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salantis [7]
3 years ago
10

Suppose a child drives a bumper car head on into the side rail, which exerts a force of 3400 N on the car for 0.400 s. (Assume t

he initial velocity is in the positive direction.) (a) What impulse (in kg · m/s) is imparted by this force? (Indicate the direction with the sign of your answer.) kg · m/s (b) Find the final velocity (in m/s) of the bumper car if its initial velocity was 3.50 m/s and the car plus driver have a mass of 200 kg. You may neglect friction between the car and floor. (Indicate the direction with the sign of your answer.) m/s
Physics
1 answer:
babunello [35]3 years ago
3 0

Answer:

a)   I = -1360 N s   b)  v_{f} = -3.3 m / s

Explanation:

a) The expression of the impulse is

      I = F t = Δp

The force applied by the rail is in the opposite direction to the movement, so it is negative

     I = -3400 0.400

     I = -1360 N s

b) We use the momentum relationship with the moment

    I = ΔP

    I = m v_{f}- mv₀

   v_{f} = (I + m v₀) / m

   v_{f} = (-1360 + 200 3.5) / 200

   v_{f} = -3.3 m / s

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56km

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168 ÷ 240 = 0.7

0.7 × 80 = 56km

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Can someone help with my physics homework? please
Murrr4er [49]

Answer:

a) 19536 joules of work are done.

b) The work is done by the engine on the structure of the cart.

c) There are three options:  (i) Keeping the engine and changing the travelled distance, (ii) Changing the engine and keeping the travelled distance, (iii) Changing the engine and the travelled distance.

d) 24442 joules of work are done.

e) We may change for a bigger engine if it allows a greater acceleration and higher peak speed.

f) The bigger engine uses more gas to go 22 meters.

g) An empty semi truck uses more gas than a car since the first has much more mass than a car and is designed for moves big loads and for being fast.

Explanation:

a) If force applied in the cart is uniform, that is, constant in magnitude and direction and is parallel to distance travelled by the car, the work done on the cart is defined by the following equation:

W = F\cdot \Delta s (1)

Where:

F - Force applied by the cart, measured in newtons.

\Delta s - Distance travelled by the car, measured in meters.

W - Work done on the cart, measured in joules.

If we know that F = 888\,N and \Delta s = 22\,m, then the work done on the cart is:

W =(888\,N)\cdot (22\,m)

W = 19536\,J

19536 joules of work are done.

b) The work is done by the engine on the structure of the cart.

c) There are three options:  (i) Keeping the engine and changing the travelled distance, (ii) Changing the engine and keeping the travelled distance, (iii) Changing the engine and the travelled distance.

d) If we know that F = 1111\,N and \Delta s = 22\,m , then the work on the cart is:

W = (1111\,N)\cdot (22\,m)

W = 24442\,N

24442 joules of work are done.

e) We may change for a bigger engine if it allows a greater acceleration and higher peak speed.

f) The gas consumption is directly proportional to the square of velocity and mass of the cart and, hence, to the work done on the cart. In consequence, we conclude that the bigger engine uses more gas to go 22 meters.

g) An empty semi truck uses more gas than a car since the first has much more mass than a car and is designed for moves big loads and for being fast.

3 0
3 years ago
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