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Goryan [66]
2 years ago
15

What is the central idea for stacking rocks vertically to test the effects of forces

Physics
1 answer:
otez555 [7]2 years ago
8 0

Answer:

You can see how forces can stack on one another by comparing the normal force of the bottom most rock to the normal force of the upper most rock.

Explanation:

Say you have 3 rocks all with mass m=3kg

These 3 rocks will all have the gravitational force of -29.4N (multiply 3kg by -9.8m/s^2)

If none of the rocks are stacked, then they all have the same normal force as well (29.4N)

However, stack them, and see how their normal forces change.

m1 will continue to have a normal force of 29.4N

m2 will now have a normal force of 58.8

m3 will now have a normal force of 88.2

Though they all retain the same gravitational force, their normal force changes depending on the forces acting down on them.

Mass 1 have no forces acting downwards other than gravity.

Mass 2 has the force of mass 1 and gravity acting down on it.

Mass 3 has the force of mass 1, mass 2, and gravity acting down on it.

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Let’s say I am in a bumper car and have a velocity of 14 m/s, driving in the positive x-direction. I and my bumped car have a ma
AlekseyPX

Answer:

160 kg

12 m/s

Explanation:

m_1 = Mass of first car = 120 kg

m_2 = Mass of second car

u_1 = Initial Velocity of first car = 14 m/s

u_2 = Initial Velocity of second car = 0 m/s

v_1 = Final Velocity of first car = -2 m/s

v_2 = Final Velocity of second car

For perfectly elastic collision

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}\\\Rightarrow m_2v_2=m_{1}u_{1}+m_{2}u_{2}-m_{1}v_{1}\\\Rightarrow m_2v_2=120\times 14+m_2\times 0-(120\times -2)\\\Rightarrow m_2v_2=1920\\\Rightarrow m_2=\frac{1920}{v_2}

Applying in the next equation

v_2=\frac{2m_1}{m_1+m_2}u_{1}+\frac{m_2-m_1}{m_1+m_2}u_2\\\Rightarrow v_2=\frac{2\times 120}{120+\frac{1920}{v_2}}\times 14+\frac{m_2-m_1}{m_1+m_2}\times 0\\\Rightarrow \left(120+\frac{1920}{v_2}\right)v_2=3360\\\Rightarrow 120v_2+1920=3360\\\Rightarrow v_2=\frac{3360-1920}{120}\\\Rightarrow v_2=12\ m/s

m_2=\frac{1920}{v_2}\\\Rightarrow m_2=\frac{1920}{12}\\\Rightarrow m_2=160\ kg

Mass of second car = 160 kg

Velocity of second car = 12 m/s

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Answer:

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