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leonid [27]
3 years ago
8

Assume that when you stretch your torso vertically as much as you can, your center of mass is 1.0 m above the floor. The maximum

force you can exert on the floor in pushing off is 2.3 times the gravitational force Earth exerts on you
How low do you have to crouch in order to jump straight up and have your center of mass be 2.0 m above the floor? Determine the lowest height of your center of mass above the floor in the jump.?

Is this crouch practical?
Physics
1 answer:
Elenna [48]3 years ago
3 0

1) 0.77 m

2) 0.23 m

Explanation:

1)

Here we want to find the time elapsed for crouching in order to jump and reach a height of 2.0 m above the floor, starting from 1.0 m above the floor.

First of all, we start by calculating the speed required to jump up to a height of 2.0 m. Since the total energy is conserved, the initial kinetic energy is converted into gravitational potential energy, so:

\frac{1}{2}mv^2 = mgh

where

m is the mass of the man

v is the speed after jumping

g=9.8 m/s^2 is the acceleration due to gravity

h = 2.0 - 1.0 = 1.0 m is the change in height

Solving for v,

v=\sqrt{2gh}=\sqrt{2(9.8)(1.0)}=4.43 m/s

In the acceleration phase, we know that the initial velocity is

u=0

And the force exerted on the floor is 2.3 times the gravitational force, so

F=2.3 mg

This means the net force on you is

F_{net} = F-mg=2.3mg-mg=1.3 mg

because we have to consider the force of gravity acting downward.

So the acceleration of the man is

a=\frac{F_{net}}{m}=\frac{1.3mg}{m}=1.3g

Now we can use the  following suvat equation to find the displacement in the acceleration phase, which is how low the man has to crouch in order to jump:

v^2-u^2=2as

where s is the quantity we want to find. Solving for s,

s=\frac{v^2-u^2}{2a}=\frac{4.43^2-0}{2(1.3g)}=0.77 m

2)

At the beginning, we are told that the height of the center of mass above the floor is

h = 1.0 m

During the acceleration phase and the crouch, the height of the center of mass of the body decreases by

\Delta h = -0.77 m

This means that the lowest point reached by the center of mass above the floor during the crouch is

h'=h+\Delta h = 1.0 - 0.77 = 0.23 m

This value seems unpractical, since it is not really easy to crouch until having the center of mass 0.23 m above the ground.

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Answer:

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Explanation:

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The centripetal acceleration is the ratio of instant velocity squared and the radius of the circle

a_c = \frac{v^2}{r} = \frac{30^2}{300} = \frac{900}{300} = 3 m/s^2

So the magnitude of the total acceleration is

a = \sqrt{a_c^2 + a_l^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 m/s^2

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When the mirror is rotated, the normal will turn as well, but will the incident Ray and reflected ray turn?
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When a mirror is rotated . . .

-- The incident ray doesn't turn.  It's just the line from the source to the mirror. 
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3 0
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4. All electromagnetic radiation moves at the speed of ________________, which is _________________ m/s.
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7 0
3 years ago
2.
Alex_Xolod [135]

Answer:

The work done on the sled by friction, W = - 4593.75 J

Explanation:

Given data,

The combined mass of sled and the boy, m = 75 kg

The displacement of the boy, S = 25 m

The coefficient of the friction, u = 0.25

The frictional force acting on the boy,

                  <em>F = u η</em>

Where,

                        η - is the normal force acting on the boy (mg)

Substituting the values,

                   F = 0.25 x 75 x 9.8

                      = 183.75 N

Since the direction of the frictional force is against the direction of motion

                      F = - 183.75 N

The work done on the sled by friction,

                         W = F x S

                             = - 183.75 x 25

                             = - 4593.75 J

Hence, the work done on the sled by friction, W = - 4593.75 J

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