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lozanna [386]
3 years ago
8

What keeps the electrons from leaving an atom in the rutherford model of the atom?

Physics
1 answer:
alexdok [17]3 years ago
7 0

A. strong electrostatic attraction to the nucleus


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What state of matter has a defined volume but undefined shape?
slavikrds [6]
LIQUID is that state............
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A traditional unit of length in Japan is the ken (1 ken 1.97 m). What are the ratios of (a) square kens to square meters and (b)
Roman55 [17]
If 1 ken is 1.97 meter, then 1 square ken is 3.8809 square meters, and one cubic ken is 7.645373. As for the cylindrical tank, the volume of it would be 10.835 times the radius of the cylinder time 1.97^2 times pi. As you didn't specify the radius, I can't give the exact answer but that would be how to get it.
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A car accelerates from 30 m/s to 50 m/s in 2 seconds. Calculate the cars acceleration
masha68 [24]

Answer:

applying 1st eq of motion vf=vi+at we have to find a=vf-vi/t here a=50-30/2=10 so we got a=10m/s²

5 0
3 years ago
A -0.00325 C charge q1 is placed 5.62 m from a second charge q2. The first charge is repelled with a 48900 N force. What is the
blagie [28]

Answer: q2 = -0.05286

Explanation:

Given that

Charge q1 = - 0.00325C

Electric force F = 48900N

The electric field strength experienced by the charge will be force per unit charge. That is

E = F/q

Substitute F and q into the formula

E = 48900/0.00325

E = 15046153.85 N/C

The value of the repelled second charge will be achieved by using the formula

E = kq/d^2

Where the value of constant

k = 8.99×10^9Nm^2/C^2

d = 5.62m

Substitutes E, d and k into the formula

15046153.85 = 8.99×10^9q/5.62^2

15046153.85 = 284634186.5q

Make q the subject of formula

q2 = 15046153.85/ 28463416.5

q2 = 0.05286

Since they repelled each other, q2 will be negative. Therefore,

q2 = -0.05286

6 0
3 years ago
What is the electric potential energy of an electron at the negative end of the cable, relative to the positive end of the cable
VashaNatasha [74]

Answer:

Electric potential energy at the negative terminal: 1.92\cdot 10^{-18}J

Explanation:

When a particle with charge q travels across a potential difference \Delta V, then its change in electric potential energy is

\Delta U = q \Delta V

In this problem, we know that:

The particle is an electron, so its charge is

q=-1.60\cdot 10^{-19}C

We also know that the positive terminal is at potential

V_+=0V

While the negative terminal is at potential

V_-=-12 V

Therefore, the potential difference (final minus initial) is

\Delta V = -12-0 = -12 V

So, the change in potential energy of the electron is

\Delta U = (-1.6\cdot 10^{-19})(-12)=1.92\cdot 10^{-18}J

This means that the electron when it is at the negative terminal has 1.92\cdot 10^{-18}J of energy more than when it is at the positive terminal.

Since the potential at the positive terminal is 0, this means that the electric potential energy of the electron at the negative end is

1.92\cdot 10^{-18}J

3 0
3 years ago
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