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ycow [4]
4 years ago
11

Which statement describes nuclear binding energy?

Physics
2 answers:
Tamiku [17]4 years ago
3 0

The answer for Plato users is C. It is the energy that is required to overcome mass defect.

From Plato "Nuclear binding energy is the energy that is required to overcome the energy that is used to keep the nucleus together, the mass defect."

Scorpion4ik [409]4 years ago
3 0

Answer: Option (B) is the correct answer.

Explanation:

As nucleus of an atom contains protons and neutrons and due to the same charge present n protons there will occur repulsion.

Hence, a binding energy will act on the nucleus of an atom which will bind the protons and neutrons together. When this binding energy is greater than the force of repulsion between the sub-atomic particles of a nucleus then the atom is stable.

Therefore, to change the nucleus of an atom high amount of energy is required. As a result, nuclear reactions tend to release large amount of energy due to changes in the nucleus of an atom.

Thus, we can conclude that the statement it is the energy that is required to bind protons and neutrons together in a nucleus, best describes nuclear binding energy.

You might be interested in
A torque of 36.5 N · m is applied to an initially motionless wheel which rotates around a fixed axis. This torque is the result
vivado [14]

Answer:

21.6\ \text{kg m}^2

3.672\ \text{Nm}

54.66\ \text{revolutions}

Explanation:

\tau = Torque = 36.5 Nm

\omega_i = Initial angular velocity = 0

\omega_f = Final angular velocity = 10.3 rad/s

t = Time = 6.1 s

I = Moment of inertia

From the kinematic equations of linear motion we have

\omega_f=\omega_i+\alpha_1 t\\\Rightarrow \alpha_1=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha_1=\dfrac{10.3-0}{6.1}\\\Rightarrow \alpha_1=1.69\ \text{rad/s}^2

Torque is given by

\tau=I\alpha_1\\\Rightarrow I=\dfrac{\tau}{\alpha_1}\\\Rightarrow I=\dfrac{36.5}{1.69}\\\Rightarrow I=21.6\ \text{kg m}^2

The wheel's moment of inertia is 21.6\ \text{kg m}^2

t = 60.6 s

\omega_i = 10.3 rad/s

\omega_f = 0

\alpha_2=\dfrac{0-10.3}{60.6}\\\Rightarrow \alpha_1=-0.17\ \text{rad/s}^2

Frictional torque is given by

\tau_f=I\alpha_2\\\Rightarrow \tau_f=21.6\times -0.17\\\Rightarrow \tau=-3.672\ \text{Nm}

The magnitude of the torque caused by friction is 3.672\ \text{Nm}

Speeding up

\theta_1=0\times t+\dfrac{1}{2}\times 1.69\times 6.1^2\\\Rightarrow \theta_1=31.44\ \text{rad}

Slowing down

\theta_2=10.3\times 60.6+\dfrac{1}{2}\times (-0.17)\times 60.6^2\\\Rightarrow \theta_2=312.03\ \text{rad}

Total number of revolutions

\theta=\theta_1+\theta_2\\\Rightarrow \theta=31.44+312.03=343.47\ \text{rad}

\dfrac{343.47}{2\pi}=54.66\ \text{revolutions}

The total number of revolutions the wheel goes through is 54.66\ \text{revolutions}.

3 0
3 years ago
You are looking down on a N = 9 turn coil in a magnetic field B = 0.5 T which points directly down into the screen. If the diame
Novay_Z [31]

Answer:

The magnitude of the voltage is 2.27\times10^{-4}\ V and the direction of the current is clockwise.

Explanation:

Given that,

Number of turns = 9

Magnetic field = 0.5 T

Diameter = 3 cm

Time t = 0.14 s

We need to calculate the flux

Using formula of flux

\phi=NAB

Put the value into the formula

\phi=9\times\pi\times(1.5\times10^{-2})^2\times0.5

\phi=0.003180

We need to calculate the emf

Using formula of emf

\epsilon=-\dfrac{d\phi}{dt}

\epsilon=-\dfrac{0.003180}{0.14}

\epsilon =-0.000227\ V

\epsilon=-2.27\times10^{-4}\ V

Negative sign shows the direction of current.

Hence, The magnitude of the voltage is 2.27\times10^{-4}\ V and the direction of the current is clockwise.

8 0
3 years ago
What is the difference between charging by contact and charging by induction in terms of electron transfer.
Veronika [31]

Answer:

the main difference between charging by contact and charging by induction is that in the first case, the two objects are touching, while in the second case, the two objects do not touch

Explanation:

There are three methods of charging an object:

- Charging by friction: this is done by rubbing an object against another object. An example is when a plastic rod is rubbed with a wool cloth. When this is done, electrons are transferred from the wool to the rod, so both objects remain charged at the end of the process

- Charging by contact: this is done by putting in contact a charged object with a neutral, conducting object. In this case, the charges are transferred from the charged object to the neutral object; at the end of the process, the neutral object will also have a net electric charge, so it will be also charged.

- Charging by induction: in this case, we take a charged object, and a neutral object, and we place the two objects close to each other, but without touching. Let's assume that the charged object is negatively charged: in this case, the positive charges in the neutral object are attracted towards the negative charges of the charged object, while the negative charges of the neutral object are repelled away. As a result, the positive and negative charges in the neutral object split apart. If the object is connected to the ground, then negative charges move away, so the neutral object will remain positively charged.

Therefore, the main difference between charging by contact and charging by induction is that in the first case, the two objects are touching, while in the second case, the two objects do not touch.

5 0
3 years ago
The a992 steel rod bc has a diameter of 50 mm and is used as a strut to support the beam. determine the maximum intensity w of t
EastWind [94]

Answer:

w = 11.211 KN/m

Explanation:

Given:

diameter, d = 50 mm

F.S = 2

L = 3

Due to symmetry, we have:

Ay = By = \frac{w * 6}{2} = 3w

P_c_r = 3w * F.S = 3w * 2.0 = 6w

I = \frac{\pi}{64} (0.05)^4 = 3.067*10^-^7

To find the maximum intensity, w, let's take the Pcr formula, we have:

P_c_r = \frac{\pi^2 E I}{(KL)^2}

Let's take k = 1

E = 200*10^9

Substituting figures, we have:

6w = \frac{\pi^2 * 200*10^9 * 3.067*10^-^7}{(1 * 3)^2}

Solving for w, we have:

w = \frac{67266.84}{6}

w = 11211.14 N/m = 11.211 KN/m

Since Area, A= pi * (0.05)²

\sigma _c_r = \frac{w}{A}

\sigma _c_r = \frac{11.211}{\pi (0.05)^2} = 1.4 MPA < \sigma y. This means it is safe

The maximum intensity w = 11.211KN/m

3 0
3 years ago
A pilot claims to have seen a UFO moving initially at a speed of about 351 m/s in an easterly direction and then, in a time inte
serious [3.7K]

Answer:

Explanation:

This problem can be solved easily if we represent velocity in the form of vector.

The velocity of 351 was towards easterly direction so

V₁ = 351 i

The velocity of 351 was towards south west making - 48° with east or + ve x direction.

V₂ = 351 Cos 48 i - 351 sin 48 j

V₂ = 234.86 i - 260.84 j

Change in velocity

= V₂ - V₁ = 234.86 i - 260.84 j - 351 i

= -116.14 i - 260.84 j

acceleration

= change in velocity / time

(-116.14 i - 260.84 j )/ 1

= -116.14 i - 260.84 j

magnitude = 285.53 ms⁻²

Direction

Tan θ = 260.84 / 116.14 = 2.246

θ = 66 degree south of west .

3 0
3 years ago
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