

{}               Man doing most work   {}    Box gaining the most energy

D   {}                 Y                                          Q

Explanation:

The given parameters of the question are;

The distance the box P is pushed by the Man X = 0

The force the Man X applies to the box P = x N

The distance the box Q is lifted by the Man Y = h > 0 meters

The minimum force the Man Y applies to the box Q = W, the weight of the box

Work done = Force × Distance

Energy gained = Potential energy + Kinetic Energy = (Mass × Gravity × Height = Weight × Height  = W × h) + 1/2 × Mass × Velocity²

The final velocity of either box = 0 m/s (The boxes are at rest on the ground or the shelf)

Therefore, Kinetic energy = 0 J

The work done by Man X = 0 × x = 0 J

The energy gained by the box P = W × 0 = 0 J

The work done by Man Y = W × h = W·h J

The energy gained by the box P = W × h = W·h J

We have, the work done by the man Y = W·h J is more than the work done by the man X = 0 J

The energy gained by the box P = W·h J is more than the energy gained by the box Q = 0 J

Therefore, the correct option is D, Man doing the most work is Y, box gaining the most energy is Q.

8 0
2 years ago
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