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3 years ago

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Two protons in an atomic nucleus are typically separated by a distance of 2 ✕ 10-15 m. The electric repulsion force between the

protons is huge, but the attractive nuclear force is even stronger and keeps the nucleus from bursting apart. What is the magnitude of the electric force between two protons separated by 2.00 ✕ 10-15 m

2 answers:

SVETLANKA909090 [29]3 years ago

6 0

The magnitude of the electric force between two protons separated by 2.00 × 10⁻¹⁵ m is about 57.6 Newton

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<h3>Further explanation</h3>

Electric charge consists of two types i.e. positively electric charge and negatively electric charge.

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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. :

$\large {\boxed {F = k \frac{Q_1Q_2}{R^2} } }$

<em>F = electric force (N)</em>

<em>k = electric constant (N m² / C²)</em>

<em>q = electric charge (C)</em>

<em>r = distance between charges (m)</em>

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

Let's tackle the problem now !

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<u>Given:</u>

distance between protons = d = 2 × 10⁻¹⁵ m

charge of proton = q = 1.6 × 10⁻¹⁹ C

<u>Unknown:</u>

electric force = F = ?

<u>Solution:</u>

$F = k \frac{q_1 q_2}{(d)^2}$

$F = k \frac{q^2}{(d)^2}$

$F = 9 \times 10^9 \times \frac{(1.6 \times 10^{-19})^2}{(2 \times 10^{-15})^2}$

$F = 57.6 \texttt{ Newton}$

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<h3>Learn more</h3>

The three resistors : brainly.com/question/9503202
A series circuit : brainly.com/question/1518810
Compare and contrast a series and parallel circuit : brainly.com/question/539204

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<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Static Electricity

castortr0y [4]3 years ago

4 0

Answer:

The magnitude of the electric force between the to protons will be 57.536 N.

Explanation:

We can use Coulomb's law to find out the force, in scalar form, will be:

$F \ = \ \frac{1}{4 \pi \epsilon_0 } \frac{q_1 q_2}{d^2}$ .

Now, making the substitutions

$d \ = \ 2.00 * 10 ^{-15} \ m$ ,

$q_1 = q_2 = 1.60 * 10 ^ {-19} \ C$ ,

$\frac{1}{4\pi\epsilon_0}=8.99 * 10^9 \frac{Nm^2}{C^2}$ ,

we can find:

$F \ = \ 8.99 * 10^9 \frac{Nm^2}{C^2} \frac{(1.60 * 10 ^ {-19} \ C)^2}{(2.00 * 10 ^{-15} \ m)^2}$ .

$F \ = 57.536 N$ .

Not so big for everyday life, but enormous for subatomic particles.

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wA₁ = 8π -αA*t₁

The angle in B is:

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Hello there!

I hope you and your family are staying safe and healthy during this unprecendented time.

A) What is the work done?

Answer: We need to use the formula

$w=-F_f(d)$

$w=-(35)(20)$

$w=-700J$

B) What is the work done on the cart by the gravitational force?

Alright, we know that the gravitional force is perpendicular to the diplacement. Therefore, we gonna use the following formula:

$w=Fdcos90$

$w=0$

C) What is the work done on the cart by the shopper?

This is the easier part, since we already know that the work done by the shopper is the same as the work done by the friction force

$W shopper + W friction = 0\\W shopper = W-friction \\W shopper = 700J$

D) Find the force the shopper exerts, using energy considerations.

$F_f+Fcos25=0\\-35+Fcos25=0\\F=38.6N$

E) What is the total work done?

You just need to add them:

$w=wshopper+wfriction\\w=0$

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3 years ago