What is one of the most common ways in which workers get hurt around machines?

How high of a column of sae 30 oil would be required to give the same pressure as 700 mm hg?

Rasek [7]

Hoiu-10,4000 mm.

<h3>Is positive pressure good for PC?</h3>

A balanced configuration is the most efficient way to cool your pc although it should tend towards a slight positive pressure if you can help it.
Tip: As much as it might seem important, the concept of heat rising doesn't have too much of an effect.

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7 0

1 year ago

For a steel alloy it has been determined that a carburizing heat treatment of 15 h duration will raise the carbon concentration

Free_Kalibri [48]

Answer:

135 hour

Explanation:

It is given that a carburizing heat treatment of 15 hour will raise the carbon concentration by 0.35 wt% at a point of 2 mm from the surface.

We have to find the time necessary to achieve the same concentration at a 6 mm position.

we know that $\frac{x_1^2}{Dt}=constant$ where x is distance and t is time .As the temperature is constant so D will be also constant

So $\frac{x_1^2}{t}=constant$

then $\frac{x_1^2}{t_1}=\frac{x_2^2}{t_2}$ we have given $x_1=2 mm\ ,t_1=15 hour\ ,x_2=6\ mm$ and we have to find $t_2$ putting all these value in equation

$\frac{2^2}{15}=\frac{6^2}{t_2}$

so $t_2=135\ hour$

5 0

3 years ago

What is the area enclosed by the cycle area of the Carnot cycle illustrating on a P-V diagram?

Inga [223]

Answer:

The work of the cycle.

Explanation:

The area enclosed by the cycle of the Pressure-Volume diagram of a Carnot engine represents the net work performed by the cycle.

The expansions yield work, and this is represented by the area under the curve all the way to the p=0 line. But the compressions consume work (or add negative work) and this is substracted fro the total work. Therefore the areas under the compressions are eliminated and you are left with only the enclosed area.

7 0

3 years ago

An airline ticket counter forecasts that 220 people per hour will need to check in. It takes an average of 2 minutes to service

Nadusha1986 [10]

A) Number of agents required to achieve a wait time of 10 minutes or less = 8 agents

B) The number of agents required on duty to reduce cost = 9 agents

Given data :

Arrival rate of customers ( β ) = 220 per hour

Service rate ( mu ) = 60 minutes / 2 minutes = 30 customer per hour

utilization ( rho ) = 220 / 30 ≈ 7

at least 8 server personnel are required for stability of the queue

A) Determine the number of agents required to achieve a wait time of 10 minutes or less per customer

waiting time = 10 - 2 = 8 minutes

number of customers waiting ( ∝ ) = 7 and required server = 8

assuming Lq = 5.2266

Hence the waiting time in line = Lq / arrival rate

= 5.2266 / 220 = 0.0238 hour

= 0.0238 * 60 = 1.428 minutes

Since the waiting time ( 1.428 minutes ) is less than the original waiting time ( 2 minutes ) the number of agents that will achieve a wait time of 10 minutes or less is = 8 agents

B) Determine the number of ticket agents that should be on duty to minimize cost

salary of ticket agent = £12 per hour

cost of customer waiting in queue = £5 per hour per customer

i) When 8 agents are used

waiting time of customers = 0.0238 * 220 = 5.236

waiting cost for customers = 5.236 * 5 = £26.18

employee cost = 8 * 12 = £96

∴ Total cost = 96 + 26.18

= £ 122.18

ii) When 9 agents are used

waiting time for customers = 0.0074 * 220 = 1.628

Wq = 1.6367 / 220 = 0.0074

waiting cost for customers = 1.6367 * 5 = £ 8.1835

assuming Lq = 1.6367

employee cost = 9 * 12 = £ 108

∴ Total cost = 108 + 8.1835 = £ 116.18

From the calculations in ( i ) and ( ii ) the Ideal number of ticket agents that should be on duty to minimize cost should be 9 agents.

Hence we can conclude that A) Number of agents required to achieve a wait time of 10 minutes or less = 8 agents and The number of agents required on duty to reduce cost = 9 agents.

Learn more about cost reduction : brainly.com/question/14115944

6 0

3 years ago

The mechanical properties of a metal may be improved by incorporating fine particles of its oxide. Given that the moduli of elas

mojhsa [17]

Answer:

A) Upper bound modulus of elasticity; E = 165.6 GPa

B) Lower bound modulus of elasticity; E = 83.09 GPa

Explanation:

A) Formula for upper bound modulus is given as;

E = E_m(1 - V_f) + E_f•V_f

We are given;

E_m = 60 GPa

E_f = 380 GPa

V_f = 33% = 0.33

Thus,

E = 60(1 - 0.33) + 380(0.33)

E = (60 x 0.67) + 125.4

E = 165.6 GPa

B) Formula for lower bound modulus is given as;

E = 1/[(V_f/E_f) + ((1 – V_f)/E_m)]

E = 1/[(0.33/380) + ((1 – 0.33)/60)]

E = 1/(0.0008684 + 0.01116667)

E = 1/0.01203507

E = 83.09 GPa

3 0

3 years ago