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Ratling [72]
3 years ago
7

13. A set of pulleys lifts an 800 N crate 4 meters in 7 seconds. What power was used?

Physics
2 answers:
kiruha [24]3 years ago
8 0

Answer:power=457.1watts

Explanation:

Vlad [161]3 years ago
5 0

Answer:

457.14 Watt

Explanation:

The information we have is:

Force: F=800N

Distance: d=4m

Time: t=7s

the formula for the power is:

P=\frac{W}{t}

where W is work, and t si time.

Also, the formula for work is:

W=F*d

Where F is the force, and d is the distance.

Substituting the formula for work into the formula for power:

W=\frac{F*d}{t}

substituting all the values to find the power:

P=\frac{(800N)(4m)}{7s}=\frac{3200Nm}{7s}\\ P=457.14W

the power that was used is 457.14 Watt

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Answer: 11369.46 m/s

Explanation:

We have the following data:

m_{1}=5.8 kg is the mass of the bowling ball

V_{1}=4.34 m/s is the velocity of the bowling ball

m_{2}=2.214 g \frac{1 kg}{1000 g}=0.002214 kg is the mass of the ping-pong ball

V_{2} is the velocity of the ping-pong ball

Now, the momentum p_{1} of the bowling ball is:

p_{1}=m_{1}V_{1} (1)

p_{1}=(5.8 kg)(4.34 m/s)  

p_{1}=25.172 kg m/s (2)

And the momentum p_{2} of the ping-pong ball is:

p_{2}=m_{2}V_{2} (3)

If the momentum of the bowling ball is equal to the momentum of the ping-pong ball:

p_{1}=p_{2} (4)

m_{1}V_{1}=m_{2}V_{2} (5)

Isolating V_{2}:

V_{2}=\frac{m_{1}V_{1}}{m_{2}} (6)

V_{2}=\frac{25.172 kg m/s}{0.002214 kg} (7)

Finally:

V_{2}=11369.46 m/s

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