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djyliett [7]
3 years ago
5

What are 3 subatomic particles of an atom?

Physics
2 answers:
Vesnalui [34]3 years ago
7 0
Protons, neutrons, and electrons<span> are the three main subatomic particles found in an atom.</span>
skad [1K]3 years ago
7 0
I think you're talking about protons, neutrons, and electrons, but I'm not sure.
You might be interested in
How long does it take to shut down a nuclear reactor?.
Sphinxa [80]

Answer:

5 months

Explanation:

3 0
2 years ago
Two ships of equal mass are 110 m apart. What is the acceleration of either ship due to the gravitational attraction of the othe
dusya [7]

Answer:

Acceleration of the ship, a=2.14\times 10^{-7}\ m/s^2

Explanation:

It is given that,

Mass of both ships, m=39000\ metric\ tons=39\times 10^6\ kg

Distance between two ships, d = 110 m

The gravitational force between two ships is given by :

F=G\dfrac{m^2}{d^2}

F=6.67\times 10^{-11}\ Nm^2/kg^2\times \dfrac{(39\times 10^6\ kg)^2}{(110\ m)^2}

F = 8.38 N

Let a is the acceleration. Now, using second law of motion as :

a=\dfrac{F}{m}

a=\dfrac{8.38\ N}{39\times 10^6\ kg}

a=2.14\times 10^{-7}\ m/s^2

So, the acceleration of either ship due to the gravitational attraction of the other is 2.14\times 10^{-7}\ m/s^2. Hence, this is the required solution.

7 0
3 years ago
Crude oil coal and peat are examples of fossil fuels name another one<br><br>​
Nikolay [14]

Answer:

petroleum and Natural gas are fossil fuels

Explanation:

3 0
3 years ago
What is the mechanical advantage of a lever that has an input arm of 6 meters and an output arm of 2 meters
Setler79 [48]

Answer:A

Explanation:I TOOK THE Test

5 0
3 years ago
The parallel plates in a capacitor, with a plate area of 7.90 cm2 and an air-filled separation of 2.70 mm, are charged by a 7.90
s2008m [1.1K]

Answer:

A) 26V

Explanation:

(a) the potential difference between the plates

Initial capacitance can be calculated using below expresion

C1= A ε0/ d1

Where d1= distance between = 2.70 mm= 2.70× 10^-3 m

ε0= permittivity of space= 8.85× 10^-12 Fm^-1

A= area of the plate = 7.90 cm2 = 7.90 ×10^-4 m^2

If we substitute the values we

C1= A ε0/ d1

=( 7.90 ×10^-4 × 8.85× 10^-12 )/2.70× 10^-3

C1=2.589 ×10^-12 F= 2.59 pF

Initial charge can be determined using below expresion

q1= C1 × V1

V1=2.589 ×10^-12 F

V1= voltage=7.90 V

If we substitute we have

q1= 2.589 ×10^-12 × 7.90

q1= 20.45×10^-12C

20.45 pC

Final capacitance can be calculated as

C2= A ε0/ d2

d2=8.80 mm= /8.80× 10^-3

7.90 ×10^-4 × 8.85× 10^-12 )/8.80× 10^-3

C1=0.794 ×10^-12 F= 0.794 pF

Final charge= initial charge

q2=q1 (since the battery is disconnected)

q2=q1= 20.45 pC

Final potential difference

V2= q/C2

= 20.45/0.794

= 26V

6 0
3 years ago
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