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sattari [20]
2 years ago
13

PLEASE HELP I WILL MARK YOU

Physics
1 answer:
DaniilM [7]2 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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An ideal monatomic gas at 275 K expands adiabatically and reversibly to six times its volume. What is its final temperature (in
Gwar [14]

The final temperature is 83 K.

<u>Explanation</u>:

For an adiabatic process,

T {V}^{\gamma - 1} = \text{constant}

\cfrac{{T}_{2}}{{T}_{1}} = {\left( \cfrac{{V}_{1}}{{V}_{2}} \right)}^{\gamma - 1}

Given:-

{T}_{1} = 275 \; K  

{T}_{2} = T \left( \text{say} \right)

{V}_{1}  = V

{V}_{2} = 6V

\gamma = \cfrac{5}{3} \;    (the gas is monoatomic)

\therefore \cfrac{T}{275} = {\left( \cfrac{V}{6V} \right)}^{\frac{5}{3} - 1}

 

\Rightarrow \cfrac{T}{275} = {\left( \cfrac{1}{6} \right)}^{\frac{2}{3}}  

T  =  275 \times 0.30

T  =  83 K.

3 0
3 years ago
10 points
grandymaker [24]
The answer is B tell me if I am wrong.
6 0
2 years ago
which of the following safety harnesses is the safest and most recommended when hunting from a treestand?
Lyrx [107]

The best tree stand safety harness is the Hunter Safety System Hybrid Flex Safety Harness, with its awesome ElimiShield Scent Control Technology.

Moreover, These stands are designed to be attached directly to the tree. Hunters using a fixed or suspended stand must choose a method of climbing up and down from the platform. The safest and most used method is sectional ladders.

You can learn more about this at:

brainly.com/question/28335498#SPJ4

4 0
10 months ago
A ball is thrown horizontally from a window that is 15.4 meters high at a speed of 3.01 m/s. How far will the ball go before hit
tia_tia [17]

The distance travelled by the ball that is thrown horizontally from a window that is 15.4 meters high at a speed of 3.01 m/s is 5.34 m

s = ut + 1 / 2 at²

s = Distance

u = Initial velocity

t = Time

a = Acceleration

Vertically,

s = 15.4 m

u = 0

a = 9.8 m / s²

15.4 = 0 + ( 1 / 2 * 9.8 * t² )

t² = 3.14

t = 1.77 s

Horizontally,

u = 3.01 m / s

a = 0 ( Since there is no external force )

s = ( 3.01 * 1.77 ) + 0

s = 5.34 m

Therefore, the distance travelled by the ball before hitting the ground is 5.34 m

To know more about distance travelled

brainly.com/question/12696792

#SPJ1

7 0
1 year ago
A ball initially at rest rolls down a hill with an acceleration of 3.3 m/s2. If it accelerates for 7.5 s, how far will it move?
goldfiish [28.3K]

<u>Answer:</u>

  Ball will move 92.8125 meter along the cliff in 7.5 seconds.

<u>Explanation:</u>

We have equation of motion , s= ut+\frac{1}{2} at^2, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In this case initial velocity = 0 m/s, acceleration = 3.3 m/s^2, we need to calculate displacement when time = 7.5 seconds.

Substituting

  s=0*7.5+\frac{1}{2} *3.3*7.5^2\\ \\ =92.8125 meter

  So ball will move 92.8125 meter along the cliff in 7.5 seconds.

5 0
3 years ago
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