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Minchanka [31]
3 years ago
5

An unwary football player collides with a padded goalpost while running at a velocity of 7.50 m/s and comes to a full stop after

compressing the padding and his body 0.350 m. (a) What is his deceleration?(b) How long does the collision last?
Physics
1 answer:
frez [133]3 years ago
7 0

Answer:

(a) the deceleration of the player is -80.36 m/s²

(b) the time the collision last is 0.093 s

Explanation:

Given;

Initial velocity of the football player, u = 7.50 m/s

Final velocity of the football player, v = 0

distance traveled = compression of the pad, s = 0.35 m

Part (a) the deceleration of the player

v² = u² + 2as

0 = 7.5² + (2 x 0.35)a

0 = 56.25 + 0.7a

- 56.25 = 0.7a

a = -56.25 / 0.7

a = -80.36 m/s²

Part (b) the time the collision last

v = u + at

t = (v - u)/a

t = (0 - 7.5)/ -80.36

t = - 7.5 / -80.36

t = 0.093 s

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<h3><u>Answer</u>;</h3>

Price floor

When the government sets a price for wheat that is above the equilibrium price, it is imposing a<u> price floor</u>.

<h3><u>Explanation</u>;</h3>
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The table shows the percentage of carbon dioxide in the Earth’s atmosphere in the years 1800 and 2013. Calculate the difference
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The difference in the mass of carbon dioxide in 500 kg of air in 2013 compared to 1800 is 0.06 Kg

<h3>Data obtained from the question</h3>
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  • Mass of air = 500 Kg
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<h3>How to determine the mass of CO₂ in 500 Kg in year 1800</h3>
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  • Mass of air = 500 Kg
  • Mass of CO₂ =?

Mass = percent × mass of air

Mass of CO₂ = 0.028% × 500

Mass of CO₂ = 0.14 Kg

<h3>How to determine the mass of CO₂ in 500 Kg in year 2013</h3>
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  • Mass of air = 500 Kg
  • Mass of CO₂ =?

Mass = percent × mass of air

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<h3>How to determine the difference</h3>
  • Mass of CO₂ in year 1800 = 0.14 Kg
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Difference = mass in 2013 - mass in 1800

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A child slides down the water slide at a swimming pool and enters the water at a final speed of 4.22 m/s. At what final speed wo
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Answer:

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mgh = \frac {1}{2} mv^2\\v =\sqrt{2gh}

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A. The flow of energy from Earth's interior to the surface is about 50 terawatt (1 terawatt = 1e12 joule/sec). This is the energ
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A) The amount of geothermal energy that reaches Earth's surface in one day is : 43.2 * 10¹⁷ joules

B) Humanity's current consumption of energy is; 71.4 GJ per individual.

C) Ratiois ; 60.5 * 10⁶ Joules

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Amount of energy flow = 50 terawatt =  50*10¹² joule/sec.

<h3>Determine the geothermal energy that reaches the earth</h3>

Since the geothermal energy reaches the earth in a single day

Energy = power * time

Time = 24 hours ( 1 day ) = 86400 secs.

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B) Humanity's current consumption of energy is; 71.4 GJ per individual.

C) Ratiois ; 60.5 * 10⁶ Joules

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