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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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A car on a roller coaster starts at zero speed at an elevation above the ground of 26 m. It coasts down a slope, and then climbs
nirvana33 [79]

The speed of the car at the top of the hill is 14m/s

<u>Explanation:</u>

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5 0
3 years ago
que 2. Why do we keep frequency constant instead of keeping vibrating length constam second law of vibrating string?​
ella [17]

Answer:

The second law of a vibrating string states that for a transverse vibration in a stretched string, the frequency is directly proportional to the square root of the string's tension, when the vibrating string's mass per unit length and the vibrating length are kept constant

The law can be expressed mathematically as follows;

f = \dfrac{1}{2\cdot l} \cdot \sqrt{\dfrac{T}{m} }

The second law of the vibrating string can be verified directly, however, the third law of the vibrating string states that frequency is inversely proportional to the square root of the mass per unit length cannot be directly verified due to the lack of continuous variation in both the frequency, 'f', and the mass, 'm', simultaneously

Therefore, the law is verified indirectly, by rearranging the above equation as follows;

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From which it can be shown that the following relation holds with the limits of error in the experiment

m₁·l₁² = m₂·l₂² = m₃·l₃² = m₄·l₄² = m₅·l₅²

Explanation:

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3 years ago
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