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N76 [4]
3 years ago
9

The potential difference between two parallel plates is 227 V. If the plates are 6.8 mm apart, what is the electric field betwee

n them? O S.0x10 N/C O 28 x 10* N/C O 4.1 x 10 N/C O 3.3 x 10 N/C
Physics
1 answer:
nalin [4]3 years ago
3 0

Answer:

E=3.3\times 10^4N/C

Option D is the correct answer.

Explanation:

Electric field, E is the ratio of potential difference and distance between them.

Potential difference, V = 227 V

Distance between plates = 6.8 mm = 0.0068 m

Substituting,

         E=\frac{V}{d}=\frac{227}{0.0068}=3.3\times 10^4N/C

Option D is the correct answer.

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saul85 [17]

Answer:

D.Outer space is answer

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Which layer of the atmosphere is directly above the troposhpere? A.troposhpere B.stratosphere C.mesosphere D.exoshpere
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Its B. i hope i helped!!!
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How long must a current of 100mA flow so as to transfer a charge of 80 C? ​
defon

Answer:

800s

Explanation:

Q=It

where,

Q=80c

I =100MA=0.1A

t=?

Q=It

800=0.1×t

t=80×0.1

t=800s

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The velocity of a 1.3 kg block sliding down a frictionless inclined plane is found to be 1.26 m/s. 1.10 s later, it has a veloci
nasty-shy [4]

Answer:

\theta = 25.3^\circ

Explanation:

The acceleration of the block can be found by the kinematics equations:

v = v_0 + at\\5.88 = 1.26 + a(1.1)\\a = 4.2~m/s^2

Since the plane is frictionless, the only force acting on the block along the motion of the block is its weight.

F = mg\sin(\theta) = ma\\g\sin(\theta) = a\\(9.8)\sin(\theta) = 4.2\\\theta = 25.3^\circ

7 0
3 years ago
An emf of 28.0 mV is induced in a 501 turn coil when the current is changing at a rate of 12.0 A/s. What is the magnetic flux th
dmitriy555 [2]

Answer:

Φ = 5.589×10⁻⁵  Wb

Explanation:

The inductance of a coil is given as

L = e/(di/dt) ..................... Equation 1

Where L = inductance of the coil, e = induced e.m.f, di/dt = rate of change of current in the coil.

Also,

The inductance of each turn of the coil when a magnetic field is step up in the coil  is

L = NΦ/i ................. Equation 2

Where N = number of turns, Φ = magnetic field, i = current.

equating equation 1 and equation 2

e/(di/dt) = NΦ/i

making Φ the subject of the equation,

Φ = (e×i)/N.(di/dt) .................. Equation 3

Given: e = 28.0 mV = 0.028 V, N = 501 turns, di/dt = 12.0 A/s, i = 4.00 a

Substitute into equation 3,

Φ = (0.028×4)/(12×501)

Φ = 0.112/2004

Φ = 5.589×10⁻⁵ Weber

Φ = 5.589×10⁻⁵ Wb

6 0
3 years ago
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