All categories

11 months ago

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]11 months 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

You might be interested in

Jane starts from her house to take a stroll in her neighborhood. After walking for 2 hours at a steady pace, she has walked 4 mi

9966 [12]

Answer:

C. 2 mph

Explanation:

speed equals total distance covered in total time of of travel

since,

total distance = 4 miles

time taken = 2 hrs

speed = 4m / 2hrs =2mph

3 0

3 years ago

Read 2 more answers

Sully uses a battery and a coil of wire to create an electromagnet. Using the same materials, if he wants to increase the streng

garik1379 [7]

An electromagnet is a type of magnet in which the magnetic field is produced using the current. The simplest form of an electromagnet is a wire wrapped around in a coil.
The strength of magnetic field of such magnet is given with this equation:
$B=\frac{NI\mu}{L}$
Where N is the number of loops in the coil, I is the strength of the current flowing through the coil, L is the length of the coil, and $\mu$ is <span>permeability of the electromagnet core material.
From this equation, we can see that increasing both the current and number of loops will increase the strength of the magnet.
Both BLANKS should be Increase. When you use the additional battery you will have more voltage and more voltage means more electricity.</span>

5 0

2 years ago

The three main types of nuclear radiation are alpha, beta, and gamma

valkas [14]

Answer:

Gamma beta alpha

Explanation:

6 0

2 years ago

Read 2 more answers

A water balloon is thrown horizontally from a tower that is 45 m high. It strikes the shoes of an unsuspecting passerby who is 4

LekaFEV [45]

Answer:

14.85 m/s

Explanation:

From the question given above, the following data were obtained:

Height (h) of tower = 45 m

Horizontal distance (s) moved by the balloon = 45 m

Horizontal velocity (u) =?

Next, we shall determine the time taken for the balloon to hit the shoe of the passerby. This is illustrated below:

Height (h) of tower = 45 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

h = ½gt²

45 = ½ × 9.8 × t²

45 = 4.8 × t²

Divide both side by 4.9

t² = 45/4.9

Take the square root of both side

t = √(45/4.9)

t = 3.03 s

Finally, we shall determine the magnitude of the horizontal velocity of the balloon as shown below:

Horizontal distance (s) moved by the balloon = 45 m

Time (t) = 3.03 s

Horizontal velocity (u) =?

s = ut

45 = u × 3.03

Divide both side by 3.03

u = 45/3.03

u = 14.85 m/s

Thus, the magnitude of the horizontal velocity of the balloon was 14.85 m/s

4 0

2 years ago

In the high jump, the kinetic energy of an athlete is transformed into gravitational potential energy without the aid of a pole.

Fiesta28 [93]

Answer:

6.0 m/s

Explanation:

According to the law of conservation of energy, the total mechanical energy (potential, PE, + kinetic, KE) of the athlete must be conserved.

Therefore, we can write:

$KE_i+PE_i =KE_f+PE_f$

$\frac{1}{2}mu^2+0=\frac{1}{2}mv^2+mgh$

where:

m is the mass of the athlete

u is the initial speed of the athlete (at the bottom)

0 is the initial potential energy of the athlete (at the bottom)

v = 0.80 m/s is the final speed of the athlete (at the top)

$g=9.8 m/s^2$ is the acceleration due to gravity

h = 1.80 m is the final height of the athlete (at the top)

Solving the equation for u, we find the initial speed at which the athlete must jump:

$u=\sqrt{v^2+2gh}=\sqrt{0.80^2+2(9.8)(1.80)}=6.0 m/s$

4 0

3 years ago