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madam [21]
3 years ago
9

A 1000-kilogram car traveling with a velocity of 20. meters per second decelerates uniformly at -5.0 meters per second2 until it

comes to rest. What is the magnitude of the impulse applied to the car to bring it to rest?
Physics
1 answer:
Elis [28]3 years ago
3 0

Answer:

-20000 kgm/s

Explanation:

Impulse: This can be defined as the product of the mass of a body and its change in velocity. The S.I unit of impulse is kgm/s.

Mathematically, impulse can be expressed as

I = m(v-u).............. Equation 1.

Where I = impulse applied to the car to bring it to rest, m = mass of the car, u = initial velocity of the car, v = final velocity of the car.

Given: m = 1000 kg, u = 20 m/s, v = 0 m/s ( to rest)

Substitute into equation 1

I = 100(0-20)

I = 1000(-20)

I = -20000 kgm/s

Hence the impulse applied to the car to bring it to rest = -20000 kgm/s

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Answer:

Explanation:

Given the height reached by a balloon after t sec modeled by the equation

h=1/2t²+1/2t

a) To calculate the height of the balloon after 40 secs we will substitute t = 40 into the modeled equation and calculate the value of t

If h(t)=1/2t²+1/2t

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b) Velocity is the change of displacement of a body with respect to time.

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when v = 0sec

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c) Velocity is the change of displacement of a body with respect to time.

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v(30) = 30+1/2

v(30) = 30.5ft/sec

The velocity of the balloon after 30 secs is 30.5 feet/sec

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