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Arturiano [62]
3 years ago
11

What is the relationship between the density of the equipotential lines, the density of the electric field lines and the strengt

h of the electric field?
Physics
1 answer:
12345 [234]3 years ago
7 0

Answer:

I dont. understand the question, maybe insert the picture?

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Which equation can be used to calculate the normal force on an object if you know the speed of the object, the coefficient of ki
tino4ka555 [31]

Answer:

D

Explanation:

The friction force is the weight force times the coefficient of friction. So diving by the coefficient gives you the weight force which is equivalent to the normal force.

3 0
2 years ago
a painting in an art gallery has height h and is hung so that its lower edge is a distance d above the eye of an observer. How f
harkovskaia [24]

Solution:

With reference to Fig. 1

Let 'x' be the distance from the wall

Then for \DeltaDAC:

tan\theta = \frac{d}{x}

⇒ \theta = tan^{-1} \frac{d}{x}

Now for the \DeltaBAC:

tan\theta = \frac{d + h}{x}

⇒ \theta = tan^{-1} \frac{d + h}{x}

Now, differentiating w.r.t x:

\frac{d\theta }{dx} = \frac{d}{dx}[tan^{-1} \frac{d + h}{x} -  tan^{-1} \frac{d}{x}]

For maximum angle, \frac{d\theta }{dx} = 0

Now,

0 = [/tex]\frac{d}{dx}[tan^{-1} \frac{d + h}{x} -  tan^{-1} \frac{d}{x}][/tex]

0 = \frac{-(d + h)}{(d + h)^{2} + x^{2}} -\frac{-d}{x^{2} + d^{2}}

\frac{-(d + h)}{(d + h)^{2} + x^{2}} = \frac{{d}{x^{2} + d^{2}}

After solving the above eqn, we get

x = \sqrt{\frac{d}{d + h}}

The observer should stand at a distance equal to x = \sqrt{\frac{d}{d + h}}

4 0
3 years ago
What internal process cause most earthquake
Alisiya [41]
Plate Tectonics cause most earthquakes.
7 0
4 years ago
Read 2 more answers
Three laws that indicate a chemical change
alexandr1967 [171]

Temperature, The highness, and the time.

Hope this helps!

C=

4 0
3 years ago
Two parallel plates are 1 cm apart and are connected to a 500 V source. What force will be exerted on a single electron half way
VladimirAG [237]

Answer: 8*10^-15 N

Explanation: In order to calculate the force applied on an electron in the middle of the two planes at 500 V we know that,  F=q*E

The electric field between  the plates is given by:

E = ΔV/d = 500 V/0.01 m=5*10^3 N/C

the force applied to the electron is: F=e*E=8*10^-15 N

3 0
3 years ago
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