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mamaluj [8]
1 year ago
14

protons in an atomic nucleus are typically 10-15 m apart. what is the electric force of repulsion between nuclear protons?

Physics
1 answer:
Aleonysh [2.5K]1 year ago
5 0

230 Newton

Electric charge consists of two types i.e. positively electric charge and negatively electric charge.There was a famous scientist who investigated about this charges. His name is Coulomb and succeeded in formulating the force of attraction or repulsion between two charges i.e. :

F = electric force (N)

k = electric constant (N m² / C²)

q = electric charge (C)

r = distance between charges (m)

The value of k in a vacuum = 9 x 10⁹ (N m² / C²)

F = k(q1 q2)/ r^2

Distance between protons = d = 10⁻¹⁵ m

charge of proton = q = 1.6 × 10⁻¹⁹ C

Here q1=q2

electric force = F =230N

Coulomb's Law. Two protons in an atomic nucleus are typically separated by a distance of 2×10−15m. The electric repulsive force between the protons is huge, but the attractive nuclear force is even stronger and keeps the nucleus from bursting apart.

2 Nuclei and the Need for an Attractive Nuclear Force. The Coulomb force also acts within atomic nucleii, whose characteristic dimension is 10 m, which is called a fermi. There are two protons in a He nucleus, which repel each other because of the Coulomb force.

Find more about electric force of repulsion between nuclear protons

brainly.com/question/8404637

#SPJ4

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An object with total mass mtotal = 16.2 kg is sitting at rest when it explodes into three pieces. One piece with mass m1 = 4.7 k
vichka [17]

Answer:

1.) 0 kgm/s

2) 6.3 kg

3) -0.0978 m/s

4)

5)

6)

An object with total mass mtotal = 16.2 kg is sitting at rest when it explodes into three pieces. One piece with mass m1 = 4.7 kg moves up and to the left at an angle of θ1 = 23° above the –x axis with a speed of v1 = 25.4 m/s. A second piece with mass m2 = 5.2 kg moves down and to the right an angle of θ2 = 28° to the right of the -y axis at a speed of v2 = 23.8 m/s.

2) What is the mass of the third piece?

3) What is the x-component of the velocity of the third piece?

4) What is the y-component of the velocity of the third piece?

5) What is the magnitude of the velocity of the center of mass of the pieces after the collision?

6) Calculate the increase in kinetic energy of the pieces during the explosion

Explanation:

Since explosions and collisions follow the law of conservation of Momentum.

1) Magnitude of the final momentum of the system = Magnitude of the initial momentum of the system

Since the body was initially at rest,

Magnitude of the initial momentum of the system = 0 kgm/s

Hence, Magnitude of the final momentum of the system is also equal to 0 kgm/s.

2) mass of the third piece

Sum of all the masses = 16.2 kg

4.7 + 5.2 + c = 16.2

c = 6.3 kg

3) doing an x-component balance on momentum

(4.7)×(-25.4 cos 23) + (5.2)×(23.8 cos 28) + (6.3)(v) = 0

-0.6163 + 6.3v = 0

v = -0.0978 m/s

Hope this Helps!!!

4 0
3 years ago
Your answer should be precise to 0.1 m/s. Use a gravitational acceleration of 10 m/s/s. At it lowest point, a pendulum is moving
saw5 [17]

Explanation:

It is given that,

Speed, v₁ = 7.7 m/s

We need to find the velocity after it has risen 1 meter above the lowest point. Let it is given by v₂. Using the conservation of energy as :

\dfrac{1}{2}mv_1^2=\dfrac{1}{2}mv_2^2+mgh

v_2^2=v_1^2-2gh

v_2^2=(7.7)^2-2\times 10\times 1

v_2=6.26\ m/s

So, the velocity after it has risen 1 meter above the lowest point is 6.26 m/s. Hence, this is the required solution.

4 0
3 years ago
Read 2 more answers
While standing atop a building 49.6 m tall, you spot a friend standing on a street corner. Using a protractor and a dangling plu
olga nikolaevna [1]

Answer:

75degree don't forget wind and gravity force pulling down

6 0
3 years ago
A 5.00 kg object oscillates back and forth at the end ofa spring whose spring constant is 49.3 N/m. An obersever istraveling at
defon

Answer:

5.571 sec

Explanation:

angular frequency = √ (k/m) = √ (49.3 / 5) = 3.14 rad/s

Period To = 2π / angular frequency

Period To = 2π/3.14 = 2 × 3.14 / 3.142 = 2.00 sec which you got

T measured by the observer = To / (√ (1 - (v²/c²))) = 2 / √( 1 - 0.871111) = 2 / 0.35901 = 5.571 sec

t=2.00/(1-√((2.80*10^8)^2/(3.00*10^8)^2))= should have been  ( To / (√ (1 - (v²/c²))).  where To = 2.00 sec

8 0
3 years ago
A dragster starts from rest and travels 1/4 mi in 6.80 s with constant acceleration. What is its velocity when it crosses the fi
Ahat [919]
<h2>Its velocity when it crosses the finish line is 117.65 m/s</h2>

Explanation:

We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = 0 m/s

        Acceleration, a = ?

        Time, t = 6.8 s    

        Displacement, s = 1/4 mi =    400 meters

     Substituting

                      s = ut + 0.5 at²

                      400 = 0 x 6.8 + 0.5 x a x 6.8²

                      a = 17.30 m/s²

Now we have equation of motion v = u + at

     Initial velocity, u = 0 m/s

     Final velocity, v = ?

     Time, t = 6.8 s

      Acceleration, a = 17.30 m/s²

     Substituting

                      v = u + at  

                      v = 0 + 17.30 x 6.8

                      v = 117.65 m/s

Its velocity when it crosses the finish line is 117.65 m/s

6 0
3 years ago
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