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Effectus [21]
3 years ago
12

20% Part (a) Use an "E Field Sensor" and move it along either equipotential. What can you say about the E field along an equipot

ential?
a. Varies in magnitude and direction, parallel to the equipotential.
b. Constant magnitude, but varying direction, perpendicular to the equipotential.
c. Varies in magnitude and direction, perpendicular to the equipotential.
d. Constant direction, but varying magnitude, perpendicular to the equipotential.
e. Constant direction, but varying magnitude, parallel to the equipotential.
f. Constant magnitude, but varying direction, parallel to the equipotential.
Physics
1 answer:
Alex3 years ago
6 0

Answer:

b. Constant magnitude, but varying direction, perpendicular to the equipotential.

Explanation:

As we know that the relation between electric field and electric potential is given as

\Delta E = -\frac{dV}{dr}

here if we say that potential is constant because electric field sensor is moving along equi-potential line.

Then we will say

V = constant

so we have

\Delta E = 0

so electric field will remain constant always in magnitude and always remains perpendicular to the surface

so we have

b. Constant magnitude, but varying direction, perpendicular to the equipotential.

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4 0
3 years ago
A ball, which has a mass of 1.25 kg, is thrown straight up from the top of a building 225 meters tall with a velocity of 52.0 m/
Anastasy [175]

First we will find the speed of the ball just before it will hit the floor

so in order to find the speed of the cart we will first use energy conservation

KE_i + PE_i = KE_f + PE_f

\frac{1}{2}mv_i^2 + mgh = \frac{1}{2}mv_f^2 + 0

\frac{1}{2}(1.25)(52)^2 + 1.25(9.8)(225) = \frac{1}{2}(1.25)v_f^2

So by solving above equation we will have

v_f = 84.3 m/s

now in order to find the momentum we can use

P = mv

P = 1.25 \times 84.3

P = 105.4 kg m/s

3 0
3 years ago
Unpolarized light with an intensity of (25.0 A) units is passed through two successive polarizing filters, the first with its po
coldgirl [10]

Answer:

the intensity of the light after passing through the two polarizing filters is 4.11 units  

 

Explanation:

Given the data in the question;

the intensity of an unpolarized light; I₀ = 25.0 units

when the unpolarized light passes through the first polarizer, its intensity reduces to half of its initial value;

⇒ I₁ = I₀/2 = 25/2 = 12.5 units

the angle between the transmission axes of two polarizers is;

∅ = 55° - 0° = 55°

The intensity of the light after passing through two polarizing filters will be;

I₂ = I₁cos²∅      

we substitute

I₂ = 12.5 × cos²(55)

I₂ = 12.5 × 0.3289899

I₂ = 4.11 units

Therefore, the intensity of the light after passing through the two polarizing filters is 4.11 units

6 0
3 years ago
Five 60 ohm resistors are connected in parallel. What is their equivalent resistance?
Mnenie [13.5K]

Answer:

i think the answer is 12 ohms

plz mark me as brainliest :)

7 0
3 years ago
Read 2 more answers
Why water in earthern pot remain cool in summer
hjlf

Answer:

Since in summer, the eastern side do not face the sunlight and hence the water in eastern pot remain cool in summer.

8 0
3 years ago
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