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3 years ago

2000-kg car decelerated from 10 m/s to 5 m/s. What is the work done on the car ?

1 answer:

3 0

A 2000-kg car decelerated from 10 m/s to 5 m/s. What is the work done on the car ?

Known :

Car’s mass (m) = 2000 kg

Initial speed (vo) = 10 m/s

Final speed (vt) = 5 m/s

Wanted: net work

Solution :

Net work :

Wnet= ΔEK

Wnet = ½ m (vt2 – vo2)

Wnet = ½ (2000)(52 – 102)

Wnet = (1000)(25 – 100)

Wnet = (1000)(-75)

Wnet = -75,000 Joule

The minus sign indicates that the direction of displacement is opposite with the direction of the net force.

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3 years ago

Hey help me and thanks in advance if you do

blsea [12.9K]

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3 years ago

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A fireman is standing on top of a building 19 m high, holding a firehose 1 m above the top of the building. She finds that if sh

DENIUS [597]

Answer:

A 2 d vector model

The acceleration function is -9.8 m/s2 which is gravity

Initial velocity on the Y axis is 0, on the X axis is 12 m/s

Inital position is 20 mts above the ground.

It takes the water 1.01 seconds to reach the other building.

THe distace from one building to the other is 12.11 meters.

Explanation:

In order to solve this you just need to carefully read the problem and the data you are given, and use the formula for height in free fall:

$h=\frac{1}{2} g*t^{2}$

So first the data, we know that the water is coming out at a height of 20 meters since the building is 19 meters tall and the fireman is holding the firehose 1 meter above it, and the water is hitting the second building at a height of 15 meters, that means that the water is travelin -5meters.

Gravity as it doesn´t say otherwise would be 9.8m/s2 since that is gravity on earth, and water is leaving the firehose at 12m/s horizontally.

We can calculate the time by using the height formula fro free fall:

$h=\frac{1}{2} g*t^{2}\\5=\frac{1}{2} 9.81*t^{2}\\t^{2}= \frac{10}{9.8} \\t^{2}=1.0193\\t=1.009 seconds$

So it takes 1.009 seconds for the water to frop from 20 to 15 meters, as the horizontal velocity remains the same we just multiply it by the time and we get the horizontal distance between the two buildings and that would be:

12.11 meters.

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4 years ago

PLEASE HELP

Gennadij [26K]

Answer:

3.675 m

Explanation:

$a_{x} =0$ $v_{xo}=100$ $a_{y} =-g$ $v_{yo}=0$

X-direction | Y-direction

$R=x_{o}+ v_{xo} t$ | $y=y_{o}+v_{yo}t+\frac{1}{2}a_{y}t^2$

$75=100t$ | $y=0+0+\frac{1}{2} (9.8)(0.75)$

$\frac{75}{100} =t$ | $y=3.675 m$

$0.75s=t$

Hope it helps

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3 years ago

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3 years ago

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