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4 years ago

13

How does the strong nuclear force hold the nucleus of an atom together?

2 answers:

son4ous [18]4 years ago

6 0

Answer: Option (b) is the correct answer.

Explanation:

An atom consists of three sub-atomic particles that is, protons, neutrons and electrons.

Protons are positively charged, neutrons have no charge and electrons are negatively charged.

Both protons and neutrons are present inside the nucleus and electrons revolve around the nucleus. Due to like charges of protons there occur electrostatic forces of repulsion which are overcome by a strong nuclear force that is known as binding energy.

Hence, we can conclude that the strong nuclear force hold the nucleus of an atom together as it overcomes the electrical force of repulsion between protons in the nucleus.

erastovalidia [21]4 years ago

3 0

The strong nuclear force holds the nucleus of an atom together.

Somehow, it overcomes the electrical force of repulsion between protons in the nucleus, which all have the same charge but still stay close together somehow. (b)

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A barbel weighing 215N is raised to a height of 2.0 M above the ground. It is then dropped from that height.

aleksandrvk [35]

Answer:

430 J
6.26 m/s

Explanation:

A. The kinetic energy is the same as the initial potential energy:

PE = mgh = (215 N)(2.0 M) = 430 J

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B. The velocity achieved by falling from a height h is given by ...

v = √(2gh)

v = √(2·9.8 m/s^2·2 m) = √(39.2 m^2/s^2)

v ≈ 6.26 m/s

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3 years ago

A block of mass m=1kg sliding along a rough horizontal surface is traveling at a speed v0=2m/s when it strikes a massless spring

topjm [15]

Here we can use the work energy theorem

$W_f + W_s = K_f - K_i$

here we know that

$K_f = 0$

as it come to rest finally

$K_i = \frac{1}{2}mv_i^2$

$K_i = \frac{1}{2}\times 1\times 2^2$

$K_i = 2 J$

now work done by friction force will be given as

$W_f = - F_f \times d = -\mu mg d$

$W_f = - \mu(1)(9.8)(0.10) = - 0.98\mu$

Work done by spring force is given as

$W_s = \frac{1}{2}k(x_i^2 - x_f^2)$

$W_s = \frac{1}{2}(10)( 0 - 0.10^2)$

$W_s = -0.05 J$

so now plug in all data above

$- 0.05 - \mu(0.98) = 0 - 2$

$\mu = 1.99$

so above is the friction coefficient

4 0

4 years ago

6.6 kg block initially at rest is pulled to theright along a horizontal, frictionless surfaceby a constant, horizontal force of

neonofarm [45]

Answer:

v = 3.04 m/s

Explanation:

given,

mass of the block, M = 6.6 Kg

horizontal force, F = 12.2 N

distance, L = 2.5 m

initial speed = 0 m/s

speed of the block,v = ?

we now

Work done is equal to change in Kinetic energy.

Work done = Force x displacement

W = Δ K E

Δ K E = Force x displacement

$\dfrac{1}{2}mv^2 - \dfrac{1}{2}mu^2= F .s$

$\dfrac{1}{2}\times 6.6 \times v^2 - 0= 12.2\times 2.5$

3.3 v² = 30.5

v² = 9.242

v = 3.04 m/s

speed of the block is equal to 3.04 m/s

4 0

3 years ago

What was the main aim behind Wegener's continental drift theory?

Aleks [24]

D.all of the above is the answer for this question

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Vaselesa [24]

Answer:

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3 years ago