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o-na [289]
3 years ago
11

Light of a single frequency falls on a photo-electric material but no electrons are emitted . Electrons may be emitted if the __

___________

Physics
1 answer:
lesya692 [45]3 years ago
8 0
Correct answer is "frequency of the light is increased".

The light that hits the photo-electric material consists of several photons, each one carrying an energy of
E=hf
where h is the Planck constant while f is the light frequency. Part of this energy is used to extract the photo-electrons from the material, and part it is converted into kinetic energy K of the electron:
hf= \phi + K
where \phi is the work function of the material, which is the amount of energy needed to extract the photo-electron.

If no photo-electrons are emitted, it means that the energy given by the photon is not enough to extract the photo-electron. But we see that the energy of the photon is directly proportional to the frequency of the light f, so in order to increase the energy, we should increase the frequency of the light.
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In 'coin on card' experiment a smooth card is used. ​
KIM [24]

Answer:

In coin card experiment smooth card is used so that the card can slide easily from glass

5 0
2 years ago
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A single loop of nickel wire, lying flat in a plane, has an area of 7.40 cm^2 and a resistance of 2.40 Ω. A uniform magnetic fie
ale4655 [162]

Explanation:

It is given that,

Area of nickel wire, A=7.4\ cm^2=7.4\times 10^{-4}\ m^2

Resistance of the wire, R = 2.4 ohms

Initial value of magnetic field, B_1=0.5\ T

Final magnetic field, B_2=3\ T

Time, t = 1.12 s

Let I is the induced current in the loop of wire over this time. Te emf induced in the wire is given by Faraday's law as :

\epsilon=-\dfrac{d\phi}{dt}

\epsilon=-\dfrac{d(BA)}{dt}

\epsilon=-A\dfrac{d(B)}{dt}

\epsilon=-A\dfrac{B_2-B_1}{t}

\epsilon=-7.4\times 10^{-4}\times \dfrac{3-0.5}{1.12}

\epsilon=1.65\times 10^{-3}\ V

Induced current in the loop of wire is given by :

I=\dfrac{\epsilon}{R}

I=\dfrac{1.65\times 10^{-3}}{2.4}

I=6.87\times 10^{-4}\ A

So, the induced current in the loop of wire over this time is 6.87\times 10^{-4}\ A. Hence, this is the required solution.

7 0
3 years ago
Students in an introductory physics lab are performing an experiment with a parallel-plate capacitor made of two circular alumin
Alex777 [14]

Answer:

0.92 μC

Explanation:

In a parallel-plate capacitor, the electric field formed is equal to the charge density divited by the vacuum permisivity e0, as there are no dielectric between the plates. e0 is equal to 8.85*10^-12 C^2/Nm^2. The charge density is the total charge of each individual plate divided by its area. Then, the maximum charge allowed will be equal to:

E = \frac{o}{e_0} = \frac{Q}{Ae_0} \\ Q = E*A*e_0 = 3*10^6 N/C * (0.25*\pi *(0.21m)^2)*8.85*10^{-12}C^2/Nm^2 = 9.196 *10^{-7} C

or 0.92 μC

7 0
2 years ago
An unknown radioactive sample is observed to decrease in activity by a factor of two in a one hour period. What is its half-life
elena55 [62]

Answer:

The half-life is t_{1/2} = 1.005 h

Explanation:

Using the decay equation we have:

A=A_{0}e^{-\lambda t}

Where:

  • λ is the decay constant
  • A(0) the initial activity
  • A is the activity at time t

We know the activity decrease by a factor of two in a one hour period (t = 1 h), it means that A = \frac{A_{0}}{2}

\frac{A_{0}}{2}=A_{0}e^{-\lambda*1 h}

0.5=e^{-\lambda*1 h}

Taking the natural logarithm on each side we have:

ln(0.5)=-\lambda

\lambda=0.69 h^{-1}

Now, the relationship between the decay constant λ and the half-life t(1/2) is:

\lambda = \frac{ln(2)}{t_{1/2}}

t_{1/2} = \frac{ln(2)}{\lambda}

t_{1/2} = \frac{ln(2)}{0.69}

t_{1/2} = 1.005 h

I hope it helps you!

6 0
2 years ago
An object travels with velocity v = 4.0 meters/second and it makes an angle of 60.0° with the positive direction of the y-axis. 
Darya [45]
Goooooooaaaaalllll , C 2.5 meters
7 0
3 years ago
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