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Kamila [148]
3 years ago
12

5. An object has a momentum of 4,000 kg-m/s and a mass of 115 kg. It crashes into another object that has a mass of 100 kg, and

the two objects stick together. If the momentum is conserved, what is the new velocity of the combined object? Round to the nearest hundredth.
A. 34.78 m/s
B. 15.82 m/s
C. 69.56 m/s
D. 18.60 m/s​
Physics
1 answer:
marishachu [46]3 years ago
4 0

Answer:

D. 18.60

Explanation:

By the law of conservation, the momentum is neither loss nor gained but instead transfered. When they crash into each other, and stick, they combine to create a total mass of 215 kg. Since the momentum is transfered, the two objects, combined, have a total momentum of 4000 kg-m/s. We know that momentum equals mass times velocity. You then divide 4000 by 215 and get approximately 18.6 m/s

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A ball is thrown down vertically with an initial speed of 31 ft/s from a height of 40 ft. (a) What is its speed just before it s
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Using energy considerations, calculate the average force (in N) a 62.0 kg sprinter exerts backward on the track to accelerate fr
slava [35]

Answer:

69.68 N

Explanation:

Work done is equal to change in kinetic energy

W = ΔK = Kf - Ki = \frac{1}{2} mv^{2} _{f}  - \frac{1}{2} mv^{2} _{i}

W = F_{total} .d

where m = mass of the sprinter

vf = final velocity

vi = initial velocity

W  = workdone

kf = final kinetic energy

ki = initial kinetic energy

d = distance traveled

Ftotal = total force

vf = 8m/s

vi= 2m/s

d = 25m

m = 60kg

inserting parameters to get:

W = ΔK = Kf - Ki = \frac{1}{2} mv^{2} _{f}  - \frac{1}{2} mv^{2} _{i}

F_{total} .d =\frac{1}{2} mv^{2} _{f}  - \frac{1}{2} mv^{2} _{i}

F_{total} = \frac{\frac{1}{2} mv^{2} _{f} - \frac{1}{2} mv^{2} _{i}}{d}

F_{total=} \frac{\frac{1}{2} X 62 X6^{2} -\frac{1}{2} X 62 X2^{2} }{25}

= 39.7

we know that the force the sprinter exerted F sprinter, the force of the headwind Fwind = 30N

F_{sprinter} = F_{total} + F_{wind}  = 39.7 + 30 = 69.68 N

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