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babunello [35]
3 years ago
15

Which of the following quantities are unknown? initial separation of the particles final separation of the particles initial spe

ed of the proton initial speed of the alpha particle final speed of the proton final speed of the alpha particle mass of the proton mass of the alpha particle charge of the proton charge of the alpha particle Enter the letters of all the correct answers in alphabetical order. Do not use commas. For instance, if A, C, and D are unknowns, enter ACD.
Physics
1 answer:
tino4ka555 [31]3 years ago
7 0

ALL of the quantities are unknown, because you haven't bothered to tell us anything that's known.

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Explain Rutherford's experiment?
Ipatiy [6.2K]

Answer:

Rutherford bombarded aluminum foil with beam of light known as alpha particles. The mass of this alpha particle is equivalent to helium atom.

Explanation:

Rutherford bombarded aluminum foil with beam of light known as alpha particles. The mass of this alpha particle is equivalent to helium atom.

When this alpha particles were made to strike the aluminum foil, some passed through the foil, some were reflected and speed others changed.

The ones reflected encountered heavier particle known as the nucleus, preventing them from passing through it. The whole observations indicated that atom is not is uniformly charged sphere as proposed by J.J Thomson.

Rutherford proposed new model known as the Planetary model of atom, which described atom as containing a nucleus which is revolved by electron, just like planets revolve round the sun. And this nucleus contains opposite charge to electron which is proton, to balance the motion.

7 0
3 years ago
An orange light (f = 5.2 * 10'4Hz) is
KonstantinChe [14]

Answer:

2.145×10^-10 V or 0.2145nV

Explanation:

From hf=eV

h= Plank's constant = 6.6×10^-34JS

f= frequency of the electromagnetic wave = 5.2×10^4 Hz

e= electronic charge= 1.6×10^-19 C

V= voltage

V= hf/e

V= 6.6×10^-34JS × 5.2×10^4 Hz/ 1.6×10^-19 C

V= 2.145×10^-10 V or 0.2145nV

Therefore the voltage created is 2.145×10^-10 V or 0.2145nV

7 0
2 years ago
An electric field of 1.32 kV/m and a magnetic field of 0.516 T act on a moving electron to produce no net force. If the fields a
Lapatulllka [165]

Answer:

The speed of the electron is 2.55\times 10^3\ m/s.

Explanation:

Given that,

The magnitude of electric field, E=1.32\ kV/m=1.32\times 10^3\ V/m

The magnitude of magnetic field, B = 0.516 T

Both the magnetic and electric fields are acting on the moving electron. Then,  the magnitude of electric field and magnetic field is balanced such that :

evB=eE\\\\v=\dfrac{E}{B}\\\\v=\dfrac{1.32\times 10^3}{0.516}\\\\v=2558.13\ m/s

or

v=2.55\times 10^3\ m/s

So, the speed of the electron is 2.55\times 10^3\ m/s. Hence, this is the required solution.

3 0
3 years ago
Choose the answer to fill in the blank, which best completes the sentence.
Allisa [31]

Answer:

A on edge 2020

Explanation:

4 0
2 years ago
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A heavy truck and a small truck roll down a hill. Neglecting friction, at the bottom of the hill the heavy truck has greater
postnew [5]

Answer:

kenetic energy

Explanation:

or potential energy

3 0
3 years ago
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