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3 years ago

In the phenomenon known as the photoelectric effect, electric current will flow when light shines on certain substances.

Physics

2 answers:

svetlana [45]3 years ago

8 0

The answer is true...............

kotegsom [21]3 years ago

7 0

Answer:

Yes, the given statement is correct.

Explanation:

In the phenomenon photoelectric effect, when photon(energy packet of light) falls on a conductor or semiconductor emits electron and when that electron passes through a closed circuit electric current flow though it. This photoelectric effect have vast application now a days such as:- solar cell, photo cell,solar pump.

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A rubber ball and a steel ball are dropped from the same height onto a concrete floor. They have the same mass, and lets assume

Mkey [24]

Answer:

The average forces would be the same

Explanation:

Both have the same velocity on impact as they fell from the same height.

Both have the same velocity after the bounce because they reach the same height.

Both have the same mass

Both will thus experience the same impulse because both have the same change in momentum.

Therefore both experience the same average force.

5 0

3 years ago

An electron and a proton are held on an x axis, with the electron at x = + 1.000 m and the proton at x = - 1.000 m. Part A How m

r-ruslan [8.4K]

PART A)

Electrostatic potential at the position of origin is given by

$V = \frac{kq_1}{r_1} + \frac{kq_2}{r_2}$

here we have

$q_1 = 1.6 \times 10^{-19} C$

$q_2 = -1.6 \times 10^{-19} C$

$r_1 = r_2 = 1 m$

now we have

$V = \frac{Ke}{r} - \frac{Ke}{r}$

$V = 0$

Now work done to move another charge from infinite to origin is given by

$W = q(V_f - V_i)$

here we will have

$W = e(0 - 0) = 0$

so there is no work required to move an electron from infinite to origin

PART B)

Initial potential energy of electron

$U = \frac{Kq_1e}{r_1} + \frac{kq_2e}{r_2}$

$U = \frac{9\times 10^9(-1.6\times 10^{-19}(-1.6 \times 10^{-19})}{19} + \frac{9\times 10^9(1.6\times 10^{-19}(-1.6 \times 10^{-19})}{21}$

$U = (2.3\times 10^{-28})(\frac{1}{19} - \frac{1}{21})$

$U = 1.15\times 10^{-30}$

Now we know

$KE = \frac{1}{2}mv^2$

$KE = \frac{1}{2}(9.1\times 10^{-31}(100)^2$

$KE = 4.55 \times 10^{-27} kg$

now by energy conservation we will have

So here initial total energy is sufficient high to reach the origin

PART C)

It will reach the origin

4 0

3 years ago

A heat engine with a thermal efficiency of 45 percent rejects 500 kj/kg of heat. how much heat does it receive

adell [148]

Alot as far as i know unless you need it in formal terms.

8 0

3 years ago

How is force related to math

dybincka [34]

Answer:

Newton's second law of motion describes the relationship between force and acceleration. They are directly proportional. If you increase the force applied to an object, the acceleration of that object increases by the same factor. In short, force equals mass times acceleration.

Explanation:

8 0

3 years ago

Read 2 more answers

PLEASE HELP ASAP!!!! A huge thanks to anyone who can help me with 14 problems. I'll do anything to return the favor. All true an

snow_lady [41]

Hello, I see you are in a jam. Lemme help.

1.) True
2.) True
3.) True
4.) True
5.) True

LOL these are all true ;)

4 0

3 years ago