Which two factors does the power of a machine depend on?

natka813 [3]

It's A they repel eachother

8 0

3 years ago

An electron having 500ev energy enters at right angle to a uniform magnetic field of 10^-4 Tesla. If its specific charge is 1.75

gtnhenbr [62]

Answer:

The correct answer might be $r = 2.8^{27}$ meters.

Explanation:

<u>The Answer Given might not be correct, I just did what my brain said.</u>

As the angle is perpendicular so Θ=90.

Putting this in the equation to calculate the magnetic force as:

F = evBsinΘ

F= evBsin90 *sin90 = 1 so,

F= evB.

Now when the electron will start to move in a circle, The necessary force that makes the electron rotate in a circle is given by Centripetal force.

So,

Magneteic Force = Centripetal Force

evB = $\frac{mv^{2} }{r}$

r = $\frac{mv}{Be}$ ......(1)

Now the problem is, We don't know " v " so we need to calculate velocity first,

Calculation of Velocity:

In order to calculate the velocity of electron, We should know the potiential difference with which the electrons are accelerated which in our case is 500ev. If "V" is the potiential difference, the energy gained by electrons during accelreation will be Ve. This appear as kinectic enrgy of electrons as,

K.E = Ve

$\frac{1}{2}$ m $v^{2}$ = Ve

v = $\sqrt{\frac{2ve}{m} }$ ................(2)

Putting value of velocity in equation 2 from 1:

r = $\frac{mv}{Be}$ $\sqrt{\frac{2ve}{m} }$

$r = \sqrt{\frac{2mev}{Be}}$

$r = \sqrt{\frac{(2)(9.1^{-31})(500) (1.6^{-19} ) }{ (1.75^{11} ) (10^{-4} ) } }$

$r = 2.8^{27}$ meters.

8 0

3 years ago

A 1.2 kg block is held at rest against the spring with a force constant k= 730 N/m. Initially, the spring is compressed a distan

Nesterboy [21]

Answer:

Compression distance: $d \approx 0.102\,m$

Explanation:

According to this statement, we know that system is non-conservative due to the rough patch. By Principle of Energy Conservation and Work-Energy Theorem, we have the following expression that represents the system having a translational kinetic energy ( $K$ ), in joules, at the expense of elastic potential energy ( $U$ ), in joules, and overcoming work losses due to friction ( $W_{l}$ ), in joules:

$K + W_{l} = U$ (1)

By definitions of translational kinetic and elastic potential energies and work losses due to friction, we expand the equation described above:

$\frac{1}{2}\cdot m \cdot v^{2} +\mu\cdot m\cdot g \cdot s = \frac{1}{2} \cdot k \cdot d^{2}$ (2)

Where:

$m$ - Mass of the block, in kilograms.

$v$ - Final velocity of the block, in meters per second.

$\mu$ - KInetic coefficient of friction, no unit.

$g$ - Gravitational acceleration, in meters per square second.

$s$ - Width of the rough patch, in meters.

$k$ - Spring constant, in newtons per meter.

$d$ - Compression distance, in meters.

If we know that $m = 1.2\,kg$ , $v = 2.3\,\frac{m}{s}$ , $\mu = 0.44$ , $g = 9.807\,\frac{m}{s^{2}}$ , $s = 0.05\,m$ and $k = 730\,\frac{N}{m}$ , then the compression distance of the spring is: