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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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A mother and daughter are on a seesaw in the park. How far from the center must the 160.9lb mother sit in order to balance the 6
Soloha48 [4]

Answer:

The mother has to sit 2.17 ft from the center on the other side of the seesaw.

Explanation:

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we need that the absolute value of the torque exerted by the mom (160.9 lb) to be the same in magnitude (and of course opposite direction). So we assume that "d" is the distance at which the mother locates to make this torque equal in magnitude to her daughter's torque:

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Two loudspeakers are placed 1.8 m apart. They play tones of equal frequency. If you stand 3.0 m in front of the speakers, and ex
Ahat [919]

The question is incomplete. The complete question is :

Two loudspeakers are placed 1.8 m apart. They play tones of equal frequency. If you stand 3.0 m in front of the speakers, and exactly between them, you hear a minimum of intensity. As you walk parallel to the plane of the speakers, staying 3.0 m away, the sound intensity increases until reaching a maximum when you are directly in front of one of the speakers. The speed of sound in the room is 340 m/s.

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Solution :

Given :

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At the point D, the speakers are out of phase and so the path difference is $=\frac{\lambda}{2}$

Therefore,

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Thus the frequency is :

$f=\frac{v}{\lambda}$

$f=\frac{340}{0.99714}$

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