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gtnhenbr [62]
3 years ago
15

A car’s engine provides a forward force of 500 N. It is opposed by a resistive force of 200 N. The car accelerates forwards for

20 m. How much work is done by each force? By how much does the car’s energy increase?​
Physics
1 answer:
wariber [46]3 years ago
5 0

Answer:

Explanation:

W = Fd

Engine 500(20) = 10000 J

friction 200(-20) = -4000 J

Energy increase is 10000 - 4000 = 6000 J

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Answer:

that best describes the process is C

Explanation:

This problem is a calorimeter process where the heat given off by one body is equal to the heat absorbed by the other.

Heat absorbed by the smallest container

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              Q_a = M ce (T_{i}-T_{f})

how

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4 years ago
An airplane is moving at 350 km/hr. If a bomb is
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Answers:

a) -171.402 m/s  

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2} (1)

x=V_{ox}t (2)

V_{f}=V_{oy}-gt (3)

Where:

y=0 m is the bomb's final height

y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m is the bomb's initial height

V_{oy}=0 m/s is the bomb's initial vertical velocity, since the airplane was moving horizontally

t is the time

g=9.8 m/s^{2} is the acceleration due gravity

x is the bomb's range

V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s is the bomb's initial horizontal velocity

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Knowing this, let's begin with the answers:

<h3>b) Time </h3>

With the conditions given above, equation (1) is now written as:

y_{o}=\frac{1}{2}gt^{2} (4)

Isolating t:

t=\sqrt{\frac{2 y_{o}}{g}} (5)

t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}} (6)

t=17.49 s (7)

<h3>a) Final velocity </h3>

Since V_{oy}=0 m/s, equation (3) is written as:

V_{f}=-gt (8)

V_{f}=-(97.22)(17.49 s) (9)

V_{f}=-171.402 m/s (10) The negative sign only indicates the direction is downwards

<h3>c) Range </h3>

Substituting (7) in (2):

x=(97.22 m/s)(17.49 s) (11)

x=1700.99 m (12)

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