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krek1111
4 months ago
7

Suppose a 0.04-kg car traveling at 2.00 m/s can barely break an egg. What is the min

Physics
1 answer:
Softa [3K]4 months ago
4 0

Answer: Option (A) is correct.

Explanation:

The momentum of the 0.04 kg car moving at 2.00 m/s is

P_1=mass\times velocity=m_1\times v_1=0.04 kg\times 2.00 m/s=8.00 kg m/s

The highest permissible speed for the other car to avoid breaking the eggs matches that of the first car:

P_1=P_2

0.08 kg m/s=m_2\times v_2=0.08 kg\times v_2

v_2=1 m/s

Any velocity just above 1 m/s will raise the second car's momentum, causing the eggs to break. Therefore, the minimum speed required by the second car from the options given is 1.42 m/s.

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5 0
2 months ago
A 25kg child sits on one end of a 2m see saw. How far from the pivot point should a rock of 50kg be placed on the other side of
Sav [3153]

Answer:

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Explanation:

τchild=τrock  

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(F)child(d)child)=(F)rock(d)rock)

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(m)childg(d)child)=(m)rockg(d)rock)

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Now employing the given masses for the rock and child. The seesaw's total length is 2 meters, with the child sitting at one end, placing them 1 meter from the center of the seesaw.

(25kg)(1m)=(50kg)drock

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drock=25kg⋅m50kg

drock=0.5m

6 0
3 months ago
When θ= 0 ̊, the assembly is held at rest, and the torsional spring is untwisted. if the assembly is released and falls downward
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The rod measures 450mm in length, while the disk has a radius of 75mm. An upward-supporting pin holds the assembly in place when Θ=0, and there exists a torsional spring with a constant of k=20N m/rad at the pin. One end of the rod connects to the pin, while the other connects to the disk.


7 0
3 months ago
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