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3 years ago

What are the similarities between strong nuclear force and weak nuclear force

1 answer:

7 0

Alike because they both act on the quarks making up the nucleons and they have very short ranges. The Strong Nuclear Force is an attractive force between protons and neutrons that keep the nucleus together and the Weak Nuclear Force is responsible for the radioactive decay of certain nuclei. Which also makes them very different

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Compute the density in g/cm? of a piece of metal that has a mass of 0.450 kg and a volume of 52 cm3

Y_Kistochka [10]

Answer:

Ro = 8.65 [g/cm³]

Explanation:

We must remember that density is defined as the ratio of mass to volume.

$Ro=m/V$

where:

m = mass = 0.450 [kg] = 450 [g]

V = volumen = 52 [cm³]

Ro = density [g/cm³]

Now replacing:

$Ro = 450/52\\Ro = 8.65 [g/cm^{3} ]$

8 0

3 years ago

What do all elements have in common?

Sindrei [870]

The common<span> feature is that the atoms of </span>all elements<span> consist of electrons, protons, and neutrons. Hope this helps!</span>

7 0

3 years ago

If a 100-N net force acts on a 50-kg car, what will the acceleration of the car be?

Tema [17]

Newton's 2nd law of motion:

   Force = (mass) x (acceleration)

Divide each side by (mass):

   Acceleration = (force) / (mass)

   = (100 N) / (50 kg)

   = 2 m/s²

5 0

3 years ago

Two objects of the same size are both perfect blackbodies. One has a temperature of 3000 K, so its frequency of maximum emission

bija089 [108]

Answer:

a) The colder body (3000k), b) hearter body c) 12000K body

Explanation:

This exercise should know the power emitted by the objects and the distribution of this emission in the energy spectrum, for this we will use Stefan's laws and that of Wien's displacement

Stefan's Law P = σ A e T⁴

Wien displacement law λ T = 2,898 10⁻³ m K

Let's calculate the power emitted for each object.

As they are perfect black bodies e = 1, they also indicate that they have the same area

T = 3000K

P₁ = σ A T₁⁴

T = 12000K

P₂ = σ A T₂⁴

P₂ / P₁ = T₂⁴ / T₁⁴

P₂ / P₁ = (12000/3000)⁴

P₂ / P₁ = 256

This indicates that the hottest body emission is 256 times the coldest body emission.

Let's calculate the maximum emission wavelength

Body 1

T = 3000K

λ T = 2,898 10-3

λ₁ = 2.89810-3 / T

λ₁ = 2,898 10-3 / 3000

λ₁ = 0.966 10-6 m

λ₁ = 966 nm

T = 12000K

λ₂ = 2,898 10-3 / 12000

λ₂ = 0.2415 10-6 m

λ₂ = 214 nm

a) The colder body (3000k) emits more light in the infrared, since the emission of the hot body is at a minimum (emission tail)

b) The two bodies have emission in this region, the body of 3000K in the part of rise of the emission and the body to 12000K in the descent of the emission even when this body emits 256 times more than the other, so this body should have the highest broadcast in this area

c) The emission of the hottest 12000K body is mainly in UV

d) The hottest body emits more energy in UV and visible

e) No body has greater emission in all zones

5 0

3 years ago

A striped billiard ball moves toward the right with speed 3 m/s. A solid billiard ball with the same mass moves toward the left

Helen [10]

Answer:

Final speed of striped ball is 3 m/s in left direction .

Explanation:

Given :

Two billiard ball with the same mass moves toward the left at the same speed 3 m/s .

Let , us assume right hand side direction to be positive and left hand side direction to be negative .

Also , let speed of ball after collision is (striped ball ) u and (solid ball) v .

It is also given that the collision is elastic .

Therefore , kinetic energy is conserved .

$\dfrac{m(3)^2}{2}+\dfrac{m(3)^2}{2}=\dfrac{mu^2}{2}+\dfrac{mv^2}{2}\\\\u^2+v^2=18$ ...... ( 1 )

Also , by conserving linear momentum .

We get :

$3m-3m=mu+mv\\u=-v$ ...... ( 2 )

Putting value of u from equation 2 to equation 1 .

We get :

$2v^2=18\\v=3\ m/s$

And , u = -3 m/s .

Therefore , final speed of striped ball is 3 m/s in left direction .

Hence , this is the required solution .

4 0

4 years ago