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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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Red light of wavelength 630 nm passes through two slits and then onto a screen that is 1.4 m from the slits. The center of the 3
VARVARA [1.3K]

Answer:

Part a)

f = 4.76 \times 10^{14} Hz

Part b)

d = 3.48 \times 10^{-4} m

Part c)

\theta = 0.311 degree

Explanation:

Part a)

As we know that the speed of light is given as

c = 3 \times 10^8 m/s

\lambda = 630 nm

now the frequency of the light is given as

f = \frac{c}{\lambda}

so we have

f = \frac{3 \times 10^8}{630 \times 10^{-9}}

f = 4.76 \times 10^{14} Hz

Part b)

Position of Nth maximum intensity on the screen is given as

y_n = \frac{n\lambda L}{d}

so here we know for 3rd order maximum intensity

y_3 = 0.76 cm

n = 3

L = 1.4 m

0.76 \times 10^{-2} = \frac{3(630 \times 10^{-9})(1.4)}{d}

d = 3.48 \times 10^{-4} m

Part c)

angle of third order maximum is given as

d sin\theta = 3 \lambda

3.48 \times 10^{-4} sin\theta = 3(630 \times 10^{-9})

\theta = 0.311 degree

8 0
3 years ago
If earth's axis was tilted 45 degrees at what latitudes
givi [52]
23.5 DEGREES is the answer. 
7 0
3 years ago
What is the necessary condition for the conservation of angular momentum
RoseWind [281]

Answer:

The conservation of momentum of a system is a fundamental principle in classical mechanics. The only condition for this principle being valid is that the system should be an isolated one, i.e. It should not be acted upon by any external force.

3 0
3 years ago
Two volumes of William Shakespeare stand on a bookshelf next to each other: volume one, then volume two. Each volume is 4 cm thi
Setler79 [48]

Answer:

L = 7 [cm]

Explanation:

To solve this problem we must analyze each of the distances mentioned and take into account the number of covers and thicknesses of these.

The worm crosses through the sheets of the first book, this distance can be determined by the following length analysis.

4 = P + 2*C

Where:

P = thicknesses of the pages [cm]

C = thicknesses of each cover [cm]

P = 4 - 2*(0.5)

P = 3 [cm]

The distance crossed was:

L = P + 2C + P "the pages of the first book + 2 covers + the pages of the second book"

L = 3 + (2*0.5) + 3

L = 7 [cm]

7 0
3 years ago
Ilya and Anya each can run at a speed of 8.90mph and walk at a speed of 3.80mph . They set off together on a route of length 5.0
jekas [21]

Answer:

a) t = 56.34 minutes

b) v = 5.32mph

c) t = 47.22 minutes

d) v = 6.35 mph

Explanation:

Given;

The speed of Ilya and Anya are

Run = 8.90mph

Walk = 3.80mph

distance = 5.00 miles

a) anya walk half the distance and run half the distance.

Speed = distance/time

time = distance/speed

t1 = 5×0.5/3.80 = 0.658hr

t2 = 5×0.5/8.90 = 0.281hr

t = t1 + t2 = 0.658+0.281 = 0.939hr × 60minutes/hr

t = 56.34 minutes

b) average speed = total distance/total time taken

v = 5.00miles/0.939hr = 5.32mph

c) Ilya walk half the time and run half the time.

t1 = t2 = t/2

Distance = speed × time

Substituting the values.

5 = 3.80(t/2) + 8.90(t/2)

5 = 12.70(t/2)

t = 10/12.70

t = 0.787 hr × 60 minutes/hr

t = 47.22 minutes

d) average speed = 5miles/0.787hr

v = 6.35 mph

5 0
3 years ago
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