PHYSICS COLLEGE

2H is a loosely bound isotope of hydrogen, called deuterium or heavy hydrogen. It is stable but relatively rare — it form only 0.015% of natural hydrogen. Note that deuterium has Z = N, which should tend to make it more tightly bound, but both are odd numbers.Required:
Calculate BE/A, the binding energy per nucleon, for 2H in megaelecton volts per nucleon

Answers

Answer 1

Answer:

Answer:

0.88 MeV/nucleon

Explanation:

The binding energy (B) per nucleon of deuterium can be calculated using the following equation:

$B = (Zm_(p) + Nm_(n) - M)/(A)*931.49 MeV/u$

Where:

Z: is the number of protons = 1

N: is the number of neutrons = 1

$m_(p)$ : is the proton's mass = 1.00730 u

$m_(n)$ : is the neutron's mass = 1.00869 u

M: is the nucleu's mass = 2.01410

A = Z + N = 1 + 1 = 2

Now, the binding energy per nucleon for ²H is:

$B = (Zm_(p) + Nm_(n) - M)/(A)*931.49 MeV/u = (1*1.00730 + 1*1.00869 - 2.01410)/(2)*931.49 MeV/u = 9.45 \cdot 10^(-4) u*931.49 MeV/u = 0.88 MeV/nucleon$

Therefore, the binding energy per nucleon for ²H is 0.88 MeV/nucleon.

I hope it helps you!

Answer 2

Answer:

Final answer:

The binding energy per nucleon for 2H (deuterium) is 1.1125 MeV per nucleon.

Explanation:

The binding energy per nucleon, or BE/A, can be calculated by dividing the total binding energy of the nucleus by the number of nucleons. To calculate the BE/A for 2H (deuterium), we need to know the total binding energy and the number of nucleons in deuterium. The total binding energy of deuterium is approximately 2.225 MeV (megaelectron volts) and the number of nucleons is 2. Therefore, the BE/A for 2H is 2.225 MeV / 2 = 1.1125 MeV per nucleon.

Learn more about Binding energy per nucleon here:

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Of the four most important pathways by which stress affects health, the first one to occuris usually related to physiology

Answers

Answer: Physiologic response to fear is very similar to that of PTSD and stress. Fear is accompanied by increased heart rate due to the release of adrenaline, sympathetic nervous system is aroused. The release of adrenaline also causes increased sweating, pulse and blood pressure. In line with this, the parasympathetic nervous system experiences reduced activity such as decrease in digestion.

The surface pressure of the atmosphere is about 14.7 psi (pounds per square inch). How many pounds per square yard does that amount to

Answers

Answer:

14.7 psi is equal to 19051.2 pounds per square yard.

Explanation:

Dimensionally speaking, a square yard equals 1296 square inches. Therefore, we need to multiply the atmospheric pressure by 1296 to obtain its equivalent in pounds per square yard. That is:

$p = 14.7\,(lbf)/(in^(2))* 1296\,(in^(2))/(yd^(2))$

$p = 19051.2\,(lbf)/(yd^(2))$

14.7 psi is equal to 19051.2 pounds per square yard.

A electromagnetic wave of light has a wavelength of 500 nm. What is the energy (in Joules) of the photon representing the particle interpretation of this light?

Answers

Answer:

Energy, $E=4.002* 10^(-19)\ J$

Explanation:

It is given that,

Wavelength of the photon, $\lambda=500\ nm=5* 10^(-7)\ m$

We need to find the photon representing the particle interpretation of this light. it is given by :

$E=(hc)/(\lambda)$

$E=(6.67* 10^(-34)* 3* 10^8)/(5* 10^(-7))$

$E=4.002* 10^(-19)\ J$

So, the energy of the photon is $4.002* 10^(-19)\ J$ . Hence, this is the required solution.

1. The resistance of an electric device is 40,000 microhms. Convert that measurement to ohms2. When an electric soldering iron is used in a 110 V circuit, the current flowing through the iron is
2 A. What is the resistance of the iron?
3. A current of 0.2 A flows through an electric bell having a resistance of 65 ohms. What must be
the applied voltage in the circuit?

Answers

Answer:

(1) 0.04 ohms (2) 55 ohms (3) 13 volt

Explanation:

(1) The resistance of an electric device is 40,000 microhms.

We need to convert it into ohms.

$1\ \mu \Omega =10^(-6)\ \Omega$

To covert 40,000 microhms to ohms, multiply 40,000 and 10⁻⁶ as follows :

$40000 \ \mu \Omega =40000 * 10^(-6)\ \Omega\n\n=0.04\ \Omega$

(2) Voltage used, V = 110 V

Current, I = 2 A

We need to find the resistance of the iron. Using Ohms law to find it as follows :

V = IR, where R is resistance

$R=(V)/(I)\n\nR=(110)/(2)\n\nR=55\ \Omega$

(3) Current, I = 0.2 A

Resistance, R = 65 ohms

We need to find the applied voltage in the circuit. Using Ohms law to find it as follows :

V=IR

V = 0.2 × 65

V = 13 volt

Answer:

1. 0.04 Ohms

2. 55 Ohms

3. 13 Volts

Explanation:

Penn Foster

A solid 0.6950 kg ball rolls without slipping down a track toward a vertical loop of radius ????=0.8950 m . What minimum translational speed ????min must the ball have when it is a height H=1.377 m above the bottom of the loop in order to complete the loop without falling off the track? Assume that the radius of the ball itself is much smaller than the loop radius ???? . Use ????=9.810 m/s2 for the acceleration due to gravity.

Answers

Answer:

The minimum transnational speed is 4.10 m/s.

Explanation:

Given that,

Mass of solid ball = 0.6950 kg

Radius = 0.8950 m

Height = 1.377 m

We need to calculate the minimum velocity of the ball at bottom of the loop to complete the track

Using formula velocity at lower point

$v_(min)=√(5gR)$

Put the value into the formula

$v_(min)=√(5*9.8*0.8950)$

$v_(min)=6.62\ m/s$

We need to calculate the velocity

Using conservation of energy

P.E at height +K.E at height = K.E at the bottom

$mgH+(1)/(2)mv^2=(1)/(2)m(√(5gR))^2$

$v^2=(√(5gR))^2-2gH$

$v^2=(6.62)^2-2*9.8*1.377$

$v^2=16.8352$

$v=√(16.8352)$

$v=4.10\ m/s$

Hence, The minimum transnational speed is 4.10 m/s.

Final answer:

The minimum translational speed the solid ball must have when it is at a height H=1.377 m above the bottom of the loop to successfully complete the loop without falling off the track is approximately 7.672 m/s. This was derived using principles of energy conservation.

Explanation:

The minimum translational speed must be sufficient enough to maintain contact with the track even at the highest point of the loop. Using the principle of energy conservation, the total energy at the height H, assuming potential energy to be zero here, should be equal to the total energy at the highest point of the loop. Here, the total energy at height H will consist of both kinetic and potential energy while at the top of the loop it consists of potential energy only. Setting these equations equal to each other: 0.5 * m * v² + m * g * H = m * g * 2R Solving the above equation for v:v = √2g (2R-H). Substituting known values henceforth gives us √2*9.81*(2*0.895-1.377) = 7.672 m/s. Hence, the ball must have a minimum translational speed of approximately 7.672 m/s at height H to complete the loop without falling.