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satela [25.4K]
3 years ago
13

.

Engineering
1 answer:
Gennadij [26K]3 years ago
3 0

Answer:

II

Explanation:

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In casting experiments performed using a certain alloy and type of sand mold, it took 170 sec for a cube-shaped casting to solid
poizon [28]

Answer:

Answer for the question is : Solidification time will be same i.e. 170. See attached file for explanation.

Explanation:

Download pdf
7 0
3 years ago
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Two wastewater treatment plant workers (one male and one female) are exposed to hydrogen sulfide in confined spaces in the treat
Vlada [557]

Answer:

Go to explaination for the details of the answer.

Explanation:

In order to determine the lifetime (75 years) chronic daily exposure for each individual, we have to first state the terms of our equation:

CDI = Chronic Daily Intake

C= Chemical concentration

CR= Contact Rate

EFD= Exposure Frequency and Distribution

BW= Body Weight

AT = Average Time.

Having names our variables lets create the equations that will be used to derive our answers.

Please kindly check attachment for details of the answer.

5 0
3 years ago
A 1020 CD steel shaft is to transmit 15 kW while rotating at 1750 rpm. Determine the minimum diameter for the shaft to provide a
vladimir2022 [97]

Answer:

diameter is 14 mm

Explanation:

given data

power = 15 kW

rotation N = 1750 rpm

factor of safety = 3

to find out

minimum diameter

solution

we will apply here power formula to find T that is

power = 2π×N×T / 60    .................1

put here value

15 ×10^{3} = 2π×1750×T / 60

so

T = 81.84 Nm

and

torsion = T / Z                        ..........2

here Z is section modulus i.e = πd³/ 16

so from equation 2

torsion = 81.84 / πd³/ 16

so torsion = 416.75 / / d³     .................3

so from shear stress theory

torsion = σy / factor of safety

so here σy = 530 for 1020 steel

so

torsion = σy / factor of safety

416.75 / d³ = 530 × 10^{6} / 3

so d = 0.0133 m

so diameter is 14 mm

3 0
3 years ago
Find the current Lx in the figure
AleksandrR [38]

Explanation:

\frac{1}{8}  +  \frac{1}{2}   \\ 1.6 + 1.4 = 3 \\  \frac{1}{3}  +  \frac{1}{9}   \\ 2.25 + 2 = 4.25 \: ohm

R total = 4.25 ohm

I total = Vt/Rt

I total= 17/4.25= 4 A

Ix= 600 mA

\frac{9}{9 + 3}  \times 4 = 3\\   \frac{2}{2 + 8} \times 3 = 0.6a \\  = 0.6 \: milli \: amper

6 0
3 years ago
Consider a W21x93. Determine the moment capacity of the beam. Assume the compression flange is not laterally braced and that the
OLga [1]

Answer:

The answer is "828.75"

Explanation:

Please find the correct question:

For W21x93 BEAM,

Z_x = 221.00 in^3 \\\\\to \frac{b_t}{2t_f} =4.53\\\\\to \frac{h}{t_w}=32.3

For A992 STREL,

F_y= 50\  ks

Check for complete section:

\to \frac{b_t}{2t_f} =4.53 < \frac{65}{\sqrt{f_y = 9.19}}\\\\\to \frac{h}{t_w} =32.3 < \frac{640}{\sqrt{f_y = 90.5}}

Design the strength of beam =\phi_b Z_x F_y\\\\

                                                =0.9 \times 221 \times 50\\\\=9945 \ in \ \ kips\\\\=\frac{9945}{12}\\\\= 828.75 \ft \ kips \\

8 0
3 years ago
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