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Temka [501]
3 years ago
10

uniform electric field exists between two parallel plates separated by 2.0 cm. The intensity of the field is 15 kN/C. What is th

e potential difference between the plates?
Physics
1 answer:
adoni [48]3 years ago
7 0

Answer:

potential difference V= 300 volts

Explanation:

Given:

d= 2.0 cm = 0.02m

E = 15 kN/C = 15 × 10³ N/C

For a uniform field between two plates, the Electric Filed Intensity (E) is proportional to the potential difference (V) and inversely  proportional to distance between the plates.

E= V/d

⇒ V= E×d = 15 × 10³ N/C × 0.02 m = 300 volts  (∴1 Nm/C = 1 J/C= 1 volts)

You might be interested in
A laser produces light at 5.32x10-7 m.
Blababa [14]

Answer:

D

Explanation:

speed = frequency x wavelength

speed of light in vacuum is 3.0 x 10^8

wavelength = 5.32 x10 ^-7

3.0 x 10 ^ 8 = 5.32 x 10^-7 x frequency

frequency = 5.63909 x 10^14

round it off = 5.64 x 10^14 Hz

thus the answer is D

hope this helps please mark it

6 0
3 years ago
The distance from the Earth to the Sun equals 1 AU. Neptune is 30 AU from the Sun. How far is Neptune from the Earth?AU
Ipatiy [6.2K]

Answer:

Depending on the relative position of the Earth the Sun and Neptune in the Earths orbit the distances are;

The closest (minimum) distance of Neptune from the Earth is 29 AU

The farthest (maximum) distance of Neptune fro the Earth is 31 AU

Explanation:

The following parameters are given;

The distance from the Earth to the Sun = 1 AU

The distance of Neptune from the Earth = 30 AU

We have;

When the Sun is between the Earth and Neptune, the distance is found by the relation;

Distance from the Earth to Neptune = 30 + 1 = 31 AU

When the Earth is between the Sun  and Neptune, the distance is found by the relation;

Distance from the Earth to Neptune = 30 - 1 = 29 AU

Therefore, the closest distance from Neptune to the Earth in the Earth's Orbit is 29 AU

The farthest distance from Neptune to the Earth in the Earth's orbit is 31 AU.

8 0
3 years ago
Which increases the work output of a machine?
Dafna11 [192]
A reducing friction
3 0
3 years ago
A rifle has a mass of 7-kg and the bullet has a mass of 0.7-kg. If the velocity of the bullet is 350-m/s after the rifle is fire
nevsk [136]

Answer:

-35 m/s

Explanation:

Momentum is conserved.

Momentum before firing = momentum after firing

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

Before the bullet is fired, the bullet and rifle have no velocity, so u₁ and u₂ are 0.

0 = m₁v₁ + m₂v₂

Given m₁ = 0.7 kg, v₁ = 350 m/s, and m₂ = 7 kg:

0 = (0.7 kg) (350 m/s) + (7 kg) v

v = -35 m/s

The rifle recoils at 35 m/s in the opposite direction.

8 0
4 years ago
A 7.50 kg bowling ball has 70.4<br> kg•m/s of momentum. What is its<br> velocity?
dolphi86 [110]

Answer:

<h2>9.39 m/s</h2>

Explanation:

The velocity of the bowling ball can be found by using the formula

v =  \frac{p}{m}  \\

p is the momentum

m is the mass

From the question we have

v =  \frac{70.4}{7.5}  \\  = 9.38666666..

We have the final answer as

<h3>9.39 m/s</h3>

Hope this helps you

3 0
3 years ago
Read 2 more answers
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