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2 years ago

Why do we not consider the strong nuclear force when we describe the forces between atoms?

Physics

1 answer:

solong [7]2 years ago

3 0

This is because, nuclear forces are short range forces.

The protons and neutrons are collectively called nucleons and they are found in the nucleus of an atom. These nucleons are held together by a strong attractive force called the nuclear force.

When atoms combine, they are held together by forces determined by electrons in the outermost shell. The influence of the nuclear force does not extend beyond the nucleus because they are short range forces.

Hence, we not consider the strong nuclear force when we describe the forces between atoms.

Learn more:brainly.com/question/3964366

You might be interested in

A pure sound wave, generated by a tuning fork, is considered a periodic wave. Which statement is true for this tuning fork sound

soldier1979 [14.2K]

Answer : Every wave has the same wave pattern.

Explanation : A pure sound wave has a single frequency and its generated by a tuning fork . The sound wave is a simple periodic wave.

When an object such as a guitar push and pull on the around it. When it pushes on the air, then the pressure increases and when it pull on the air, then the pressure decreases and sound waves are formed.

Hence, sound wave has the same wave pattern.

4 0

3 years ago

Read 2 more answers

Derive the equation of motion of the block of mass m1 in terms of its displacement x. The friction between the block and the sur

Alenkasestr [34]

Answer:

the equivalent mass : $m_e = m_1+m_2+\frac{I}{R^2}$

the equation of the motion of the block of mass $m_1$ in terms of its displacement is = $(m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)$

Explanation:

Let use m₁ to represent the mass of the block and m₂ to represent the mass of the cylinder

The radius of the cylinder be = R

The distance between the center of the pulley to center of the block to be = x

Also, the angles of inclinations of the cylinder and the block with respect to the ground to be $\phi$ and $\beta$ respectively.

The velocity of the block to be = v

The equivalent mass of the system = $m_e$

In the terms of the equivalent mass, the kinetic energy of the system can be written as:

$K.E = \frac{1}{2} m_ev^2$ --------------- equation (1)

The angular velocity of the cylinder = $\omega$ : &

The inertia of the cylinder about its center to be = I

The angular velocity of the cylinder can be written as:

$v = \omega R$

$\omega =\frac{v}{R}$

The kinetic energy of the system in terms of individual mass can be written as:

$K.E = \frac{1}{2}m_1v^2+\frac{1}{2} m_2v^2+\frac{1}{2}I\omega^2$

By replacing $\omega$ with $\frac{v}{R}$ ; we have:

$K.E = \frac{1}{2}m_1v^2+\frac{1}{2} m_2v^2+\frac{1}{2}I(\frac{v}{R})^2$

$K.E = \frac{1}{2}(m_1+ m_2+ \frac{I}{R} )v^2$ ------------------ equation (2)

Equating both equation (1) and (2); we have:

$m_e = m_1+m_2+\frac{I}{R^2}$

Therefore, the equivalent mass : $m_e = m_1+m_2+\frac{I}{R^2}$ which is read as;

The equivalent mass is equal to the mass of the block plus the mass of the cylinder plus the inertia by the square of the radius.

The expression for the force acting on equivalent mass due to the block is as follows:

$f_{block }=m_1gsin \beta$

Also; The expression for the force acting on equivalent mass due to the cylinder is as follows:

$f_{cylinder} = m_2gsin \phi$

Equating the above both equations; we have the equation of motion of the equivalent system to be

$m_e \bar x = f_{cylinder}-f_{block}$

which can be written as follows from the previous derivations

$(m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)$

Finally; the equation of the motion of the block of mass $m_1$ in terms of its displacement is = $(m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)$

8 0

3 years ago

Explain how you think an asteroid impact could affect the tilt of Earth’s axis. Explain how this effect would change Earth’s sea

a_sh-v [17]

Answer:

An asteroid impact could affect the tilt of the Earth due to the force it applies onto the planet. This would change Earth's seasons due to the fact that Earth's tilt causes seasons.

4 0

3 years ago

Water enters the constant 130-mm inside-diameter tubes of a boiler at 7 MPa and 65°C and leaves the tubes at 6 MPa and 450°C wit

snow_lady [41]

The inlet velocity is 1.4 m/s and inlet volume is 0.019 m³/s.

Explanation:

When water entering the tube of constant diameter flows through the tube, it exhibits continuity of mass in the hydrostatics. So the mass of water moving from the inlet to the outlet tend to be same, but the velocity may differ.

As per mass flow equality which states that the rate of flow of mass in the inlet is equal to the product of area of the tube with the velocity of the water and the density of the tube.

Since, the inlet volume flow is measured as the product of velocity with the area.

Inlet volume flow=Inlet velocity*Area*time

And the mass flow rate is

Mass flow rate in the inlet=density*area*inlet velocity*time

Mass flow rate in the outlet=density*area*outlet velocity*time

Since, the time and area is constant, the inlet and outlet will be same as

(Mass inlet)/(density*inlet velocity)=Area*Time

(Mass outlet)/(density*outlet velocity)=Area*Time

As the ratio of mass to density is termed as specific volume, then

(Specific volume inlet)/(Inlet velocity)=(Specific volume outlet)/(Outlet velocity)

Inlet velocity= (Specific volume inlet)/(Specific volume outlet)*Outlet velocity

As, the specific volume of water at inlet is 0.001017 m³/kg and at outlet is 0.05217 m³/kg and the outlet velocity is given as 72 m/s, the inlet velocity

$Inlet velocity = \frac{0.001017}{0.05217}*72 =1.4035 m/s$

So, the inlet velocity is 1.4035 m/s.

Then the inlet volume will be

$Inlet volume = inlet velocity*area of circle=\pi r^{2}*inlet velocity$

As the diameter of tube is 130 mm, then the radius is 65 mm and inlet velocity is 1.4 m/s

$Inlet volume = 1.4*3.14*65*65*10^{-6} =0.019 \frac{m^{3} }{s}$

So, the inlet volume is 0.019 m³/s.

Thus, the inlet velocity is 1.4 m/s and inlet volume is 0.019 m³/s.

4 0

3 years ago

The total blood volume of a human is 75 ml/kg. If the volumetric flow rate is the same in adults and children, how much more blo

Tema [17]

The amount of blood that flows through the venae cavea of the adult is 3750 ml more than that of the child.

This is the volume of fluid flowing in a vessel per unit time.

First, we need to get the volume of blood flowing through the venae cavae of the adult and the child in 1 hour, then subtract the volume of blood for the child from that of the adult.

Using,

V = F'm................ Equation 1

Where:

V = Volume of the blood
F' = Flow rate
m = mass of blood.

For the Adult,

Given:

m = 70 kg
F = 75 ml/kg

Substitute these values into equation 1

V = 70(75)
V = 5250 ml.

For the child,

Given:

m = 20 kg
F = 75 ml/kg

Substitute these values into equation 1 also

V = 75(20)
V = 1500 ml.

The amount of blood that flows more through the adult than the child is

A = 5250-1500
A = 3750 ml.

Hence, the amount of blood that flows through the venae cavea of the adult is 3750 ml more than that of the child.

Learn more about flow rate here: brainly.com/question/21630019

5 0

2 years ago