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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

5 0

3 years ago

Series circuits are characterized by the fact that there is a single pathway by which charge can travel. True or false?

Ulleksa [173]

I Think Its True My Dude Or Dudette
.
Hope this helps
.
Zane

6 0

3 years ago

Read 2 more answers

1.)State which definition of power below is correct.

Natalija [7]

C

kW, watts, kilowatts, W

6 0

2 years ago

Envision holding the end of a ruler with one hand and deforming it with the other. When you let go, you can see the oscillations

lapo4ka [179]

Answer: To increase the rigidity of the system you could hold the ruler at its midpoint so that the part of the ruler that oscillates is half as long as in the original experiment.

Explanation:

When a rule is displaced from its vertical position, it oscillates back and forth because of the restoring force opposing the displacement. That is, when the rule is on the left there is a force to the right.

By holding a ruler with one hand and deforming it with the other a force is generated in the opposite direction which is known as the restoring force. The restoring force causes the ruler to move back toward its stable equilibrium position, where the net force on it is zero. The momentum gained causes the ruler to move to the right leading to opposite deformation. This moves the ruler again to the left. The whole process is repeated until dissipative forces reduce the motion causing the ruler to come to rest.

The relationship between restoring force and displacement was described by Hooke's law. This states that displacement or deformation is directly proportional to the deforming force applied.

F= -kx, where,

F= restoring force

x= displacement or deformation

k= constant related to the rigidity of the system.

Therefore, the larger the force constant, the greater the restoring force, and the stiffer the system.

5 0

2 years ago

Which statement is true for a car that first goes around a curve of radius r at a constant speed v, and then goes around the sam

sasho [114]

Centripetal force is equal to (mv^2)/r
The way I use to answer these question is to set every variable to 1
m=1
v=1
r=1
so centripetal force =1
then change the variable we're looking at
and since we're find when it's half we could either change it to 1/2 or 2, but 2 is easier to use
m=1
v=2
r=1
((1)×(2)^2)/1=4
So the velocity in the 1st part is half the velocity in the 2nd part and the centripetal force is 4× less
The answer is the centripetal force is 1/4 as big the second time around

3 0

3 years ago