______________ help protect the lower legs and feet from heat hazards like molten metal and welding sparks.

Air is contained in a vertical piston–cylinder assembly such that the piston is in static equilibrium. The atmosphere exerts a p

oee [108]

Answer:

a) 24 kg

b) 32 kg

Explanation:

The gauge pressure is of the gas is equal to the weight of the piston divided by its area:

p = P / A

p = m * g / (π/4 * d^2)

Rearranging

p * (π/4 * d^2) = m * g

m = p * (π/4 * d^2) / g

m = 1200 * (π/4 * 0.5^2) / 9.81 = 24 kg

After the weight is added the gauge pressure is 2.8kPa

The mass of piston plus addded weight is

m2 = 2800 * (π/4 * 0.5^2) / 9.81 = 56 kg

56 - 24 = 32 kg

The mass of the added weight is 32 kg.

5 0

3 years ago

A biotechnology company produced 225 doses of somatropin, including 11 which were defective. Quality control test 15 samples at

Radda [10]

Answer:

0.59

Explanation:

The batch is rejected if any of the random samples are found defective, or, what is the same, it will be accepted only if all 15 samples are good.

The probability that none be defective is the same probability that all the samples are good. Thus, start to calculate the probability that the batch is accepted.

The probability that the first sample is good is 214 /225, because there are 225 - 11 = 214 good samples in 225 doses.

The probability that the second samples is good too is 213/224, because there is 1 less good sample, in the 224 remaining samples.

By the same process, you conclude that the consecutive probabilities of selecting a good sample are: 212/223, 211/222, 210/221, . . . up to 199/211.

The joint probability of all the samples are good is the product of each probability:

$\frac{214}{225}\cdot\frac{213}{224}\cdot\frac{212}{223}\cdot\frac{211}{222}\cdot\frac{210}{221}\cdot\frac{209}{220}\cdot\frac{208}{219}\cdot\frac{207}{218}\cdot\frac{206}{217}\cdot\frac{205}{216}\cdot\frac{204}{215}\cdot\frac{203}{214}\cdot\frac{202}{213}\cdot\frac{201}{212}\cdot\frac{200}{211}\cdot\frac{199}{210}$

The result is: 0.41278 ≈ 0.41

The conclusion is that the probability that all the samples are good and the batch is accepted is 0.41.

Therefore, the probability that the batch is rejected is 1 - 0.41 = 0.59.

4 0

3 years ago

.a. What size vessel holds 2 kg water at 80°C such that 70% is vapor? What are the pressure and internal energy? b. A 1.6 m3 ves

vesna_86 [32]

Answer:

Part a: The volume of vessel is 4.7680 $m^3$ and total internal energy is 3680 kJ.

Part b: The quality of the mixture is 90.3% or 0.903, temperature is 120 °C and total internal energy is 4660 kJ.

Explanation:

Part a:

As per given data

m=2 kg

T=80 °C =80+273=353 K

Dryness=70% vapour =0.7

From the steam tables at 80 °C

Specific volume of saturated vapours=v_g=3.40527 $m^3/kg$

Specific volume of saturated liquid=v_f=0.00102 $m^3/kg$

Now the relation of total specific volume for a specific dryness value is given as

$v=v_f+x(v_g-v_f)$

Substituting the values give

$v=v_f+x(v_g-v_f)\\v=0.00102+0.7(3.40527-0.00102)\\v_f=2.38399 m^3/kg$

Now the volume of vessel is given as

$v=\frac{V}{m}\\V=v \times m\\V=2.38399 \times 2\\V=4.7680 m^3$

So the volume of vessel is 4.7680 $m^3$ .

Similarly for T=80 and dryness ratio of 0.7 from the table of steam

Pressure=P=47.4 kPa

Specific internal energy is given as u=1840 kJ/kg

So the total internal energy is given as

$u=\frac{U}{m}\\U=u \times m\\U=1840 \times 2\\U=3680 kJ$

The total internal energy is 3680 kJ.

So the volume of vessel is 4.7680 $m^3$ and total internal energy is 3680 kJ.

Part b

Volume of vessel is given as 1.6

mass is given as 2 kg

Pressure is given as 0.2 MPa or 200 kPa

Now the specific volume is given as

$v=\frac{V}{m}\\v=\frac{1.6}{2}\\v=0.8 m^3/kg$

So from steam tables for Pressure=200 kPa and specific volume as 0.8 gives

Temperature=T=120 °C

Quality=x=0.903 ≈ 90.3%

Specific internal energy =u=2330 kJ/kg

The total internal energy is given as

$u=\frac{U}{m}\\U=u \times m\\U=2330 \times 2\\U=4660 kJ$

So the quality of the mixture is 90.3% or 0.903, temperature is 120 °C and total internal energy is 4660 kJ.

5 0

3 years ago

2. What is the original length of the rectangular bar if the deformation is 0.005 in with a force of 1000 lbs and an area of 0.7

Ugo [173]

Answer:

18.75in

Explanation:

Modulus of elasticity = Stress/Strain

Since stress = Force/Area

Given

Force = 1000lb

Area = 0.75sqin

Stress = 1000/0.75

Stress = 1333.33lbsqin

Strain

Strain = Stress/Modulus of elasticity

Strain = 1333.33/5,000,000

Strain = 0.0002667

Also

Strain = extension/original length

extension = 0.005in

Original length = extension/strain

Original length = 0.005/0.0002667

Original length = 18.75in

Hence the original length of the rectangular bar is 18.75in

6 0

3 years ago

Technician A says that a way to prevent galvanic corrosion is to duplicate the original installation method. Technician B says t

sertanlavr [38]

Answer:

Technician A

Explanation:

Galvanic corrosion is not on only one metal alone but caused when two metals are interacting. Thus, Duplicating the original installation method is a better option because re-using a coated bolt doesn't prevent galvanic corrosion because both materials must be coated and not just the bolt and in technician B's case he is coating just the bolt. Thus, technician B's method will not achieve prevention of galvanic corrosion but technician A's method will achieve it.

8 0

2 years ago