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Rudik [331]
3 years ago
7

when you jump from an elevated position you usually bend your knees upon reaching the ground. by doing this, you make the time o

f the impact about 10 times as great as for a stiff legged landing. in this way, the average force your body experiences is...
Physics
1 answer:
never [62]3 years ago
8 0

about 1/10th as great by bending your legs 

Time extended decreases acceleration value. 

solve using Newton's 2nd and Uniform Acceleration Laws. plug in values for descending. 

Vf = 0.0 m/s 

Vi = 10 m/s 

t = 1 s 

Acceleration is given by 

a = [Vf - Vi] / t 

a = [ (0.0 m/s) - (10 m/s) ] / (1 s) 

a = [ -10 m/s ] / (1 s) 

a = -10 m/s^2 

Solve the same thing, but with time 10x, so t = 10 s 

a = [ (0.0 m/s) - (10 m/s) ] / (10 s) 

a = [ -10 m/s ] / (10 s) 

a = -1 m/s^2 

Solve for first force and second force, using a mass of 100 kg 

F = m * a 

F = (100 kg) * (-10 m/s^2) 

F = -1,000 N 

F = (100 kg) * (-1 m/s^2) 

F = -100 N 

So divide long-force by short-force 

(-100 N) / (-1,000 N) = 0.1 (which is 1/10)

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The speed at which a light aircraft can take off is 120 km/h. (A) What is the minimum constant acceleration required for the pla
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A) a = 2.31[m/s^2]; B) t = 14.4 [s]

Explanation:

We can solve this problem using the kinematic equations, but firts we must identify the data:

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Vi= initial velocity = 0, because the plane starts to move from the rest.

dx= distance to run = 240 [m]

v_{f} ^{2} =v_{i} ^{2}+2*g*dx\\where:\\v_{f}=120[\frac{km}{h} ]*\frac{1hr}{3600sg} * \frac{1000m}{1km} =33.33[m/s]\\\\Replacing\\33.33^{2}=0+2*a*(240)\\ a=\frac{11108.88}{2*240}\\  a=2.31[m/s^2]\\

To find the time we must use another kinematic equation.

v_{f} =v_{i} +a*t\\replacing:\\33.33=0+(2.31*t)\\t=\frac{33.33}{2.31}\\ t=14.4[s]

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