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V125BC [204]
3 years ago
5

Why does light of a certain frequency need to be used to produce a current in the photoelectric effect?

Physics
1 answer:
olchik [2.2K]3 years ago
3 0
The reason to why does light of a certain frequency need to be used to produce a current in the photoelectric effect is <span>A. The frequency of light must be the same as the frequency of vibration of the atoms in the substance to cause a current.</span>
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A point charge with a charge q1 = 2.30 μC is held stationary at the origin. A second point charge with a charge q2 = -5.00 μC mo
Alla [95]

Answer:

W = 2.74 J

Explanation:

The work done by the charge on the origin to the moving charge is equal to the difference in the potential energy of the charges.

This is the electrostatic equivalent of the work-energy theorem.

W = \Delta U = U_2 - U_1

where the potential energy is defined as follows

U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}

Let's first calculate the distance 'r' for both positions.

r_1 = \sqrt{(x_1 - x_0)^2 + (y_1 - y_0)^2} = \sqrt{(0.170 - 0)^2 + (0 - 0)^2} = 0.170~m\\r_2 = \sqrt{(x_2 - x_0)^2 + (y_2 - y_0)^2} = \sqrt{(0.250 - 0)^2 + (0.250 - 0)^2} = 0.353~m

Now, we can calculate the potential energies for both positions.

U_1 = \frac{kq_1q_2}{r_1^2} = \frac{(8.99\times 10^9)(2.3\times 10^{-6})(-5\times 10^{-6})}{(0.170)^2} = -3.57~J\\U_2 = \frac{kq_1q_2}{r_2^2} = \frac{(8.99\times 10^9)(2.3\times 10^{-6})(-5\times 10^{-6})}{(0.3530)^2} = -0.829~J

Finally, the total work done on the moving particle can be calculated.

W = U_2 - U_1 = (-0.829) - (-3.57) = 2.74~J

4 0
3 years ago
Read 2 more answers
Does gravity exist between a pencil on a coffee mug
DIA [1.3K]

Answer:

Yes it does

Explanation:

Gravity is pushing down on the pencil but the coffee mug is also pushing the pencil up with the same amount of force so they both don't move

5 0
3 years ago
The distance between two consecutive nodesof a standing wave is 20.9cm.Thehandgen-erating the pulses moves up and down throughac
irina1246 [14]

Answer:

Velocity, v = 0.239 m/s

Explanation:

Given that,

The distance between two consecutive nodes of a standing wave is 20.9 cm = 0.209 m

The hand generating the pulses moves up and down through a complete cycle 2.57 times every 4.47 s.

For a standing wave, the distance between two consecutive nodes is equal to half of the wavelength.

\dfrac{\lambda}{2}=0.209\ m\\\\\lambda=0.418\ m

Frequency is number of cycles per unit time.

f=\dfrac{2.57}{4.47}\\\\f=0.574\ Hz

Now we can find the velocity of the wave.

Velocity = frequency × wavelength

v = 0.574 × 0.418

v = 0.239 m/s

So, the velocity of the wave is 0.239 m/s.

4 0
3 years ago
8. Three grams of Bismuth-218 decay to 0.375 grams in one hour. What is the half-
Evgen [1.6K]

Answer: 0.333 h

Explanation:

This problem can be solved using the <u>Radioactive Half Life Formula</u>:  

A=A_{o}.2^{\frac{-t}{H}} (1)  

Where:  

A=0.375 g is the final amount of the material  

A_{o}=3 g is the initial amount of the material  

t=1 h is the time elapsed  

H is the half life of the material (the quantity we are asked to find)  

Knowing this, let's substitute the values and find h from (1):

0.375 g=(3 g)2^{\frac{-1h}{H}} (2)  

\frac{0.375 g}{3 g}=2^{\frac{-1h}{H}} (3)  

Applying natural logarithm in both sides:

ln(\frac{0.375 g}{3 g})=ln(2^{\frac{-1 h}{H}}) (4)  

-2.079=-\frac{1 h}{H}ln(2) (5)  

Clearing H:

H=\frac{-1h}{-2.079}(0.693) (6)  

Finally:

h=0.333 h This is the half-life of the Bismuth-218 isotope

4 0
3 years ago
A piano tuner sounds two strings simultaneously. One has been previously tuned to vibrate at 293.0 Hz. The tuner hears 3.0 beats
ololo11 [35]

Answer:

Part a)

f_B = 290 Hz

Part B)

percentage increase is

percentage = 1.38%

Explanation:

Part a)

As we know that the beat frequency is

f_A - f_B = 3

after increasing the tension the beat frequency is decreased and hence the tension in string B will increase

So we have

293 - f_B = 3

f_B = 290 Hz

Part B)

percentage increase in the tension of the string will be given as

f_A - f_B' = 1

f_B' = 292 Hz

now we have

f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}

so we have

T_1 = C (290)^2

T_2 = C(292)^2

so we have

\frac{\Delta T}{T} = \frac{292^2 - 290^2}{290^2}

percentage increase is

percentage = 1.38

4 0
3 years ago
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