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Viktor [21]
3 years ago
15

When a wheel is rotated through an angle of 21◦ , a point on the circumference travels through an arc length of 2.1 m. When the

wheel is rotated through angles of 21 rad and 21 rev, the same point travels through arc lengths of 120 m and 7.6×102 m, respectively. What is the radius of the wheel? Answer in units of m.
Physics
1 answer:
LiRa [457]3 years ago
6 0

Answer:

R = 5.73 m

Explanation:

For an angle of rotation through 21 degree we know that

arc length is given as

angle \times Radius = Arc

now we know that

Arc = 2.1 m

Angle = 21 degree

Angle = 21\times \frac{\pi}{180}

so now we have

21\times \frac{\pi}{180} = \frac{2.1}{R}

R = \frac{2.1 \times 180}{21 \times \pi}

R = 5.73 m

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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
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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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yKpoI14uk [10]

Explanation:

Hope its helps u

if wrong then sry

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