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USPshnik [31]
3 years ago
5

As a stands, her entire weight is momentarily placed on the heels of her high-heeled shoes. Calculate the pressure exerted on th

e floor by the heels if the heels are circular with a radius of 0.578 cm, and the woman’s mass is 65.0 kg. Express the pressure in Pa.
Physics
1 answer:
Lena [83]3 years ago
5 0

Answer:

Pressure will be 6072449.952Pa

Explanation:

We have given mass of the women m = 65 kg

Radius of the heels r = 0.578 cm = 0.00578 m

We have to find the pressure

We know that pressure is given by

P=\frac{F}{A}=\frac{mg}{A}

So force F = mg = 65×9.8 = 637 N

Area A=\pi r^2=3.14\times 0.00578^2=1.049\times 10^{-4}m^2

So pressure p=\frac{637}{1.049\times 10^{-4}}=6072449.952Pa

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they stay shifted the same amount to the red

Explanation:

Redshift is given by

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Where,

\lambda_o = Wavelength observed

\lambda_e = Wavelength emitted

Also

Transverse redshift is given by

1+z=\dfrac{1}{\sqrt{1-v^2/c^2}}

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7) The acceleration due to gravity near the surface of Mars is about one-third of the value
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How does the north pole of a magnet respond to the poles of other magnets?
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Answer:

Explanation:

A magnet has a magnetic field around it which originates at the north pole and enters through the south pole.

In a magnet, like poles will repel each other and unlike poles will attract.

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3 years ago
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You serve a volleyball with a mass of 2.1 kg. The ball leaves your hand with a speed of 2.1 m/s. The
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Answer:

4.6 Joules

Explanation:

K=1/2*MV^2

1/2 * 2.1kg * 2.1^2m/s

==4.6305  Joules

simplified to 4.6 Joules

7 0
2 years ago
A converging lens of focal length 20 cm is used to form a real image 1.0 m away from the lens. How far from the lens is the obje
Galina-37 [17]

Answer:

0.25 m

Explanation:

We can solve the problem by using the lens equation:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where

f is the focal length

p is the distance of the object from the lens

q is the distance of the image from the lens

In this problem, we have

f = +20 cm=+0.20 m (the focal length is positive for a converging lens)

q = +1.0 m (the image distance is positive for a real image)

Solving the equation for p, we find

\frac{1}{p}=\frac{1}{f}-\frac{1}{q}=\frac{1}{0.20 m}-\frac{1}{1 m}=4 m^{-1}\\p=\frac{1}{4 m^{-1}}=0.25 m

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