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4 years ago

Who clarified the photoelectric effect?

1 answer:

7 0

When visible light, X rays, gamma rays, or other forms of electromagnetic radiation are shined on certain kinds of matter, electrons are ejected. That phenomenon is known as the photoelectric effect. The photoelectric effect was discovered by German physicist Heinrich Hertz (1857–1894) in 1887. You can imagine the effect as follows: Suppose that a metal plate is attached by two wires to a galvanometer. (A galvanometer is an instrument for measuring the flow of electric current.) If light of the correct color is shined on the metal plate, the galvanometer may register a current. That reading indicates that electrons have been ejected from the metal plate. Those electrons then flow through the external wires and the galvanometer. HOPE THIS HELPED

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What is the force of gravity for a 12 kg turkey?<br><br> Please help asap

pogonyaev

Answer: 117.6N

Explanation:

By the second Newton's law, we know that:

F = m*a

F = force

m = mass

a = acceleration

We know that in the surface of the Earth, the gravitational acceleration is g = 9.8m/s^2.

Then we just can input that acceleration in the above equation, and also replace m by 12kg, and find that the force due the gravity is:

F = 12kg*9.8m/s^2 = 117.6N

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3 years ago

Help ASAP <br> Just answer the first question for me please!

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Answer:

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3 years ago

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3 years ago

One electron collides elastically with a second electron initially at rest. After the collision, the radii of their trajectories

ch4aika [34]

Answer:

114.92749 keV

Explanation:

r = Radius of trajectory

m = Mass of electron = $9.11\times 10^{-31}\ kg$

B = Magnetic field = 0.044 T

q = Charge of electron = $1.6\times 10^{-19}\ C$

The centripetal force and the magnetic forces are conserved

$m\frac{v^2}{r}=Bqv\\\Rightarrow v=\frac{Bqr}{m}$

Velocity of first electron

$v=\frac{Bqr_1}{m}\\\Rightarrow v=\frac{0.044\times 1.6\times 10^{-19}\times 0.01}{9.11\times 10^{-31}}\\\Rightarrow v_1=77277716.79473\ m/s$

Velocity of second electron

$v=\frac{Bqr_2}{m}\\\Rightarrow v_2=\frac{0.044\times 1.6\times 10^{-19}\times 0.024}{9.11\times 10^{-31}}\\\Rightarrow v_2=185466520.30735\ m/s$

Total kinetic energy is given by

$K=K_1+K_2\\\Rightarrow K=\frac{1}{2}mv_1^2+\frac{1}{2}mv_2^2\\\Rightarrow K=\frac{1}{2}m(v_1^2+v_2^2)\\\Rightarrow K=\frac{1}{2}\times 9.11\times 10^{-31}(77277716.79473^2+185466520.30735^2)\\\Rightarrow K=1.83884\times 10^{-14}\ J$

Converting to eV

$1\ J=\frac{1}{1.6\times 10^{-19}}\ eV$

$1.83884\times 10^{-14}\ J=1.83884\times 10^{-14}\times \frac{1}{1.6\times 10^{-19}}\ eV\\ =114927.49\ ev=114.92749\ keV$

The energy of incident electron is 114.92749 keV

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IgorC [24]

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