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2 years ago

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a force of 35N is exerted over a cylinder with an area of 5m^2. What pressure,in pascals, will be transmitted in the hydraulic s

ystem?

1 answer:

gayaneshka [121]2 years ago

8 0

Answer:

<h3>The answer is 7 Pa</h3>

Explanation:

The pressure transmitted in the hydraulic system can be found by using the formula

$p = \frac{f}{a} \\$

f is the force

a is the area

From the question we have

$p = \frac{35}{5} \\$

We have the final answer as

<h3>7 Pa</h3>

Hope this helps you

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What properties does a gas have in terms of shape and volume

insens350 [35]

Gases, liquids and solids are all made up of atoms, molecules, and/or ions, but the behaviors of these particles differ in the three phases. The following figure illustrates the microscopic differences.

Microscopic view of a gas Microscopic view of a liquid. Microscopic view of a solid.
Microscopic view of a gas. Microscopic view of a liquid. Microscopic view of a solid.
Note that:

Particles in a:
gas are well separated with no regular arrangement.
liquid are close together with no regular arrangement.
solid are tightly packed, usually in a regular pattern.
Particles in a:
gas vibrate and move freely at high speeds.
liquid vibrate, move about, and slide past each other.
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Liquids and solids are often referred to as condensed phases because the particles are very close together.
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3 years ago

Paul and Ivan are riding a tandem bike together. They’re moving at a speed of 5 meters/second. Paul and Ivan each have a mass of

VMariaS [17]

<span>If Paul and Ivan has a speed of 5 meters/second in which their combined mass is 50 kg. To increase the bike's kinetic energy, Paul must increase its speed as well. Increasing his speed allows an increase in momentum of them running the bike. The kinetic energy equation is KE = 0.5mv</span>² where m is mass, v is speed and KE is kinetic energy.

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3 years ago

An electron is released from rest on the axis of a uniform positively charged ring, 0.200 m from the ring's center. If the linea

melisa1 [442]

Answer:

Velocity of the electron at the centre of the ring, $v=1.37\times10^7\ \rm m/s$

Explanation:

<u>Given:</u>

Linear charge density of the ring= $0.1\ \rm \mu C/m$
Radius of the ring R=0.2 m
Distance of point from the centre of the ring=x=0.2 m

Total charge of the ring

$Q=0.1\times2\pi R\\Q=0.1\times2\pi 0.4\\Q=0.251\ \rm \mu C$

Potential due the ring at a distance x from the centre of the rings is given by

$V=\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\$

The potential difference when the electron moves from x=0.2 m to the centre of the ring is given by

$\Delta V=\dfrac{kQ}{R}-\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\\Delta V={9\times10^9\times0.251\times10^{-6}} \left( \dfrac{1}{0.4}-\dfrac{1}{\sqrt{(0.4^2+0.2^2)}} \right )\\\Delta V=5.12\times10^2\ \rm V$

Let $\Delta U$ be the change in potential Energy given by

$\Delta U=e\times \Delta V\\\Delta U=1.67\times10^{-19}\times5.12\times10^{2}\\\Delta U=8.55\times10^{-17}\ \rm J$

Change in Potential Energy of the electron will be equal to the change in kinetic Energy of the electron

$\Delta U=\dfrac{mv^2}{2}\\8.55\times10^{-17}=\dfrac{9.1\times10^{-31}v^2}{2}\\v=1.37\times10^7\ \rm m/s$

So the electron will be moving with $v=1.37\times10^7\ \rm m/s$

5 0

2 years ago

Compute the kinetic energy of a proton (mass 1.67×10−27kg ) using both the nonrelativistic and relativistic expressions for spee

JulsSmile [24]

Answer:

The non-relativistic kinetic energy of a proton is $6.76\times10^{-12}\ J$

The relativistic kinetic energy of a proton is $7.25\times10^{-12}\ m/s$

Explanation:

Given that,

Mass of proton $m=1.67\times10^{-27}\ kg$

Speed $v= 9.00\times10^{7}\ m/s$

We need to calculate the kinetic energy for non relativistic

Using formula of kinetic energy

$K.E=\dfrac{1}{2}mv^2$

Put the value into the formula

$K.E=\dfrac{1}{2}\times1.67\times10^{-27}\times(9.00\times10^{7})^2$

$K.E=6.76\times10^{-12}\ J$

We need to calculate the kinetic energy for relativistic

Using formula of kinetic energy

$K.E=mc^2(\sqrt{(\dfrac{1}{1-\dfrac{v^2}{c^2}})}-1)$

$K.E=1.67\times10^{-27}\times(3\times10^{8})^{2}\cdot\left(\sqrt{\frac{1}{1-\frac{\left(9.00\times10^{7}\right)^{2}}{(3\times10^{8})^{2}}}}-1\right)$

$K.E=7.25\times10^{-12}\ m/s$

Hence, The non-relativistic kinetic energy of a proton is $6.76\times10^{-12}\ J$

The relativistic kinetic energy of a proton is $7.25\times10^{-12}\ m/s$

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2 years ago

Any one please give me the correct answer of this question???<br> I hope you can help me...

Phantasy [73]

Answer:

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2 years ago