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2 years ago

8

A low-temperature physicist wishes to publish his experimental result that the heat capacity of a nonmagnetic dielectric materia

l between 0.05K and 0.5K varies as
A√ T + BT^3

As editor of the journaL should you accept the paper for publication?

1 answer:

tatiyna2 years ago

6 0

Answer:

Yes, this because the specific heat does not violate the third law of thermodynamics.

Explanation:

The third law of thermodynamic is usually used for a closed system to relate the thermodynamic properties of the system at equilibrium conditions. For this law, a body that exists at a temperature of 0 K will cease to move or stop moving. It can be inferred that heat capacity at (T = 0 K) is equivalent to 0. Therefore, the equation is valid for the given temperature range.

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A 50 mm diameter shaft is subjected to a static axial load of 160 kN. If the yield stress of the material is 350 MPa, the ultima

zvonat [6]

In order to develop this problem it is necessary to take into account the concepts related to fatigue and compression effort and Goodman equation, i.e, an equation that can be used to quantify the interaction of mean and alternating stresses on the fatigue life of a materia.

With the given data we can proceed to calculate the compression stress:

$\sigma_c = \frac{P}{A}$

$\sigma_c = \frac{160*10^3}{\pi/4*0.05^2}$

$\sigma_c = 81.5MPa$

Through Goodman's equations the combined effort by fatigue and compression is expressed as:

$\frac{\sigma_a}{S_e}+\frac{\sigma_c}{\sigma_u}=\frac{1}{Fs}$

Where,

$\sigma_a=$ Fatigue limit for comined alternating and mean stress

$S_e =$ Fatigue Limit

$\sigma_c=$ Mean stress (due to static load)

$\sigma_u =$ Ultimate tensile stress

$Fs =$ Security Factor

We can replace the values and assume a security factor of 1, then

$\frac{\sigma_a}{320}+\frac{81.5}{400}=\frac{1}{1}$

Re-arrenge for $\sigma_a$

$\sigma_a = 254.8Mpa$

We know that the stress is representing as,

$\sigma_a = \frac{M_c}{I}$

Then,

Where $M_c$ =Max Moment

I= Intertia

The inertia for this object is

$I=\frac{\pi d^4}{64}$

Then replacing and re-arrenge for $M_c$

$M_c = \frac{\sigma_a*\pi*d^3}{32}$

$M_c = \frac{260.9*10^6*\pi*0.05^3}{32}$

$M_c = 3201.7N.m$

Thereforethe moment that can be applied to this shaft so that fatigue does not occur is 3.2kNm

5 0

2 years ago

Five batch jobs A through E arrive at a computer center in the order A to E at almost the same time. They have estimated running

Nikolay [14]

Answer:

Explanation:

The Turnaround time is the amount of time that elapses between the job arriving and completing. We assume that all jobs arrive at time 0, the turnaround time will simply be the time that they complete.

Round Robin:

we assume that the time quantum of the scheduler is 1 second.The table below gives a break down of which jobs will be processed during each time quantum. A asterisk(*) indicates that the job completes during that quantum.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

A B C D E A B C* D E A B D E A B D* E A B E A B* E A E A E* A A

C* = 8

D*=17

B*=23

E*=28

AVERAGE TURNAROUND = (8+17+23+28+30)/5 =106/5 = 21.2 MINUTES

B) PRIORITY SCHEDULING:

1-6 7-14 15-24 25-26 27-30

B E A C D

AVERAGETURNAROUND =(6+14+24+26+30)/5 = 100/5 = 20 MINUTES.

C)FCFS

1-10 11-16 17-18 19-22 23-30

A B C D E

AVERAGE TURNAROUND =(10+16+18+22+30)/5 = 96/5=19.2 MINTUES

D)SJF

1-2 3-6 7-12 13-20 21-30

C D B E A

AVERAGE TURNAROUND - (2+6+12+20+30)/5 =70/5 =14 MINUTES.

3 0

2 years ago

An npn BJT has emitter, base, and collector doping levels of 1019 cm????3, 5 1018 cm????3, and 1017 cm????3, respectively. It is

Darina [25.2K]

Answer:

Explanation:

The answer to the given problem is been solved in the fine attached below.

8 0

2 years ago

(SI units) Molten metal is poured into the pouring cup of a sand mold at a steady rate of 400 cm3/s. The molten metal overflows

maxonik [38]

Answer:

diameter of the sprue at the bottom is 1.603 cm

Explanation:

Given data;

Flow rate, Q = 400 cm³/s

cross section of sprue: Round

Diameter of sprue at the top $d_{top}$ = 3.4 cm

Height of sprue, h = 20 cm = 0.2 m

acceleration due to gravity g = 9.81 m/s²

Calculate the velocity at the sprue base

$V_{base}$ = √2gh

we substitute

$V_{base}$ = √(2 × 9.81 m/s² × 0.2 m )

$V_{base}$ = 1.98091 m/s

$V_{base}$ = 198.091 cm/s

diameter of the sprue at the bottom will be;

Q = AV = (π $d_{bottom}^2$ /4) × $V_{base}$

$d_{bottom}$ = √(4Q/π $V_{base}$ )

we substitute our values into the equation;

$d_{bottom}$ = √(4(400 cm³/s) / (π×198.091 cm/s))

$d_{bottom}$ = 1.603 cm

Therefore, diameter of the sprue at the bottom is 1.603 cm

6 0

2 years ago

Exercise 5.Water flows in a vertical pipe of 0.15-m diameter at a rate of 0.2 m3/s and a pressureof 275 kPa at an elevation of 2

wariber [46]

Answer:

a. Pressure head: 33.03,

Velocity Head: 6.53

b. Pressure Head: -1.97,

Velocity Head: 6.53

Explanation:

a.

Given

Diameter = 0.15-m, radius = 0.075

rate = 0.2 m3/s

Pressure =275 kPa

elevation =25 m.

We'll consider 3 points as the water flow through the pipe

1. At the entrance

2. Inside the pipe

3. At the exit

At (1), the velocity can be found using continuity equation.

V1 = ∆V/A

Where A = Area = πr² = π(0.075)² = 0.017678571428571m²

V1 = 0.2/0.017678571428571

V1 = 11.32 m/s

The value of pressure at point 1, is given by Bernoulli equation between point 1 and 2:

P1/yH20 + V1²/2g + z1 = P2/yH20 + V2²/2g + z2

Substitute in the values

P1/yH20 + 20 = (275 * 10³Pa)/yH20 + 25

P1/yH20 = (275 * 10³Pa)/yH20 + 25 - 25

=> P1/yH20 = (275/9.81 + 5)

P1/yH20 = 33.03

The velocity head at point one is then given by

V2²/2g = 11.32²/2 * 9.8

V2²/2g = 6.53

b.

The value of pressure at point 1, is given by Bernoulli equation between point 1 and 3:

P1/yH20 + V1²/2g + z1 = P3/yH20 + V3²/2g + z3

Substitute in the values

33.03 + 20 = P3/yH20 + 55

P3/yH20 = 33.03 + 20 - 55

=> P1/yH20 = -1.97

The velocity head at point three is then given by

V2²/2g = V3²/2g = 6.53

4 0

2 years ago

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