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Lemur [1.5K]
2 years ago
12

(e) For photons of energy 7.10 eV, what stopping potential would be required to arrest the current of photoelectrons

Physics
1 answer:
Masja [62]2 years ago
8 0

2.37eV stopping potential would be required to arrest the current of photoelectrons.

<h3 /><h3>What is stopping potential ?</h3>

The minimal negative voltage that must be provided to the anode to halt the photocurrent is known as stopping potential. When expressed in electron volts, the maximal kinetic energy of the electrons is equal to the stopping voltage.

Kmax = eV₀

2.37eV = eV₀

V₀ = 2.37eV

to learn more about stopping potential go to - brainly.com/question/4655588

#SPJ4

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What charge accumulates on the plates of a 2.0-μF air-filled capacitor when it is charged until the potential difference across
enot [183]

Answer:

0.0002 C.

Explanation:

Charge: This can be defined as the ratio of current to time flowing in a circuit. The S.I unit of charge is Coulombs (C)

Mathematically, charge can be expressed as

Q = CV ................................. Equation 1

Where Q = amount of charge, C = capacitance of the capacitor, V = potential difference across the plates.

Given: C = 2.0-μF = 2×10⁻⁶ F, V = 100 V.

Substitute into equation 1

Q = 2×10⁻⁶× 100

Q = 2×10⁻⁴ C

Q = 0.0002 C.

The amount of charge accumulated = 0.0002 C

7 0
3 years ago
Read 2 more answers
Give 5 examples of physical properties that could be used to describe a solid
ycow [4]
1.) appearance
2.)texture
3.)color
4.)melting point
5.)odor
8 0
3 years ago
How much heat is needed to vaporize 10.00 grams of water at 100.0°C? The latent heat of vaporization of water is 2,259 J/g
Nata [24]

Answer:

Heat of vaporization will be 22.59 j    

Explanation:

We have given mass m = 10 gram

And heat of vaporization L = 2.259 J/gram

We have to find the heat required to vaporize 10 gram mass

We know that heat of vaporization is given by Q=mL, here m is mass and L is latent heat of vaporization.

So heat of vaporization Q will be = 10×2.259 = 22.59 J

8 0
3 years ago
Read 2 more answers
Two parallel plate capacitors 1 and 2 are identical except that capacitor 1 has charge +q on one plate and charge −q on the othe
Grace [21]

Answer:

a) the capacitance is the same for both capacitors.

b) The potential difference between the plates for the capacitor with charge +2q, is double of the one for the capacitor  with charge +q.

c) The electric field magnitude between the plates for the capacitor with charge +2q, is double of the one for the capacitor with charge +q

d) The energy stored between the plates for the capacitor with charge +2q, is 4 times the value for the one with charge +q  

Explanation:

a) The capacitance of a capacitor, by definition, is as follows:

C = \frac{q}{V}

Appying Gauss' Law to one of plates, it can be showed, that the capacitance (for a parallel plates capacitor) can be  expressed as follows:

C = ε*A / d

As it can be seen, it does not depend on the charge. so we conclude that the capacitance must be the same for both capacitors, due to they are identical except for the value of the charge on the plates.

b) By definition, as we said above, the capacitance is equal to the proportion between the charge of one of the plates, and the potential difference between them.

If this proportion must remain the same, and one of the capacitors has the double of  the charge than the other, the potential difference must be the double also.

c) Applying Gauss' law, to the surface of one of  the plates, and assuming a constant surface charge density σ, it can be  showed that the  electric field can be calculated as follows:

E*A = Q/ε₀ as σ=Q/A

⇒ E = σ/ε₀

As σ is directly proportional to the charge (being the area A the same), we conclude that the electric field for the capacitor with charge +2q must be the double than the one for the capacitior with charge +q.

c) The electric potential energy, stored between plates of a capacitor, can be written as follows:

Ue = \frac{1}{2} *\frac{q^{2}}{C}

If the capacitance remains the same, we can conclude that the electric potential energy for the capacitor with charge +2q, as the charge is raised to the 2nd power, must be 4 times the one for the capacitor with charge +q.

4 0
3 years ago
How long does it take a microwave of power 0.2kW to sue 10000 J of energy
olasank [31]

Answer:

50s .

Explanation:

\frak{\pink{Given}}\begin{cases}\textsf{ The power of microvave is 0.2kW .}\\\textsf{ Amount of energy is 10000 J .}\end{cases}

Here the power of the microwave is 0.2kW . And as we know that ,

  • Rate of doing work is called power .

So from the definition , we have ;

\sf\longrightarrow Power =\dfrac{Work}{time}

  • Here the work done is equal to the energy consumed by the microwave i.e. 10000 J .So we can write it as ,

\sf\longrightarrow Power =\dfrac{Energy}{time}

\sf\longrightarrow 0.2kW = \dfrac{10^4 J }{t} \\

\sf\longrightarrow 0.2 * 1000 W =  \dfrac{10^4 J }{t}

Cross multiply ,

\sf\longrightarrow t = \dfrac{ 10^4 }{ 0.2 * 10^3}s=\dfrac{10^4}{2*10^2} s

Simplify ,

\sf\longrightarrow \boxed{\bf t = 50s}

<h3>Hence the time taken is 50s .</h3>
5 0
2 years ago
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