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STatiana [176]
3 years ago
8

A 45.0-kg girl stands on a 13.0-kg wagon holding two 18.0-kg weights. She throws the weights horizontally off the back of the wa

gon at a speed of 6.5 m/s relative to herself . Assuming that the wagon was at rest initially, what is the speed of the girl relative to the ground after she throws both weights at the same time
Physics
1 answer:
topjm [15]3 years ago
4 0

Answer:

v = 4.0 m/s

Explanation:

  • Assuming no external forces acting during the instant that the girl throws the weights, total momentum must be conserved.
  • Since all the masses at rest initially, the initial momentum must be zero.
  • So, due to momentum must keep constant, final momentum must be zero too, as follows:

       p_{f} = m_{w} * v_{w} + m_{g+w} *v_{g+w} = 0 (1)

  • Assuming the direction towards the back of the wagon as positive, and replacing the masses in (1), we can solve for vg, as follows:

       v_{g+w} =- \frac{m_{w} *v_{w}}{m_{g+w} } =  - \frac{36.0kg *6.5m/s}{58.0kg } = -4.0m/s  (2)

  • This means that the girl (along with the wagon on she is standing) will move at a speed of 4.0 m/s in an opposite direction to the one she threw the weights.
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Answer:

Question not completed, so I analysed the question first

Tony drove to the mountains last weekend. there was heavy traffic on the way there, and the trip took 6 hours. when tony drove home, there was no traffic and the trip only took 4 hours. if his average rate was 22 miles per hour faster on the trip home, how far away does tony live from the mountains?

Explanation:

Let use variables to solve the problems

Let the first trip to be mountain take x hours

Let the trip back home take y hours

Let the speed to while going to the mountain be a miles/hour

Then, while going home it was b miles/hour faster than while going to the mountain.

Then, speed going home is (a+b)miles / hour

The formula for speed is given as

Speed=distance/time

The constant through out the journey is distance, the two journey has the same distance.

Then,

Distance =speed×time

For first journey going to the mountain

Distance = a×x=ax miles

For the second journey going home

Distance =y×(a+b)

Distance Mountain= distance home

ax=y(a+b)

Make a subject of the formula

ax=ya+yb

ax-ya=yb

a(x-y)=yb

a=yb/(x-y)

Therefore, distance from mountain is

Distance=speed ×time

Distance= a×x=ax

Now, applying the questions

So from the questions

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Also, b=22miles/hour

Then,

a=yb/(x-y)

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a=44miles/hour

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