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Black_prince [1.1K]
3 years ago
8

A rubber ball and a steel ball are dropped from the same height onto a concrete floor. They have the same mass, and lets assume

that they both rebound to the same height after hitting the ground. How would the average forces experienced by the balls compare. Group of answer choices The steel ball would experience the greater average force The average forces would be the same The rubber ball would experience the greater average force
Physics
1 answer:
Mkey [24]3 years ago
5 0

Answer:

The average forces would be the same

Explanation:

Both have the same velocity on impact as they fell from the same height.

Both have the same velocity after the bounce because they reach the same height.

Both have the same mass

Both will thus experience the same impulse because both have the same change in momentum.

Therefore both experience the same average force.

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A satellite is in a circular orbit around a planet. A second satellite is placed in a different circular orbit that is farther a
irinina [24]

The speed of the second satellite is less than the speed of the first satellite.

<h3>What is speed?</h3>

The speed of any moving object is the ratio of the distance covered and the time taken to cover that distance.

Given is a satellite is in a circular orbit around a planet. A second satellite is placed in a different circular orbit that is farther away from the same planet.

When the distance from the center of the orbit increases, the time to complete the orbit will be greater.

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Learn more about speed.

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2 years ago
How do nerves tie the other parts of the nervous system together?
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They do it by followinng the centeral nervous system
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3 years ago
A mercury thermometer has a glass bulb of interior volume 0.100 cm3 at 10°c. the glass capillary 10) tube above the bulb has an
Nadya [2.5K]
Initial volume of mercury is
V = 0.1 cm³

The temperature rise is 35 - 5 = 30 ⁰C = 30 ⁰K.

Because the coefficient of volume expansion is 1.8x10⁻⁴ 1/K, the change in volume of the mercury is 
ΔV = (1.8x10⁻⁴ 1/K)*(30 ⁰K)(0.1 cm³) = 5.4x10⁻⁴ cm³

The cross sectional area of the tube is
A = 0.012 mm² = (0.012x10⁻² cm²).
Therefore the rise of mercury in the tube is
h = ΔV/A
   = (5.4x10⁻⁴ cm³)/(0.012x10⁻² cm²)
   = 4.5 cm

Answer: 4.5 cm
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Answer:

B. 6.00 cm

Explanation:

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5.625 hours and it is 450 divided by 80
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