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3 years ago

15

How long would it take a drag racer to increase her speed from 10.m/s to 20 m/s if her car accelerates at a uniform rate of 15 m

/s

1 answer:

garik1379 [7]3 years ago

5 0

It would take about 2 thirds of a second or .66666666 repeating of a second. please give brainliest?

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A. State whether the acceleration is positive, negative or zero for each section in the speed versus

bazaltina [42]

Answer:

a=positive

b=0

c=positive

d=negative

Explanation:

a=acceleration depends on the speed and time. if the speed and time are increasing at the same rate, the acceleration value will be positive as the vehicle is speeding up.

b=the speed and time are not increasing, therefore the vehicle is either stationary or travelling at a steady pace.

c=same explanation as a

d=the speed and time are not increasing at the same rate as the speed is decreasing. this means that the car is slowing down

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Velocity is a vector quantity which has both magnitude and direction. ... Net force is also a vector quantity which has both mag

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4 years ago

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At which position (1,2,3, or 4) does the roller coaster car have the most potential energy? Why?

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3 years ago

A high-jumper, having just cleared the bar, lands on an air mattress and comes to rest. Had she landed directly on the hard grou

Hoochie [10]

Answer:

e. the air mattress exerts the same impulse, but a smaller net avg force, on the high-jumper than hard-ground.

Explanation:

This is according to the Newton's second law and energy conservation that the force exerted by the hard-ground is more than the force exerted by the mattress.

The hard ground stops the moving mass by its sudden reaction in the opposite direction of impact force whereas the mattress takes a longer time to stop the motion of same mass in a longer time leading to lesser average reaction force.

<u>Mathematical expression for the Newton's second law of motion is given as:</u>

$F=\frac{dp}{dt}$ ............................................(1)

where:

dp = change in momentum

dt = time taken to change the momentum

We know, momentum:

$p=m.v$

Now, equation (1) becomes:

$F=\frac{d(m.v)}{dt}$

<em>∵mass is constant at speeds v << c (speed of light)</em>

$\therefore F=m.\frac{dv}{dt}$

and, $\frac{dv}{dt} =a$

where: a = acceleration

$\Rightarrow F=m.a$

also

$F\propto \frac{1}{dt}$

so, more the time, lesser the force.

<em>& </em><u><em>Impulse:</em></u>

$I=F.dt$

$I=m.a.dt$

$I=m.\frac{dv}{dt}.dt$

$I=m.dv=dp$

∵Initial velocity and final velocity(=0), of a certain mass is same irrespective of the stopping method.

So, the impulse in both the cases will be same.

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3 years ago

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