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2 years ago

20.What factor is more important in determining electrical force

Physics

1 answer:

Agata [3.3K]2 years ago

6 0

Answer:

20.The first factor is the amount of charge on each object. The greater the charge, the greater the electric force. The second factor is the distance between the charges. The closer together the charges are, the greater the electric force is

Explanation:

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A solid sphere has a temperature of 614 K. The sphere is melted down and recast into a cube that has the same emissivity and emi

krok68 [10]

Answer:581.87 K

Explanation:

Given

Sphere is melted to form a square

Let the radius of sphere be r and square has a side a

Therefore

$\frac{4\pi}{3}r^3=a^3$

Surface area of sphere $A_s=4\pi r^2$

Surface area of cube $A_c=6a^2$

Total emmisive remains same

Thus $P=A\epsilon \sigma T^4$

$A_sT_s^4=A_cT_c^4$

$\frac{T_c^4}{T_s^4}=\frac{A_s}{A_c}$

$\frac{T_c^4}{T_s^4}=\frac{1}{2}\times \left ( \frac{4\pi}{3}\right )^{\frac{1}{3}}$

$\frac{T_c}{T_s}=\frac{1}{2^{0.25}}\times \left ( \frac{4\pi}{3}\right )^{\frac{1}{12}}$

$T_c=T_s\times \frac{1}{2^{0.25}}\times \left ( \frac{4\pi}{3}\right )^{\frac{1}{12}}$

$T_c=614\times \frac{1.12679}{1.189}$

$T_c=581.87 K$

3 0

3 years ago

Which moon shows evidence of rainfall and erosion by some liquid substance?.

attashe74 [19]

Answer:

Titan.It is the largest moon of Saturn

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2 years ago

The impulse experienced by a body is equivalent to the body’s change in?

vovangra [49]

<span>impulse =force*time=mass*acceleration*time=mass*... in momentum , I hope this helps you out!! Also have an amazing day and good luck on any further work !!!

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3 years ago

Determine the energy required to accelerate an electron between each of the following speeds. (a) 0.500c to 0.900c MeV (b) 0.900

Aleonysh [2.5K]

Answer:

The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.

Explanation:

We know that,

Mass of electron $m_{e}=9.11\times10^{-31}\ kg$

Rest mass energy for electron = 0.511 Mev

(a). The energy required to accelerate an electron from 0.500c to 0.900c Mev

Using formula of rest,

$E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}$

$E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.500c)^2}{c^2}}}$

$E=0.582\ Mev$

(b). The energy required to accelerate an electron from 0.900c to 0.942c Mev

Using formula of rest,

$E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}$

$E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.942c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}$

$E=0.350\ Mev$

Hence, The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.

4 0

3 years ago

Physic help????????????????

maxonik [38]

My personal understanding and opinion is that ALL of those questions
should be part of an assessment of Physical Activity Readiness.

4 0

3 years ago

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