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2 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...

1 answer:

never [62]2 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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Equilibrium price will fall and quantity demanded for jelly will also fall

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A 3-m-high, 7-m-wide rectangular gate is hinged at the top edge and is restrained by a fixed ridge. Determine the hydrostatic fo

Shalnov [3]

Answer:

The Hydrostatic force is $F = 137.2 kN$

The location of pressure center is $Z = 1.333 \ m$

Explanation:

From the question we are told that

The height of the gate is $h = 3 \ m$

The weight of the gate is $w = 7 \ m$

The height of the water is $h_w = 2 \ m$

The density of water is $\rho_w = 1000 \ kg/m^3$

Note used $h_w$ for height of water and height of gate immersed by water since both have the same value

The area of the gate immersed in water is mathematically represented as

$A = h_w * w$

substituting values

$A = 2* 7$

$A = 14 \ m^2$

The hydrostatic force is mathematically represented as

$F = \rho_w * g * h_f * A$

Where

$h_f =h- h_w$

$h_f =3 -2$

$h_f = 1\ m$

So

$F = 1000 * 9.8 * 1 * 14$

$F = 137.2 kN$

The center of pressure is mathematically represented as

$Z = h_f + \frac{I_g}{h_f * A}$

Where $I_g$ is the moment of inertia of the gate which mathematically represented as

$I_g = \frac{w * h_w^2}{12}$

The $h_w$ is the height of gate immersed in water

$I_g = \frac{7 * 2^2 }{12}$

$I_g = 4.667\ kg m^2$

Thus

$Z = 1 + \frac{4.66}{1 * 14}$

$Z = 1.333 \ m$

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2 years ago

A block is pulled across a flat surface at a constant speed using a force of 50 newtons at an angle of 60 degrees above the hori

vladimir2022 [97]

The magnitude of the friction force is 25 N

Explanation:

To solve this problem, we just have to analyze the forces acting on the block along the horizontal direction. We have:

The horizontal component of the pulling force, $F cos \theta$ , where F = 50 N is the magnitude and $\theta=60^{\circ}$ is the angle between the direction of the force and the horizontal; this force acts in the forward direction
The force of friction, $F_f$ , acting in the backward direction

According to Newton's second law, the net force acting on the block in the horizontal direction must be equal to the product between the mass of the block and its acceleration:

$\sum F_x = ma_x$

where

m is the mass of the block

$a_x$ is the horizontal acceleration

However, the block is moving at constant speed, so the acceleration is zero:

$a_x = 0$

So the equation becomes

$\sum F_x = 0$ (1)

The net force here is given by

$\sum F_x = F cos \theta - F_f$ (2)

And so, by combining (1) and (2), we find the magnitude of the friction force:

$F cos \theta - F_f = 0\\F_f = F cos \theta = (50)(cos 60^{\circ})=25 N$

Learn more about force of friction:

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What is the definition of newton

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