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Lunna [17]
3 years ago
9

A man (mass=68 kg) on a parachute is falling at terminal velocity (v=59 m/s)

Physics
1 answer:
raketka [301]3 years ago
3 0

Explanation:

It is given that,

Mass of the man, m = 68 kg

Terminal velocity of the man, v = 59 m/s

We need to find the rate at which the internal energy of the man and of the air around him increase. The gravitational potential energy of the man is given by :

E=mgh

Differentiating equation (1) wrt t as :

\dfrac{dE}{dt}=mg\dfrac{dh}{dt}

Since, v=\dfrac{dh}{dt}=59\ m/s

\dfrac{dE}{dt}=68\times 9.8\times 59

\dfrac{dE}{dt}=39317.6\ J/s

So, the internal energy of the man and the air around him is increasing at the rate of 39317.6 J/s. Hence, this is the required solution.

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A 1.0 kg football is given an initial velocity at ground level of 20.0 m/s [37 above horizontal]. It gets blocked just after re
stepan [7]
Refer to the diagram shown below.

Neglect air resistance.
The horizontal component of the launch velocity is
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The vertical component of the launch velocity is
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The acceleration due to gravity is g =9.8 m/s².
The time, t s, for the ball to reach a height of 3 m is given by 
(12.036 m/s)*(t s) - (1/2)*(9.8 m/s²)*(t s)² = (3 m)
12.036t - 4.9t² - 3 = 0
2.4543t - t² - 0.6122 = 0
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Solve with the quadratic formula.
t = (1/2)[2.4563 +/- √(6.0334 - 2.4488)]
t = 2.1748 or 0.2815 s
The ball reaches a height of 3 m twice.
The first time it reaches 3 m height is 0.2815 s.

Part a.
The vertical velocity when t = 0.2815 s is
Vy  = 12.036 - 9.8*0.2815
   = 9.2773 m/s
The horizontal component of velocity is Vx = 15.973 m/s
The resultant velocity is 
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Answer:
The velocity at a height of 3.0 m  is 18.5 m/s (nearest tenth)

Part b.
The horizontal distance traveled is 
d = (15.973 m/s)*(0.2815 s) = 4.4964 m
Answer:
The horizontal distance traveled is 4.5 m (nearest tenth)

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3 years ago
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