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3 years ago

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Technician A says that ultraviolet light is bright, and very hot. Technician B says that ear protection does not need to be used

when welding. Who is correct?
a. Technician A only
b. Technician B only
c. Both Technicians A and B
d. Neither Technician A nor B

1 answer:

-Dominant- [34]3 years ago

3 0

Answer:

a. technician a only

Explanation:

I took that worksheet

You might be interested in

A steel plate has a hole drilled through it. The plate is put into a furnace and heated. What happens to the size of the inside

djverab [1.8K]

Answer:

The diameter increases

Explanation:

The expansion in the metal is uniform in every dimension

4 0

3 years ago

A parallel plates capacitor is filled with a dielectric of relative permittivity ε = 12 and a conductivity σ = 10^-10 S/m. The c

monitta

Answer:

t = 1.06 sec

Explanation:

Once disconnected from the battery, the capacitor discharges through the internal resistance of the dielectric, which can be expressed as follows:

R = (1/σ)*d/A, where d is is the separation between plates, and A is the area of one of the plates.

The capacitance C , for a parallel plates capacitor filled with a dielectric of a relative permittivity ε, can be expressed in this way:

C = ε₀*ε*A/d = 8.85*10⁻¹² *12*A/d

The voltage in the capacitor (which is proportional to the residual charge as it discharges through the resistance of the dielectric) follows an exponential decay, as follows:

V = V₀*e(-t/RC)

The product RC (which is called the time constant of the circuit) can be calculated as follows:

R*C = (1/10⁻¹⁰)*d/A*8.85*10⁻¹² *12*A/d

Simplifying common terms, we finally have:

R*C = 8.85*10⁻¹² *12 / (1/10⁻¹⁰) sec = 1.06 sec

If we want to know the time at which the voltage will decay to 3.67 V, we can write the following expression:

V= V₀*e(-t/RC) ⇒ e(-t/RC) = 3.67/10 ⇒ -t/RC = ln(3.67/10)= -1

⇒ t = RC = 1.06 sec.

3 0

3 years ago

Please answer the questions !

gizmo_the_mogwai [7]

Answer:

120

Explanation:

6 0

3 years ago

1.8 A water flow of 4.5 slug/s at 60 F enters the condenser of steam turbine and leaves at 140 F. Determine the heat transfer ra

Ann [662]

Answer:

$Hr=4.2*10^7\ btu/hr$

Explanation:

From the question we are told that:

Water flow Rate $R=4.5slug/s=144.78ib/sec$

Initial Temperature $T_1=60 \textdegree F$

Final Temperature $T_2=140 \textdegree F$

Let

Specific heat of water $\gamma= 1$

And

$\triangle T= 140-60$

$\triangle T= 80\ Deg.F$

Generally the equation for Heat transfer rate of water $H_r$ is mathematically given by

Heat transfer rate to water= mass flow rate* specific heat* change in temperature

$H_r=R* \gamma*\triangle T$

$H_r=144.78*80*1$

$H_r=11582.4\ btu/sec$

Therefore

$H_r=11582.4\ btu/sec*3600$

$Hr=4.2*10^7\ btu/hr$

3 0

3 years ago

Chlorine is one of the important commodity chemicals for the global economy. Before the advent of large scale

artcher [175]

The composition of gas in the feed, the percentage conversion and the

theoretical yield are combined to give the product stream composition.

Response:

The composition of gas in the product stream are;

HCl: 0.4 kmol/h, Cl₂: 1.6 kmol/h, H₂O: 1.6 kmol/h, O₂: 0.5 kmol/h

<h3>How can percentage conversion give the contents of the product stream?</h3>

The amount of oxygen used = 30% exceeding the theoretical amount

Number of moles of hydrochloric acid = 4 kmol/h

Percentage conversion = 80%

Required:

The composition of the gas in the product feed.

Solution;

The given reaction is; 4HCl + O₂ $\longrightarrow$ 2Cl₂ + 2H₂O

$Percentage \ conversion = \mathbf{ \dfrac{Moles \ of \ limiting \ reactant \ reacted}{Moles \ of \ limiting \ reactant \ supplied \ in \ the \, feed}}$

Which gives;

$80 \% = \mathbf{ \dfrac{Moles \ of \ limiting \ reactant \ reacted}{4 \, kmol/h}}$

Moles of limiting reactant reacted = 4 kmol/h × 0.80 = 3.6 kmol/h

Which gives;

Number of moles of HCl in the stream = 4 kmol/h - 3.6 kmol/h = 0.4 kmol/h

Number of moles of Cl₂ produced = 2 kmol/h × 0.8 = 1.6 kmol/h

Similarly;

Number of moles of H₂O produced = 2 kmol/h × 0.8 = 1.6 kmol/h

Number of moles of O₂ in the product stream = 30% × 1 kmol/h + 20% × 1 kmol/h = 0.5 kmol/h

The composition of the production stream is therefore;

<u>HCl: 0.4 kmol/h</u>

<u>Cl₂: 1.6 kmol/h</u>

<u>H₂O: 1.6 kmol/h</u>

<u>O₂: 0.5 kmol/h</u>

Learn more about theoretical and actual yield here:

brainly.com/question/14668990

brainly.com/question/82989

7 0

3 years ago