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s2008m [1.1K]
3 years ago
13

Ohms law is A.(R=E/W). B.(R=E/1). C.(E/Z). D.none of them

Physics
1 answer:
nikklg [1K]3 years ago
6 0

Answer:

D. none of them.

Explanation:

This is because Ohm's law is:

Voltage = Current × Resistance

or,

V = IR

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What is lever means,types of lever.. <br>will give brainles​
serious [3.7K]

Explanation:

A lever is a rigid bar which moves freely about a fixed point called fulcrum....

The types of lever are :

  • First class lever
  • Second class lever
  • Third class lever....

3 0
1 year ago
50 points Answer fast please
erma4kov [3.2K]

Answer:

22 hours

Explanation:

6 0
3 years ago
A parallel plate capacitor is created by placing two large square conducting plates of length and width 0.1m facing each other,
borishaifa [10]

Answer:

8.854 pF

Explanation:

side of plate = 0.1 m ,

d = 1 cm = 0.01 m,

V = 5 kV = 5000 V

V' = 1 kV = 1000 V

Let K be the dielectric constant.

So, V' = V / K

K = V / V' = 5000 / 1000 = 5

C = ε0 A / d = 8.854 x 10^-12 x 0.1 x 0.1 / 0.01 = 8.854 x 10^-12 F

C = 8.854 pF

5 0
3 years ago
The electric field in a region is uniform (constant in space) and given by E-( 148.0 1 -110.03)N/C. An additional charge 10.4 nC
enyata [817]

Answer:

The y-component of the electric force on this charge is F_y = -1.144\times 10^{-6}\ N.

Explanation:

<u>Given:</u>

  • Electric field in the region, \vec E = (148.0\ \hat i-110.0\ \hat j)\ N/C.
  • Charge placed into the region, q = 10.4\ nC = 10.4\times 10^{-9}\ C.

where, \hat i,\ \hat j are the unit vectors along the positive x and y axes respectively.

The electric field at a point is defined as the electrostatic force experienced per unit positive test charge, placed at that point, such that,

\vec E = \dfrac{\vec F}{q}\\\therefore \vec F = q\vec E\\=(10.4\times 10^{-9})\times (148.0\ \hat i-110.0\ \hat j)\\=(1.539\times 10^{-6}\ \hat i-1.144\times 10^{-6}\ \hat j)\ N.

Thus, the y-component of the electric force on this charge is F_y = -1.144\times 10^{-6}\ N.

3 0
3 years ago
A refrigerator is used to remove 84 kj/min of heat from a tank. If the electric power consumed by the refrigerator is 1. 2 kw, w
zvonat [6]

Answer:

84 kj/min = 1.4 kj/sec

Power Out / Power In = Heat Out / Heat In - Coefficient of Performance

1.4 kj/sec / 1.2 kj/sec = 1.17 = COP

3 0
2 years ago
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