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3 years ago

How does the electrostatic force compare with the strong nuclear force in the

Physics

2 answers:

soldier1979 [14.2K]3 years ago

7 0

Answer:

The electrostatic force is weaker and acts over longe distances, is the correct answer for Appex/

Explanation:

diamong [38]3 years ago

3 0

Answer:

Strong nuclear force is 1-2 order of magnitude larger than the electrostatic force

Explanation:

There are mainly two forces acting between protons and neutrons in the nucleus:

- The electrostatic force, which is the force exerted between charged particles (therefore, it is exerted between protons only, since neutrons are not charged). The magnitude of the force is given by

$F_E=\frac{kq_1 q_2}{r^2}$

where k is the Coulomb's constant, q1 and q2 are the charges of the two particles, r is the separation between the particles.

The force is attractive for two opposite charges and repulsive for two same charges: therefore, the electrostatic force between two protons is repulsive.

- The strong nuclear force, which is the force exerted between nucleons. At short distance (such as in the nucleus), it is attractive, therefore neutrons and protons attract each other and this contributes in keeping the whole nucleus together.

At the scale involved in the nucleus, the strong nuclear force (attractive) is 1-2 order of magnitude larger than the electrostatic force (repulsive), therefore the nucleus stays together and does not break apart.

You might be interested in

In an RLC series circuit that includes a source of alternating current operating at fixed frequency and voltage, the resistance

maw [93]

Answer:

Capacitive Reactance is 4 times of resistance

Solution:

As per the question:

R = $X_{L} = j\omega L = 2\pi fL$

where

R = resistance

$X_{L} = Inductive Reactance$

f = fixed frequency

Now,

For a parallel plate capacitor, capacitance, C:

$C = \frac{\epsilon_{o}A}{x}$

where

x = separation between the parallel plates

Thus

C ∝ $\frac{1}{x}$

Now, if the distance reduces to one-third:

Capacitance becomes 3 times of the initial capacitace, i.e., x' = 3x, then C' = 3C and hence Current, I becomes 3I.

Also,

$Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}$

Also,

Z ∝ I

Therefore,

$\frac{Z}{I} = \frac{Z'}{I'}$

$\frac{\sqrt{R^{2} + (R - X_{C})^{2}}}{3I} = \frac{\sqrt{R^{2} + (R - \frac{X_{C}}{3})^{2}}}{I}$

${R^{2} + (R - X_{C})^{2}} = 9({R^{2} + (R - \frac{X_{C}}{3})^{2}})$

${R^{2} + R^{2} + X_{C}^{2} - 2RX_{C} = 9({R^{2} + R^{2} + \frac{X_{C}^{2}}{9} - 2RX_{C})$

Solving the above eqn:

$X_{C} = 4R$

6 0

3 years ago

A 12.0-g bullet is fired horizontally into a 109-g wooden block that is initially at rest on a frictionless horizontal surface a

kykrilka [37]

Answer:

v₀ = 280.6 m / s

Explanation:

we have the shock between the bullet and the block that we can work with at the moment and another part where the assembly (bullet + block) compresses a spring, which we can work with mechanical energy,

We write the mechanical energy when the shock has passed the bodies

Em₀ = K = ½ (m + M) v²

We write the mechanical energy when the spring is in maximum compression

$Em_{f} = K_{e} \\= \frac{1}{2} kx^2\\ Em_0 = Em_{f}$

½ (m + M) v² = ½ k x²

Let's calculate the system speed

v = √ [k x² / (m + M)]

v = √[152 ×0.78² / (0.012 +0.109) ]

v = 27.65 m / s

This is the speed of the bullet + Block system

Now let's use the moment to solve the shock

Before the crash

p₀ = m v₀

After the crash

$p_{f} = (m + M) v$

The system is formed by the bullet and block assembly, so the forces during the crash are internal and the moment is preserved

$p_0 = p_{f}$

m v₀ = (m + M) v

v₀ = v (m + M) / m

let's calculate

v₀ = 27.83 (0.012 +0.109) /0.012

v₀ = 280.6 m / s

4 0

3 years ago

a block with length 1.5m width 1m height 0.5m and mass 300kg lays on the table.what is the pressure at the bottom surface of the

Nesterboy [21]

Answer:

your answer will be 320kg that would be the pressure at the bottom surface of the block

6 0

3 years ago

What is the minimum number of vectors lying in the same plane required to givs zero resulant

alexgriva [62]

Well the trivial answer is zero, since there is indeed a "zero vector". Assuming you aren't allowed to use the zero vector you would need at least two. They would be antiparallel and of equal magnitude. (That is be pointing in opposite directions and have the same length)

7 0

3 years ago

Read 2 more answers

. A vertical electric field is set up in space to compensate for the gravitational force on a point charge. What is the required

Delicious77 [7]

(a) The required magnitude of the electric field when the point charge is an electron is 5.57 x 10⁻¹¹ N/C.

(b) The required magnitude of the electric field when the point charge is an proton is 1.02 x 10⁻⁷ N/C.

<h3>Magnitude of electric field </h3>

The magnitude of electric field is given by the following equation.

F = qE

But F = mg

mg = qE

E = mg/q

where;

E is the electric field
m is mass of the particle
g is acceleration due to gravity
q is charge of the particle

<h3>For an electron</h3>

E = (9.11 x 10⁻³¹ x 9.8)/(1.602 x 10⁻¹⁹)

E = 5.57 x 10⁻¹¹ N/C

<h3>For proton</h3>

E = (1.67 x 10⁻²⁷ x 9.8)/(1.602 x 10⁻¹⁹)

E = 1.02 x 10⁻⁷ N/C

Thus, the required vertical electric field is greater when the charge is proton.

Learn more about electric field here: brainly.com/question/14372859

#SPJ1

4 0

2 years ago