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Ronch [10]
3 years ago
8

What is the SI unit of energy

Physics
1 answer:
Anettt [7]3 years ago
5 0
The SI unit of energy is joule(J). Hope this helps!
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The plates of a parallel-plate capacitor are 2.50 mm apart, and each carries a charge of magnitude 80.0 nC. The plates are in va
bonufazy [111]

Answer:

10000 V

0.00225988700565 m²

8\times 10^{-12}\ F

Explanation:

E = Electric field = 4\times 10^6\ V/m

d = Gap = 2.5 mm

Q = Charge = 80 nC

\epsilon_0 = Permittivity of free space = 8.85\times 10^{-12}\ F/m

Potential difference is given by

V=Ed\\\Rightarrow V=4\times 10^6\times 2.5\times 10^{-3}\\\Rightarrow V=10000\ V

The potential difference between the plates is 10000 V

Area is given by

A=\dfrac{Q}{\epsilon_0E}\\\Rightarrow A=\dfrac{80\times 10^{-9}}{8.85\times 10^{-12}\times 4\times 10^6}\\\Rightarrow A=0.00225988700565\ m^2

The area of the plate is 0.00225988700565 m²

Capacitance is given by

C=\dfrac{\epsilon_0A}{d}\\\Rightarrow C=\dfrac{8.85\times 10^{-12}\times 0.00225988700565}{2.5\times 10^{-3}}\\\Rightarrow C=8\times 10^{-12}\ F

The capacitance is 8\times 10^{-12}\ F

4 0
3 years ago
1. A heat lamp is designed to keep food and other things warm. Would it also make a good tanning lamp? Why or why not?
NemiM [27]
1. becuase it is so hot that it could tan or be used to keep warm.
4 0
3 years ago
Read 2 more answers
What is the car's acceleration from 0 to 1 second?
dusya [7]
10 mph/s because there is 60 seconds in a minute then divide by 6 which is 10.
3 0
3 years ago
What is the electric field strength inside the capacitor after insulating handles are used to pull the electrodes away from each
Bezzdna [24]
This question apparently comes after an EARLIER one,
where you were told either the voltage across the same
capacitor or the total charge stored in it.  You can't answer
THIS one without that information.
3 0
3 years ago
Through what potential difference would you need to accelerate an alpha particle, starting from rest, so that it will just reach
Svetlanka [38]

Answer:

\Delta V    = 1.8 \times 10^7 V

Explanation:

GIVEN

diameter = 15 fm  =15 \times 10^{-15}m

we use here energy conservation

K_{i}+U_{i} =K_{f}+U_{f}

there will be some initial kinetic  energy but after collision kinetic energy will zero

K_{i} + 0 = 0 + \frac{1}{4 \pi \epsilon _{0}} \frac{(2e)(92e)}{7.5 \times 10^{-15}}

on solving these equations we get kinetic energy initial

KE_{i} = 5.65\times 10 ^{-12} \times \frac  {1 eV}{1.6 \times 10^{-19}}

KE_{i} = 35.33 J ..............(i)

That is, the alpha particle must be fired with 35.33 MeV of kinetic energy. An alpha particle with charge q = 2  e

and gains kinetic energy K  =e∆V  ..........(ii)

 by accelerating through a potential difference ∆V

Thus the alpha particle will

just reach the {238}_U nucleus after being accelerated through a potential difference  ∆V

equating (i) and second equation we get

e∆V  = 35.33 Me V

\Delta V = \frac{35.33}{2}  MV\\\Delta V    = 1.8 \times 10^7 V

7 0
3 years ago
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