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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.

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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Why do cells undergo mitosis?

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3 years ago

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A block is released from rest, at a height h, and allowed to slide down an inclined plane. There is friction on the plane. At th

GREYUIT [131]

Here the block has two work done on it

1. Work done by gravity

2. Work done by friction force

So here it start from height "h" and then again raise to height hA after compressing the spring

So work done by the gravity is given as

$W_g = m_A g(h - h_A)$

Now work done by the friction force is to be calculated by finding total path length because friction force is a non conservative force and its work depends on total path

$W_f = -(\mu m_A g cos\theta)(\frac{h}{sin\theta} + \frac{h_A}{sin\theta})$

$W_f = -\mu m_A g cot\theta(h + h_A)$

Total work done on it

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So answer will be

None of these

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3 years ago

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A 75 kg baseball player runs at a velocity of 6 m/s before sliding to a stop at second base. a. What is the kinetic energy of th

lana [24]

Answer:

a. $\displaystyle k_o=1350\ J$

b. $\displaystyle k_1=0\ J$

c. $\Delta k=-1350\ J$

d. $W=-1350\ J$

e. $F=-675\ N$

Explanation:

<u>Work and Kinetic Energy </u>

When an object moves at a certain velocity v0 and changes it to v1, a change in its kinetic energy is achieved:

$\Delta k=k_1-k_0$

Knowing that

$\displaystyle k=\frac{mv^2}{2}$

We have

$\displaystyle \Delta k=\frac{mv_1^2}{2}-\frac{mv_0^2}{2}$

The work done by the force who caused the change of velocity (acceleration) is

$\displaystyle W=\frac{mv_1^2}{2}-\frac{mv_0^2}{2}$

If we know the distance x traveled by the object, the work can also be calculated by

$W=F.x$

Being F the force responsible for the change of velocity

The 75 kg baseball player has an initial velocity of 6 m/s, then he slides and stops

a. Before the slide, his initial kinetic energy is

$\displaystyle k_o=\frac{mv_0^2}{2}$

$\displaystyle k_o=\frac{(75)6^2}{2}$

$\boxed{\displaystyle k_o=1350\ J}$

b. Once he reaches the base, the player is at rest, thus his final kinetic energy is

$\displaystyle k_1=\frac{(75)0^2}{2}$

$\boxed{\displaystyle k_1=0\ J}$

c. The change of kinetic energy is

$\Delta k=k_1-k_0=0\ J-1350\ J$

$\boxed{\Delta k=-1350\ J}$

d. The work done by friction to stop the player is

$W=\Delta k=k_1-k_0$

$\boxed{W=-1350\ J}$

e. We compute the force of friction by using

$W=F.x$

and solving for x

$\displaystyle F=\frac{W}{x}$

$\displaystyle F=\frac{-1350\ J}{2\ m}$

$\boxed{F=-675\ N}$

The negative sign indicates the force is against movement

6 0

2 years ago

Suppose our apparatus will only allow us to distinguish the first order bright fringe from the central bright spot if they are s

ozzi

Answer:

λ = 8.716 mm

Explanation:

Given:

- d = 10 cm

- Q >= 5 degrees

Find:

- Find the shortest wavelength of light for which this apparatus is useful

Solution:

- The formula that relates the split difference and angle of separation between successive fringes is given by:

d*sin(Q) = n*λ

Where,

λ: wavelength

d: split separation

Q: angle of separation between successive fringes

m: order number.

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- the shortest possible wavelength corresponds to:

d*sin(Q) = λ

λ = 0.1*sin(5)

λ = 8.716 mm

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3 years ago