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alexgriva [62]
2 years ago
10

A racquetball with a mass of 42 g is moving with a horizontal speed of 7 m/s to the right (+x direction). It hits the wall of th

e court and rebounds to the hitter with a horizontal speed of 7m/a to the left (-x direction).what is the magnitude of the racquetball's change in momentum?
Physics
1 answer:
zheka24 [161]2 years ago
5 0

The magnitude of the racquetball's change in momentum is 0.59 kgm/s approximately.

Given that a racquetball with a mass of 42 g is moving with a horizontal speed of 7 m/s to the right (+x direction).

mass m  = 42g = 42/1000 = 0.042kg

initial velocity before collision u = 7 m/s

It hits the wall of the court and rebounds to the hitter with a horizontal speed of 7m/s to the left (-x direction). That is,

velocity after collision v = 7 m/s

To calculate the magnitude of the racquetball's change in momentum, we will use the formula below

Change in momentum = Mv - Mu

Since momentum is a vector quantity, we will consider the direction.

Change in momentum = 0.042 x 7 - ( 0.042 x - 7)

Change in momentum = 0.294 + 0.294

Change in momentum = 0.588 kgm/s

Therefore, the magnitude of the racquetball's change in momentum is 0.59 kgm/s approximately.

Learn more on momentum here: brainly.com/question/402617

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zimovet [89]

Answer: a. 85km/hr b.82.3km/hr

c. 84km/hr

Explanation: first let take the total time from San Antonio to Houston to be 2hr.

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Distance = 98km

Total distance from Houston to San Antonio is 98+72 =170km

a. Average speed from San Antonio to Houston is

S1 =170/2

=85km/hr

b.half distance from Houston to San Antonio which is 170km/2

= 85km was covered with speed of 72km/hr first half, so time

t = dist/speed

t = 85/72 = 1hr 12 mins

Remaining 85 km covered with a speed of 98km/hr

Time = 85/98 = 0.88*60min

= 52 mins

Total time = 1hr +12mins +52mins

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Using reciprocal law

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C. Average speed to and fro(entire tripe)

= (85+82.3)/2

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g= \frac{GM}{r^2}
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r is the distance measured from the centre of the planet.

In our problem, we are located at 300 km above the surface. Since the Earth radius is R=6370 km, the distance from the Earth's center is:
r=R+h=6370 km+300 km=6670 km= 6.67 \cdot 10^{6} m

And now we can use the previous equation to calculate the field strength at that altitude:
g= \frac{GM}{r^2}= \frac{(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2})(5.97 \cdot 10^{24} kg)}{(6.67 \cdot 10^6 m)^2}  = 8.95 m/s^2

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what are the speeds of (a) a proton that is accelerated from rest through a potential difference of −1000 v−1000 v and (b) an el
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Answer:

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