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Klio2033 [76]
3 years ago
8

A 75Kg man jumps from a window 1.0m above the sidewalk. If the man falls with his knees and ankles locked, the only cushion for

his fall is an approximately 0.49 cm give in the pads of his feet.
Calculate the average force exerted on him by the ground during this 0.49 cm of travel. This average force is sufficient to cause damage to cartilage in the joints or to break bones.
Physics
1 answer:
liq [111]3 years ago
8 0

Answer:

150153.06122 N

Explanation:

m = Mass of person = 75 kg

h = Height of fall = 1 m

g = Acceleration due to gravity = 9.81 m/s²

F = Force

s = Displacement = 0.49 cm

Potential energy is given by

P=mgh\\\Rightarrow P=75\times 9.81\times 1\\\Rightarrow P=735.75\ J

Work is given by

W=Fs\\\Rightarrow F=\frac{W}{s}\\\Rightarrow F=\frac{735.75}{0.0049}\\\Rightarrow F=150153.06122\ N

The average force exerted is 150153.06122 N

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Ksivusya [100]

Answer:

40 J

Explanation:

From the question given above, the following data were obtained:

Force (F) = 10 N

Distance (s) = 4 m

Workdone (Wd) =?

Work done is simply defined as the product of force and distance moved in the direction of the force. Mathematically, we can express the Workdone as:

Workdone = force × distance

Wd = F × s

With the above formula, we can obtain the workdone as follow:

Force (F) = 10 N

Distance (s) = 4 m

Workdone (Wd) =?

Wd = F × s

Wd = 10 × 4

Wd = 40 J

Thus, 40 J of work was done.

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On the Apollo 14 mission to the moon, astronaut Alan Shepard hit a golf ball with a golf club improvised from a tool. The free-f
aliya0001 [1]

Answer:

15.3 s and 332 m

Explanation:

With the launch of projectiles expressions we can solve this problem, with the acceleration of the moon

    gm = 1/6 ge

    gm = 1/6  9.8 m/s² = 1.63 m/s²

We calculate the range

    R = Vo² sin 2θ  / g

    R = 25² sin (2 30) / 1.63

    R= 332 m

We will calculate the time of flight,

   Y = Voy t – ½ g t2  

   Voy = Vo sin θ

When the ball reaches the end point has the same initial  height Y=0

0 = Vo sin  t – ½  g t2

0 = 25 sin (30)  t – ½ 1.63 t2

0= 12.5 t –  0.815 t2

We solve the equation

0= t ( 12.5 -0.815 t)

 t=0 s

t= 15.3 s

The value of zero corresponds to the departure point and the flight time is 15.3 s

Let's calculate the reach on earth

R2 = 25² sin (2 30) / 9.8

R2 = 55.2 m

R/R2 = 332/55.2

R/R2 = 6

Therefore the ball travels a distance six times greater on the moon than on Earth

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Answer:

The introduction of machines to make work done more easier and faster

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