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Luden [163]
4 years ago
12

A 12.0 V car battery is being used to power the headlights of a car. Each of the two headlights has a power rating of 41.8 Watts

. The number of electrons that pass through the car battery each second is ___ x 1019. Assume three significant digits in your answer. You do not need to use scientific notation when you enter your answer.
Physics
1 answer:
Marina CMI [18]4 years ago
5 0

Answer:

n = 4.35 x 10¹⁹          

Explanation:

Given that

Voltage V= 12 V

Power rating of headlights = 41.8 W

We know that headlights of the car is connected in parallel connection.

We know that

Power = Current x Voltage

P = V I

I=\dfrac{41.8}{12}\ A

I=3.48  A

Therefore the total current will be

I'= 3.48 + 3.48 A

I = 6.96 A

We know that

Charge = Current x time

q= I' t

q= 6.96 x 1

q= 6.96 C

The charge on electron ,e= 1.6 x 10⁻¹⁹

q= n e

n=Number of electron

n=\dfrac{q}{e}

n=\dfrac{6.96}{1.6\times 10^{-19}}

n = 4.35 x 10¹⁹

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ICE Princess25 [194]

Answer:

(a) decrease

Explanation:

Viscosity is the resistance which occur to flow of the fluid.

More the inter molecular forces between particles of the liquid, more the viscosity of liquid.

<u>Effect of temperature on viscosity:-</u>

Viscosity decreases with the increase in the temperature as forces among the particles decrease on increasing temperature. The kinetic energy of the particles of the liquid increases causing to move in more random motions and thus weaker inter molecular forces and this offer less resistance to the flow.

<u>Hence, viscosity of the liquids decrease with the increasing temperature.</u>

3 0
3 years ago
BRainliest if correct
kolbaska11 [484]

Answer:

C.

hope this helps

have a good day :)

Explanation:

4 0
3 years ago
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What is the electric field at a location b → = &lt;-0.5, -0.4, 0&gt; m, due to a particle with charge +2 nC located at the origi
Naddika [18.5K]

Answer:

E = \frac{8.99 x10^{9} \frac{Nm^2}{C^2} * 2x10^{-9} C}{(0.64m)^2} =43.85N

Explanation:

For this case we assume that we want to find the electrical field at the point P as we can see on the figure attached.

The electrical field wormula is given by:

E = \frac{K Q}{d^2}

Where r is the distance from the point and the charge. On this case we can use the Pythagoras theorem and we got:

d^2 = (-0.5m)^2 +(-0.4)^2 = 0.41m^2

d =\sqrt{0.41}= 0.64m

And now we can replace into the formula since we know that Q = 2nC= 2x10^{-9}C and K = 8.99 x10^{9} \frac{Nm^2}{C^2}, and we got:

E = \frac{8.99 x10^{9} \frac{Nm^2}{C^2} * 2x10^{-9} C}{(0.64m)^2} =43.85N

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3 years ago
H=K+log(a/c) isolate a
Soloha48 [4]
H=k+log(a/c)
H-k = log(a/c)
e^{h-k}=a/c
a = ce^{h-k}
3 0
3 years ago
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A force p is 46.67N. if it inclined at an angle of 50°, find its horizontal component​
Ivanshal [37]

Explanation:

F_{x} = 46.67 N(cos 50°) = 30.0 N

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3 years ago
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