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3 years ago

13

6. how do the number of electrons flowing into a light bulb compare to the number of electrons flowing out of it?

1 answer:

N76 [4]3 years ago

5 0

If there's no leak in the bottom of the light bulb where electrons can
fall out onto the floor, and no magic electron generator inside the bulb
to create electrons out of thin air, then the number flowing in must be
the same as the number flowing out.

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Which kind of air front forms when a warm air mass meets the area of a cooler air mass?

mihalych1998 [28]

<span>Answer: "a cold front" .
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3 years ago

The force of attraction between a -165.0 uC and +115.0 C charge is 6.00 N. What is the separation between these two charges in m

Simora [160]

Answer:

The distance between the charges is 5,335.026 m

Explanation:

To obtain the forces between the particles, we can use Coulomb's Law in scalar form, this is, the force between the particles will be:

$F = k \frac{q_1 q_2}{d^2}$

where k is Coulomb's constant, $q_1$ and $q_2$ are the charges and d is the distance between the charges.

Working a little the equation, we can take:

$d^2 = k \frac{q_1 q_2}{F}$

$d = \sqrt{ k \frac{q_1 q_2}{F}}$

And this equation will give us the distance between the charges. Taking the values of the problem

$k= 9.00 \ 10^9 \frac{N \ m^2}{C^2} \\q_1 = 165.0 \mu C \\q_2 = 115.0 C\\F=- 6.00$

(the force has a minus sign, as its attractive)

$d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}$

$d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}$

$d = \sqrt{ 28,462,500 \ m^2}}$

$d = 5,335.026 m$

And this is the distance between the charges.

3 0

4 years ago

Which of these is exhibiting kinetic energy?1) a rock on a mountain ledge2) a space station orbiting Earth 3) a person sitting o

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The answer would be:

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3 years ago

Read 2 more answers

A wave has a wavelength of 10 mm and a frequency of 4 hertz to what is it’s speed ?

nordsb [41]

Convert 10mm to metres = 0.01m

Wave speed = frequency *wavelength
= 4Hz*0.01m
= 0.04 m/s

Hope this helps

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3 years ago

You are designing a delivery ramp for crates containing exercise equipment. The 1890 N crates will move at 1.8 m/s at the top of

Mashcka [7]

Answer:

K = 588.3 N/m

Explanation:

From a forces diagram, and knowing that for the maximum value of K, the crate will try to rebound back up (Friction force will point downward):

Fe - Ff - W*sin(22) = 0 Replacing Fe = K*X and then solving for X:

$X = \frac{Ff + W*sin(22)}{K}=\frac{1223}{K}$

By conservation of energy:

$\frac{K*X^{2}}{2}-mg*d*sin(22)-\frac{m*V^{2}}{2}=-Ff*d$

Replacing our previous value for X and solving the equation for K, we get maximum value to prevent the crate from rebound:

K = 588.3 N/m

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4 years ago