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SVEN [57.7K]
3 years ago
5

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
A) less than 1/10 as great.
B) more than 1/10 as great.
C) about 1/10 as great.
D) about 10 times as great.
Physics
1 answer:
Leona [35]3 years ago
3 0

Answer:

C) about 1/10 as great.

Explanation:

We use the relation between Impulse, I, and momentum, p:

I=\Delta p\\ F*t=m(v_{f}-v{o})\\ \\

F=m(v_{f}-v{o})/t=-mv{o}/t\\ \\     the final speed is zero

We can see that the average Force is inversely proportional to the time, so if the time is 10 times bigger, the average Force is 1/10 as great

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Gravity, or gravitation, is a natural phenomenon by which all things with mass or energy-including planets,stars,galaxies, and even light-are brought toward one another.

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True or false: the coefficient of thermal expansion governs the way a substance expands or contracts when it is heated or cooled
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Which of the following diagrams best represents the gravitational force W. the frictional force f, and the normal force N that a
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It's D.

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3 years ago
An airplane starts from rest and reaches a takeoff velocity of 60.0m/s due north in 4.0 s. How far has it traveled before takeof
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3 years ago
A person with myopia (near-sightedness) has a far point of 45.0cm while their near point is 15.0cm. Upon wearing glasses, they h
Crank

Answer:

p = 22.5 cm

Explanation:

For this exercise we must use the equation of the constructor

       \frac{1}{f} =  \frac{1}{p} + \frac{1}{q}

where f is the focal length, p and q are the distance to the object and the image respectively.

Let's start with the far point, the object is very far away (p = ∞) and the image must be formed at the far point of view of the person q = 45.0 cm

since the image is on the same side as the object according to the sign convention the distance is negative

         \frac{1}{f} = \frac{1}{\infty }  + \frac{1}{-45}

          f = -45.0 cm

now let's use the near point (q = 15.0 cm) at what distance the object should be

          \frac{1}{p} = \frac{1}{f} - \frac{1}{q}

          \frac{1}{p} = \frac{1}{-45} - \frac{1}{-15}1 / p = 1 / -45 - 1 / -15

         \frac{1}{p} = - \frac{1}{45} + \frac{1}{15} 1 / p = -1/45 + 1/15

         \frac{1}{p} = 0.0444

          p = 22.5 cm

this is the closest distance you can see an object clearly

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3 years ago
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