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sattari [20]
3 years ago
13

PLEASE HELP I WILL MARK YOU

Physics
1 answer:
DaniilM [7]3 years ago
7 0

We want to see if the Trust SSC can archive a higher maximum speed.

Remember the second Newton's law:

Force = mass*acceleration.

And acceleration is the rate of change of the velocity.

Assuming the mass of the Trust SSC does not change, increasing the force would mean that we increase the acceleration, and thus, <u>larger </u><u>velocities </u><u>can be reached.</u>

<u />

Now there are some problems.

How we do increase the force?.

It is hard to ve precise, mostly because we don't really know how the system works, we just got some given random data that we can't use without context.

Another thing we need to remember, this velocity is measured in only one mile. So because we have a<em> fixed distance to measure it</em>, we only can get larger velocities by increasing the acceleration.

And to increase the acceleration we can increase the force, as discussed, or, decrease the mass.

How to decrease the mass? Using lighter materials.

Now let's see an example on how decreasing the mass would affect the acceleration.

  • Mass = 10,700kg
  • Acceleration = 20.8 m/s^2

Then an aproximate force is given by:

F = 10,700kg*20.8 m/s^2 = 222,560 N

Now, if we decrease the mass a little bit, for example to 10,000kg, and we do not change the force, the new acceleration will be given by:

10,000kg*a = 222,560 N

a = (222,560 N)/(10,000kg) = 22.3 m/s^2

So decreasing the mass a little, increases the acceleration, which would make a larger maximum velocity.

Concluding, there are two ways of increasing the maximum velocity:

Increasing the force.

Decreasing the mass of the car.

If you want to learn more, you can read:

brainly.com/question/12550364

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AlekseyPX

Answer:

160 kg

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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

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v_2 = Final Velocity of second car

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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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Explanation:

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Answer:

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