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ladessa [460]
3 years ago
14

a 645 N person is wearing stiletto shoes. If she lifts her left leg, and rocks back onto the heel under her right leg, the press

ure the heel exerts on the floor is 9000000 PA. what is the diameter of the circular base of the heel in centimeters?
Physics
1 answer:
fgiga [73]3 years ago
7 0

Answer:

0.96cm

Explanation:

Given parameters:

Weight of person = 645N

Pressure of the heel = 9000000PA

Unknown

Diameter of the circular base = ?

Solution

Pressure is the force per unit area of a body.

  Pressure = \frac{Force}{Area}

To solve this problem, find the unknown area of the circle;

             Pressure x Area = Force

         Area     = \frac{Force}{Pressure}

           Area  = \frac{645}{9000000} = 7.17 x 10⁻⁵m²

Note: 1Pa = 1Nm⁻²

Now;

  Area of a circle = π r²

         r = \frac{d}{2}

    Area =  π x \frac{d^{2} }{4}

d is the diameter

                    d²  = 4 x Area  x (1/π)

                d² = 4 x 7.17 x 10⁻⁵ x  (1/3.14) = 9.13 x 10⁻⁵

              d = 0.0096m

To cm;

                100cm = 1m

   0.0096m; 0.0096 x 100=  0.96cm

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Determine the CM of a rod assuming its linear mass density λ (its mass per unit length) varies linearly from λ = λ0 at the left
Dahasolnce [82]

Answer:

x_c= \dfrac{5}{9}L

I=\dfrac {7}{12}\lambda_ 0 L^3

Explanation:

Here mass density of rod is varying so we have to use the concept of integration to find mass and location of center of mass.

At any  distance x from point A mass density

\lambda =\lambda_0+ \dfrac{2\lambda _o-\lambda _o}{L}x

\lambda =\lambda_0+ \dfrac{\lambda _o}{L}x

Lets take element mass at distance x

dm =λ dx

mass moment of inertia

dI=\lambda x^2dx

So total moment of inertia

I=\int_{0}^{L}\lambda x^2dx

By putting the values

I=\int_{0}^{L}\lambda_ ox+ \dfrac{\lambda _o}{L}x^3 dx

By integrating above we can find that

I=\dfrac {7}{12}\lambda_ 0 L^3

Now to find location of center mass

x_c = \dfrac{\int xdm}{dm}

x_c = \dfrac{\int_{0}^{L} \lambda_ 0(1+\dfrac{x}{L})xdx}{\int_{0}^{L} \lambda_0(1+\dfrac{x}{L})}

Now by integrating the above

x_c=\dfrac{\dfrac{L^2}{2}+\dfrac{L^3}{3L}}{L+\dfrac{L^2}{2L}}

x_c= \dfrac{5}{9}L

So mass moment of inertia I=\dfrac {7}{12}\lambda_ 0 L^3 and location of center of mass  x_c= \dfrac{5}{9}L

8 0
3 years ago
2. Do you think the density of the ice affected the melting rate of the ice, or do you think adding the objects affected the mel
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The density of ice does not affect its melting rate. Adding objects will affect the melting rate.

  • A physical process called melting or fusing causes a substance to change its phase from a solid to a liquid. This happens when the solid's internal energy rises, usually as a result of heat or pressure being applied, which raises the substance's temperature to the melting point.
  • The term "density" refers to an extensive quality, which means that it is independent of the substance's concentration. Every substance in the world demonstrates its distinctive density. Since it does not fluctuate, it would not affect the rate of melting. The addition of the objects could speed up the process, though, as each one generates heat that could act as the mediating force for the melting process.

To learn more about density, visit :

brainly.com/question/15164682

#SPJ9

3 0
1 year ago
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