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IceJOKER [234]
2 years ago
10

An electron gun shoots electrons at a metal plate that is 4.0 mm away in a vacuum. The plate is 5.0 V lower in potential than th

e gun. How fast must the electrons be moving as they leave the gun if they are to reach the plate?
Physics
1 answer:
Nadya [2.5K]2 years ago
5 0

The speed of the electron is 1.3 * 10^6 m/s

<h3>What is the velocity?</h3>

We know that when the electron gun is shot, the potential energy of the electron is converted into kinetic energy. The mass of the electron is given as 9.11 * 10^-31 Kg.

The energy of the electron is;

eV = 1e * 5V = ev or 8 * 10^-19 J

Given that E = 1/2mv^2

8 * 10^-19  = 0.5 *  9.11 * 10^-31 * v^2

v = √ 8 * 10^-19/0.5 *  9.11 * 10^-31

v = √1.75 * 10^12

v = 1.3 * 10^6 m/s

Learn ore about speed of electron:brainly.com/question/13130380

#SPJ1

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Electron groups could be considered which of the following?
PSYCHO15rus [73]

Electron groups could be considered as  Lone pair electrons and bonded pairs of electrons.

Answer: Option D & B

<u>Explanation:</u>

The two or more electrons can be bonded by single bond, double bond, covalent bond of electrons can simply be lone pair of electrons. Unshared pair of electrons are generally termed as lone pair of electrons in an atom which are generally present in the outermost shell of atoms. Hence electron groups can be determined by bonded pairs and lone pairs of electrons.

3 0
4 years ago
Consider Compton Scattering with visible light.A photon with wavelength 500nm scatters backward(theta=180degree) from a free ele
JulijaS [17]

Answer: 4.86(10)^{-12}m

Explanation:

The Compton Shift \Delta \lambda in wavelength when photons are scattered is given by the following equation:

\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos\theta) (1)  

Where:  

\lambda'=500 nm=500(10)^{-9} m is the wavelength of the scattered photon

\lambda_{o}  is the wavelength of the incident photon

\lambda_{c}=2.43(10)^{-12} m is a constant whose value is given by \frac{h}{m_{e}.c}, being h=4.136(10)^{-15}eV.s the Planck constant, m_{e} the mass of the electron and c=3(10)^{8}m/s the speed of light in vacuum.  

\theta=180\° the angle between incident phhoton and the scatered photon.  

\Delta \lambda=2.43(10)^{-12} m (1-cos(180\°)) (2)

\Delta \lambda=4.86(10)^{-12}m (3)  This is the shift in wavelength

5 0
3 years ago
What would increase the force of gravity between two objects?
yulyashka [42]
By Newton's Law of Universal Gravitation.

F =  GMm/r²

Where F is Force of Gravitation, M = Mass of first object, m = mass of second object, r = distance of separation

From the formula, you can see that if the masses, M and m, increased, the value of F would definitely increase as well.

And if r increased the value of F would be reduced because you would be dividing by a bigger number, but when the value of r is decreased the value of F would be increased, because you would then be dividing by something smaller. Note the r is at the denominator of the formula.

So F would increase if there was increase in Masses and decrease in distance.

So the answer is C. a and b.
3 0
3 years ago
What is used as evidence for sea-floor spreading?
raketka [301]

Answer:

Sea-floor spreading occurs in the oceanic ridges. In there, volcanic activity, together with the gradual movement of the bottom, form new oceanic crust. This allows a better understanding of the continental drift explained by the theory of plate tectonics.

The greatest evidence for Sea-floor spreading is the oceanic trenches, the oceanic ridges, the magma protruding to the surface and the new seafloor.

In previous theories,  continents were assumed to be transported across the sea. Harry Hess, in the 1960s, proposed the idea that the seabed itself moves  as it expands from a central point. The theory is now accepted, and the phenomenon is thought to be caused by convection currents in the upper layer of the mantle.

4 0
4 years ago
Read 2 more answers
*PHYSICS HELP*
sveta [45]
My calculator is about 1cm thick, 7cm wide, and 13cm long.

Its volume is (length) (width) (thick) = (13 x 7 x 1) = 91 cm³ .

The question wants me to assume that the density of my calculator
is about  the same as the density of water.  That doesn't seem right
to me.  I could check it easily.  All I have to do is put my calculator
into water, watch to see if sinks or floats, and how enthusiastically. 
I won't do that.  I'll accept the assumption.

If its density is actually 1 g/cm³, then its mass is about 91 grams.

The choices of answers confused me at first, until I realized that
the choices are actually 1g, 10² g, 10⁴ g, and 10⁶ g.

My result of 91 grams is about 100 grams ... about 10² grams.

Your results could be different.
3 0
3 years ago
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