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1 year ago

How to calculate area with force and pressure

Physics

1 answer:

solmaris [256]1 year ago

4 0

Answer:

Area = Force/Pressure

Explanation:

The relation of interest is ...

Pressure = Force/Area

This can be rearranged to ...

Area = Force/Pressure

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A long, thin straight wire with linear charge density λ runs down the center of a thin, hollow metal cylinder of radius R. The c

netineya [11]

Answer:

$E=\frac{\lambda}{2\pi r\epsilon_0}$

Explanation:

We are given that

Linear charge density of wire= $\lambda$

Radius of hollow cylinder=R

Net linear charge density of cylinder= $2\lambda$

We have to find the expression for the magnitude of the electric field strength inside the cylinder r<R

By Gauss theorem

$\oint E.dS=\frac{q}{\epsilon_0}$

$q=\lambda L$

$E(2\pi rL)=\frac{L\lambda}{\epsilon_0}$

Where surface area of cylinder= $2\pi rL$

$E=\frac{\lambda}{2\pi r\epsilon_0}$

8 0

1 year ago

Suppose that a rectangular toroid has 2,000 windings and a self-inductance of 0.060 H. If the height of the rectangular toroid i

andrey2020 [161]

Answer:

0.01154 A

Explanation:

We have given the energy in the magnetic field $U=4\times 10^{-6}J$

Value of inductance L =0.060 H

Energy stored in magnetic field is given by $U=\frac{1}{2}Li^2$

$i=\sqrt{\frac{2U}{L}}$

$i=\sqrt{\frac{2\times 4\times 10^{-6}}{0.06}}=0.01154\ A$

So the current flowing through rectangular toroid will be 0.01154 A

3 0

2 years ago

if m represents mass in kg, v represents speed in m/s, and r represents radius in m, show that the force F in the equation F=mv^

Zarrin [17]

This approach is called the dimensional analysis which involves only the units of measurement without their magnitudes. You simply have to do the operations by using variables. Cancel out like items that may appear both in the numerator and denominator side. The solution is as follows:

F = mv²/r = [kg][m/s]²/[m] = [kg][m²⁻¹][1/s²] = [kg·m/s²]

4 0

2 years ago

An object with a heat capacity of 345J∘C experiences a temperature change from 88.0∘C to 45.0∘C. How much heat is released in th

pogonyaev

Answer:

There is 148.35 Joules of heat is released in the process.

Explanation:

Given that,

Heat capacity of the object, $c=345J/^oC$