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N76 [4]
3 years ago
13

In the nucleus of an atom, two protons are separated by a distance of 10^-13 m. What is the Coloumb force between them? The char

ge of a proton is + 1.602 x 10^-19 C.
Physics
1 answer:
sleet_krkn [62]3 years ago
7 0

Answer:

F=0.0231\ N

Explanation:

<u>Electrostatic Force </u>

When two point charges q_1 and q_2 are placed at a distance d, they exert a force to each other whose magnitude can be computed by the Coulomb's formula

\displaystyle F=\frac{k\ q_1\ q_2}{d^2}

Where k is the constant of proportionality

k=9.10^9\ Nw.m^2/c^2

The information provided is

d=10^{-13}\ m

q_1=q_2=+ 1.602 x 10^{-19}\ C

Computing the Coloumb force between them

\displaystyle F=\frac{9x10^9\ 1.602 x 10^{-19} \ 1.602 x 10^{-19} }{(10^{-13})^2}

\boxed{F=0.0231\ N}

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At what speed should a ball of mass 2 kg be rolled in order to reach the other side of
Veronika [31]

Answer:

M g H = 1/2 M v^2       potential energy = kinetic energy

v^2 = 2 g H = 2 * 9.80 * 6 = 117.6 m/s^2

v = 10.8 m/s    

(C)

6 0
2 years ago
Each driver has mass 79.0 kg. Including the masses of the drivers, the total masses of the vehicles are 800 kg for the car and 4
Mademuasel [1]

Answer:

Force exerted on the car driver by the seatbelt = 8139.4 N = 8.14 kN

Force exerted on the truck driver by the seatbelt = 1628.2 N = 1.63 kN

It is evident that the driver of the smaller vehicle has it worse. The car driver is in way more danger in this perfectly inelastic head-on collision with a bigger vehicle (the truck).

Explanation:

First of, we calculate the velocity of the vehicles after collision using the law of conservation of Momentum

Momentum before collision = Momentum after collision

Since the collision of the two vehicles was described as a head-on collision, for the sake of consistent convention, we will take the direction of the velocity of the bigger vehicle (the truck) as the positive direction and the direction of the car's velocity automatically is the negative direction.

Velocity of the truck before collision = 6.80 m/s

Velocity of the car before collision = -6.80 m/s

Let the velocity of the inelastic unit of vehicles after collision be v

Momentum before collision = (4000)(6.80) + (800)(-6.80) = 27200 - 5440 = 21,760 kgm/s

Momentum after collision = (4000 + 800)(v) = (4800v) kgm/s

Momentum before collision = Momentum after collision

21760 = 4800v

v = (21760/4800)

v = 4.533 m/s (in the direction of the big vehicle (the truck)

So, we then apply Newton's second law of motion which explains that the magnitude change in momentum is equal to the magnitude of impulse.

|Impulse| = |Change in momentum|

But Impulse = (Force exerted on each driver by the seatbelt) × (collision time) = (F×t)

Change in momentum = (Momentum after collision) - (Momentum before collision)

So, for the driver of the truck

Initial velocity = 6.80 m/s (the driver moves with the velocity of the truck)

Final velocity = 4.533 m/s

Change in momentum of the truck driver = (79)(6.80) - (79)(4.533) = 179.1 kgm/s

(F×t) = 179.1

F × 0.110 = 179.1

F = (179.1/0.11)

F = 1628.2 N = 1.63 kN

So, for the driver of the car

Initial velocity = -6.80 m/s (the driver moves with the velocity of the car)

Final velocity = 4.533 m/s

Change in momentum of the car driver = (79)(-6.80) - (79)(4.533) = -895.3 kgm/s

(F×t) = |-895.3|

F × 0.110 = 895.3

F = (895.3/0.11)

F = 8139.4 N = 8.14 kN

Hope this Helps!!!

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Carbon dioxide and water :) hope this helped!
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According to ohm's law what is the resistance of a light bulb if the applied voltage is 21 volts and the current is 3 amps?
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Resistance = voltage/current = 21/3 = 7Ω
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