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astra-53 [7]
3 years ago
5

A small rubber ball is thrown at a heavier, larger basketball that is still. The small ball bounces off the basketball. Assume t

here are no outside forces acting on the balls.
A. How does the force on the small ball compare to the force on the basketball?
B. Compare the total momentum of the two balls before and after the collision?
C. The mass of the basketball is 600 grams and its velocity before the small ball hits is 0 m/s. The mass of the small ball is 100 grams and its velocity is +5 m/s before the collision and -4 m/s afterward. What is the velocity of the basketball after the collision?
Physics
2 answers:
svp [43]3 years ago
6 0
The forces are the same for part A 
lina2011 [118]3 years ago
6 0

A 100-gram rubber ball was launched at a wall with different amounts of force. The speed of the ball was measured after it hit the wall and bounced off. The table below shows the data collected during the investigation.

Force Investigation Data

Launch Force on Ball Return Speed after Bounce

0.5 N 5 m/s

1.5 N 9 m/s

2.0 N 18 m/s

Which of the following best explains the trend shown by the data?

A 100-gram rubber ball was launched at a wall with different amounts of force. The speed of the ball was measured after it hit the wall and bounced off. The table below shows the data collected during the investigation.

Force Investigation Data

Launch Force on Ball Return Speed after Bounce

0.5 N 5 m/s

1.5 N 9 m/s

2.0 N 18 m/s

Which of the following best explains the trend shown by the data?

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Steam is leaving a pressure cooker whose operating pressure is 20 psia. It is observed that the amount of liquid in the cooker h
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Answer:

a) 34.05ft/s

b).1156.2BTU/lbm

c) 2.04BTU/s

Explanation:

Amount of liquid that has evaporated, m = ◇Vliq/ Vf

We replace the values to make conversion

m = (0.6gal/ 0.01683ft^3/lbm) × (0.13368ft^3/1gal)

m = 4.755lb

The mass flow rate of exit steam is given by:

m' = m/◇t

We replace values to make conversion

m' =( 4.766lb/45min) = 0.1059lb/min × 1min/60s

m' = 0.001765lb/s

The exit velocity V = m'/pA = m'Vg/A

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V =[ (0.001765lbm/s)(20.093ft^3/lbm) /(0.15 in^2)]× (144in^2/1ft^2)

V = 34.05ft/s

b) The total and flow energies per unit mass is given by:

Eflow= Pv = h - u

We replace the values to make conversion

Eflow = 1156.2 - 1081.8

Eflow = 74.4BTU/lbm

Therefore theta= h + ke + pe

Theta approximately =h = 1156.2BTU/lbs

c) The rate at which energy is leaving the cooler by steam is given by:

Emass = m'theta

Emass = (0.001765)×(1156.2)

Emass = 2.04BTU/s

6 0
3 years ago
Which part of the microscope is the circular area on the stage that light passes through?
patriot [66]

Answer: The part of the microscope that is the circular area is the APERTURE

I hope this helped!

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A crate slides down a ramp that makes a 20o angle with the ground. To keep the crate moving at a steady speed, Paige pushes back
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Answer:

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Explanation:

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Work done is given by

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Which philosopher suggested that the mind and body are separate but that a link exists between them?
hammer [34]
<span>René Descartes suggests this.</span>
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