What does abbreviation vom stand for

Refer to Figure 9-18. A #_____ electrode lead and workpiece lead should be used to carry 150 amperes of electricity 75 feet to t

Anni [7]

Answer:

AC or

Explanation:

BC

8 0

2 years ago

Help I need to know if it’s true or false

e-lub [12.9K]

Answer: False

explanation: for a bloodborne pathogen to spread you would have to have an open wound as well as the blood would have to get in it.

3 0

3 years ago

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

4 years ago

A 2-m3 rigid tank initially contains air at 100 kPa and 22°C. The tank is connected to a supply line through a valve. Air is flo

Finger [1]

Answer:

9.58 Kg of air has entered the tank.

heat entered=3483.76 Kilo.Joule

Explanation:

(A) R=287 Kilo.J/Kg.K

as per initial conditions P=100 Kilo.Pa ,V=2 cubic meter, T=22 C=295.15 K,

using the relation P*V=m*R*T

m=(100*1000*2)/(287*295.15)=2.36 Kg this is the mass that is already present in tank.

after filling tank at 600 Kilo.Pa.

P=600 Kilo Pa T=77 C=350.15 K

P*V=m*R*T

m=(600*1000*2)/(287*350.15)=11.94 Kg

mass that has entered=11.94-2.36=9.58 Kg

(b) using air psychometric property table

specific heat content initial 100 KILO Pa and 22 C=295.576 Kilo.Joule/Kg

specific heat content final 600 Kilo Pa and 77 C=350.194 Kilo.Joule/Kg

heat at initial stage=295.576*2.36=697.56 Kilo.Joule

heat at final stage=350.194*11.94=4181.32 Kilo.Joule

heat entered=4181.32-697.56=3483.76 Kilo.Joule

3 0

3 years ago

A common procedure for measuring the velocity of an air stream involves insertion of an electrically heated wire (called a hot-w

timurjin [86]

Answer:

V = 6.33 m/s

Explanation:

Given:

- The length of the wire L = 0.02 m

- The diameter of the wire D = 0.0005 m

- The calibration expression V = 0.0000625*h^2

- Environment temperature T_inf = 298 K

- Surface temperature T_s = 348 K

- The voltage drop dV = 5 V

- The electric current I = 0.1 A

Find:

- the velocity of Air

Solution:

- Calculate the surface area of the wire:

A = pi*D*L

A = pi*(0.0005)*(0.02) = 0.00003142 m^2

- The rate of energy in the wire P:

P = I*dV = 0.1*5 = 0.5 W

- Apply Newton's Law of Cooling:

P = h*A*(T_s - T_inf)

h = P /A*(T_s - T_inf)

Plug in the values:

h= 0.5/ 0.00003142*(348 - 298)

h = 318.27 W /m^2K

- Using the calibration relationship given, compute the velocity of air:

V = 6.25*10^-5 * h^2

V = 6.25*10^-5 * (318.27)^2

V = 6.33 m/s

5 0

3 years ago