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Sloan [31]
3 years ago
10

in order to generate electricity, nuclear powerplants take advantage of this part of the electromagnetic spectrum

Physics
1 answer:
antiseptic1488 [7]3 years ago
8 0
Bit of an odd question. Power Plants are known to use water-powered turbines to generate electricity, but can also make use of nuclear fission.
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Mike enters a revolving door that is not moving. explain where and how mike should push to produce a torque with the least amoun
Bingel [31]

Mike enters a revolving door that is not moving. Mike should push at the edge of the door where it is largest distance from the pivot point in order to produce a torque with the least amount of force. Torque is equal to t = force x distance.

7 0
4 years ago
Read 2 more answers
un cubo que mide 5 in por lado. ¿cual es el volumen del cubo en unidades del SI y en unidades del SUEU
Pavel [41]

Answer:

125 in^3

Explanation:

Volumen de un cubo = lado ^ 3

Lado = 5

Volumen = 5 ^ 3

5^3 = 5\times 5\times 5\\\\= 125 in^3

8 0
3 years ago
How much work is done to move a 10,000-N car 20 meters?
Helga [31]

f = 10000N
distance moved = 20 m
work done = 10000*20 = 200000
6 0
4 years ago
Electrolysis can be used to split up ionic compounds into their constituent<br> Science
tankabanditka [31]

Answer:

Electrolysis is the process by which ionic substances are decomposed (broken down) into simpler substances when an electric current is passed through them. ... Electricity is the flow of electrons or ions. For electrolysis to work, the compound must contain ions.

4 0
3 years ago
A 0.290 cm diameter plastic sphere, used in a static electricity demonstration, has a uniformly distributed 30.0 pC charge on it
mojhsa [17]

Answer:

Therefore,

The potential (in V) near its surface is 186.13 Volt.

Explanation:

Given:

Diameter of sphere,

d= 0.29 cm

radius=\dfrac{d}{2}=\dfrac{0.29}{2}=0.145\ cm

r = 0.145\ cm = 0.145\times 10^{-2}\ m

Charge ,

Q = 30.0\ pC=30\times 10^{-12}

To Find:

Electric potential , V = ?

Solution:

Electric Potential at point surface is given as,

V=\dfrac{1}{4\pi\epsilon_{0}}\times \dfrac{Q}{r}

Where,  

V= Electric potential,  

ε0 = permeability free space = 8.85 × 10–12 F/m

Q = Charge  

r = Radius  

Substituting the values we get

V=\dfrac{1}{4\times 3.14\times 8.85\times 10^{-12}}\times \dfrac{30\times 10^{-12}}{0.145\times 10^{-2}}

V=\dfrac{30}{16.117\times 10^{-2}}=186.13\ Volt

Therefore,

The potential (in V) near its surface is 186.13 Volt.

3 0
4 years ago
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