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3 years ago

Explain how fluids exert pressure

1 answer:

4 0

In a fluid, all the forces exerted by the individual particles combine to make up the pressure exerted by the fluid
Due to fundamental nature of fluids, a fluid cannot remain at rest under the presence of shear stress. However, fluids can exert pressure normal to any contacting surface. If a point in the fluid is thought of as a small cube, then it follows from the principles of equilibrium that the pressure on every side of this unit of fluid must be equal. but if this were not a case, the fluid would move in the directions of the resulting force, So the pressure on a fluid at rest is isotropic.
Hope This Helps :D

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When materials absorb heat, the kinetic energy of the particles _________?

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Explanation:

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3 years ago

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Cold air holds less water than warm air. True False

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Explanation:

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3 years ago

A block with mass of 10 kg is on a frictionless surface. One hand on the left side of the block is pushing it to the right. A se

igomit [66]

Answer:

$W_2=-12J$

Explanation:

The work of force 2 will be given by the vectorial equation $W_2=F_2.d$ . We know the value of $F_1$ and have information about its movement, which relates to the net force $F=F_1+F_2$ .

About this movement we can obtain the acceleration using the equation $v_f^2=v_i^2+2ad$ . Since it departs from rest we have $a=\frac{v_f^2}{2d}$ .

And then using Newton's 2dn Law we can obtain the net force F=ma, thus we will have $F_2=F-F_1=ma-F1=\frac{mv_f^2}{2d}-F_1$

And we had the work done by force 2 as:

$W_2=F_2.d=\frac{mv_f^2}{2}-F_1d$

(The sign will be given algebraically since we take positive the direction to the right.)

With our values:

$W_2=\frac{(10kg)(2m/s)^2}{2}-(8N)(4m)=-12J$

Another (shorter but maybe less intuitive way for someone who is learning) way of doing this would have been to say that the work done by both forces would be equal to the variation of kinetic energy:

$W_1+W_2=K_f-K_i=K_f=\frac{mv_f^2}{2}$

Which leads us to the previous equation straightforwardly.

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3 years ago

A superball with a mass m = 61.6 g is dropped from a height h = [02]____________________ m. It hits the floor and then rebounds

aleksklad [387]

There is not enough data to solve the problem, but I'm assuming the initial height as h = 10 m for the question to have a valid answer and the student can have a reference to solve their own problem

Answer:

(a) $\Delta P=1.67 \ kg.m/s$

(b) $\Delta P=0.86\ m/s$

Explanation:

Change of Momentum

The momentum of a given particle of mass m traveling at a speed v is given by

$P=m.v$

When this particle changes its speed to a value v', the new momentum is

$P'=m.v'$

The change of momentum is

$\Delta P=m.v'-m.v$

$\Delta P=m.(v'-v)$

Defining upward as the positive direction, we'll compute the change of momentum in two separate cases.

(a) The initial height of the superball of m=61.6 gr = 0.0616 Kg is set to h= 10 m. This information leads us to have the initial potential energy of the ball just after it's dropped to the floor:

$U=m.g.h=0.0616\cdot 9.8\cdot 10 =6.0368\ J$

This potential energy is transformed into kinetic energy just before the collision occurs, thus

$\displaystyle \frac{1}{2}mv^2=6.0368$

Solving for v

$\displaystyle v=\sqrt{\frac{6.0368\cdot 2}{0.0616}}$

$v=-14\ m/s$

This is the speed of the ball just before the collision with the floor. It's negative because it goes downward. Now we'll compute the speed it has after the collision. We'll use the new height and proceed similarly as above. The new height is

$h'=88.5\% (10)=8.85\ m$