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Tatiana [17]
4 years ago
7

a defensive tackle picks up the 0.5kg football to a height of 0.8m in 0.25s .... 1.calculate the work done 2.calculate the power

generated
Physics
1 answer:
IrinaVladis [17]4 years ago
7 0
1).  The work done is the potential energy the football has after it is raised.

Weight of the football = (Mass) x (gravity) = (0.5) x (9.8) = 4.9 newtons

Work= (force) x (distance) = (4.9 newtons) x (0.8 meter) = <em>3.92 joules</em>

2). Power = (work) / (time)  (You know this from the car problem.)

Power = (3.92 joules) / 0.25 s) = 15.68 joules per second = <em>15.68 watts</em>.
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g Two masses are involved in a collision on an axis (one dimensional). One mass is six times the mass of the second. Both masses
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Answer:

v₁f = 0.5714 m/s   (→)

v₂f = 2.5714 m/s   (→)

e = 1  

It was a perfectly elastic collision.

Explanation:

m₁ = m

m₂ = 6m₁ = 6m

v₁i = 4 m/s

v₂i = 2 m/s

v₁f = ((m₁ – m₂) / (m₁ + m₂)) v₁i +  ((2m₂) / (m₁ + m₂)) v₂i

v₁f = ((m – 6m) / (m + 6m)) * (4) +  ((2*6m) / (m + 6m)) * (2)  

v₁f = 0.5714 m/s   (→)

v₂f = ((2m₁) / (m₁ + m₂)) v₁i +  ((m₂ – m₁) / (m₁ + m₂)) v₂i

v₂f = ((2m) / (m + 6m)) * (4) + ((6m -m) / (m + 6m)) * (2)

v₂f = 2.5714 m/s   (→)

e = - (v₁f - v₂f) / (v₁i - v₂i)   ⇒   e = - (0.5714 - 2.5714) / (4 - 2) = 1  

It was a perfectly elastic collision.

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