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2 years ago

7

A 45 kg merry go round worker stands on the rides platform 6.3 m from the center. If her speed as she goes around the circle is

4.1 m/s, what is the force of friction necessary to keep her from falling off the platform?

1 answer:

grin007 [14]2 years ago

7 0

I’ve got 120 n, hope it helps!

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A friend rides, in turn, the rims of three fast merry-go-rounds while holding a sound source that emits isotropically at a certa

Aliun [14]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

The Ranking of the curve according to their speed would be equal Rank because $v_1 =v_2 =v_3$

b

The first frequency would have a higher rank compared to the other two which will have the same ranking when ranked with respect to their angular velocities because

$w_1 >w_2 = w_3$

c

The ranking of the second third frequency would be the same but their ranking would be greater than that of the first frequency because

$r_2 =r_3 >r_1$

Explanation:

Mathematically Frequency can be represented as

$F = \frac{v}{\lambda}$

Where $\lambda$ is the wavelength and v is the velocity

Now looking at the diagram we see that

For the first frequency we have

Let the wavelength be $\lambda_1 = \lambda$ , and the frequency $F_1 = F$

For the second frequency

Let the wavelength be $\lambda_2 = 2 \lambda$ , and the frequency $F_2 = \frac{F}{2}$

For the third frequency

Let the wavelength be $\lambda_3 = 2\lambda$ , and the frequency $F_3 = \frac{F}{2}$

To obtain v for each of the frequency we make v the subject in the equation above for each frequency

So,

For the first frequency we have

$v_1 = \lambda_1 F_1 = \lambda F$

For the second frequency

$v_2 = \lambda_2 F_2 = 2 \lambda*\frac{F} {2} = \lambda F$

For the third frequency

$v_3 = \lambda_3 F_3 = 2 \lambda*\frac{F} {2} = \lambda F$

Hence

The Ranking of the curve according to their speed would be equal Rank because $v_1 =v_2 =v_3$

Mathematically angular speed can be represented as

$w = 2 \pi f$

For the first frequency we have

$w_1 = 2\pi F_1 = 2 \pi F$

For the second frequency

$w_2 = 2 \pi F_2 = 2 \pi \frac{F}{2} = \pi F$

For the third frequency

$w_3 = 2 \pi F_3 = 2 \pi \frac{F}{2} = \pi F$

Hence

The first frequency would have a higher rank compared to the other two which will have the same ranking when ranked with respect to their angular velocities because

$w_1 >w_2 = w_3$

Mathematically the relationship between the angular velocity and the linear velocity can be represented as

$v = wr$

=> $r = \frac{v}{w}$

Since the linear velocity is constant we have that

$r \ \alpha \ \frac{1}{w}$

This means that r varies inversely to the angular velocity ,What this means for ranking due to the radius is that the ranking of the second third frequency would be the same but their ranking would be greater than that of the first frequency because

$r_2 =r_3 >r_1$

5 0

3 years ago

What is lower body strength

Lerok [7]

Lower body strength is the ability of the body to exert a maximum force against an object external to the body in one maximum effort of the lower body muscles. Gluteals are muscles that make up the buttocks. ... Their roles are to facilitate hip extension and lift the leg to the side.

3 0

2 years ago

Read 2 more answers

Two mirrors are at right angles to one another. A light ray is incident on the first at an angle of 30° with respect to the norm

marta [7]

Answer:

reflected angle - secod mirror = 60°

Explanation:

I attached an image with the solution to this problem below.

In the solution the reflection law, incident angle = reflected angle, is used. Furthermore some trigonometric relation is used.

You can notice in the image that the angle of reflection in the second mirror is 60°

7 0

3 years ago

What type of material is found in the asthenosphere?

IceJOKER [234]

Answer:

The correct answer is option C.

Explanation:

The layer below the lithosphere , the upper layer of the mantle of the earth with high heat pressure .It is made up of almost solid rocks but also flows not which indicates that it consist of viscoelastic fluid of molten rocks.This is because in some regions there can be presence of molten rocks.

6 0

3 years ago

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The state of strain at a point is plane strain with εx = ε0, εy = –2ε0, γxy = 0, where ε0 is a positive constant. What is the no

Marat540 [252]

Answer:

The normal strain along an axis oriented 45° from the positive x axis in the clockwise direction is -ε₀/2

Explanation:

Given that

$\epsilon_{x}=\epsilon_{o}\\\\\epsilon_{y}=-2\epsilon_{o}\\\\\gamma_{xy}=0\\\\\theta=-45^{o}\\\\\epsilon_{x_{1}}=?$

From equation of normal strain in x direction:

$\epsilon_{x_{1}}=\epsilon_{x}cos^{2}\theta+\epsilon_{y}sin^{2}\theta+\gamma_{xy{ sin\theta cos\theta$

Substituting the values:

$\epsilon_{x_{1}}=\epsilon_{o}cos^{2}(-45)-2\epsilon_{o}sin^{2}(-45)+0\\\\\epsilon_{x_{1}}=\frac{\epsilon_{o}}{2}-2\frac{\epsilon_{o}}{2}\\\\\epsilon_{x_{1}}=-\frac{\epsilon_{o}}{2}$

6 0

3 years ago