Ext

A glass tube is inserted into a flowing stream of water with one opening directed upstream and the other end vertical. If the wa

Furkat [3]

Answer:

$h=0.46m$

Explanation:

From the question we are told that:

Velocity of water $V=3m/s$

Height=?

Generally, the equation for Water Velocity is mathematically given by

$V=\sqrt{2gh}$

Therefore Height h is given as

$h=\frac{v}{2g}$

$h=\frac{3^2}{2*9.81}$

$h=0.46m$

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

A consolidation test was performed on a sample of fine-grained soil sample taken from a depth such that the vertical effective s

Scorpion4ik [409]

Answer:

The settlement that is expected is 1.043 meters.

Explanation:

Since the pre-consolidation stress of the layer is equal to the effective stress hence we conclude that the soil is normally consolidated soil

The settlement due to increase in the effective stress of a normally consolidated soil mass is given by the formula

$\Delta H=\frac{H_oC_c}{1+e_o}log(\frac{\bar{\sigma_o}+\Delta \bar{\sigma }}{\bar{\sigma_o}})$

where

'H' is the initial depth of the layer

$C_c$ is the Compression index

$e_o$ is the inital void ratio

$\bar{\sigma_o}$ is the initial effective stress at the depth

$\Delta \bar{\sigma_o}$ is the change in the effective stress at the given depth

Applying the given values we get

$\Delta H=\frac{8\times 0.3}{1+0.87}log(\frac{154+28}{154})=1.04$

3 0

3 years ago

What are two factors that determine the thermal energy of a substance

Sav [38]

Mass and chemical composition

6 0

3 years ago