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Katyanochek1 [597]
3 years ago
15

In the lab, a container of saturated soil had a mass of 113.27 g before it was placed in the oven and100.06 g after the soil had

dried. The container alone had a mass of 49.31 g. If the specific gravity ofsolids is 2.80, determine the void ratio and water content of the original soil sample.
Engineering
1 answer:
kipiarov [429]3 years ago
5 0

Answer:

  • Moisture/ water content w = 26%
  • Void ratio , e =  0.73

Explanation:

  • Initial mass of saturated soil w1 = mass of soil - weight of container

                                                 = 113.27 g - 49.31 g = 63.96 g

  • Final mass of soil after oven w2 = mass of soil - weight of container

                                                  = 100.06 g - 49.31 g = 50.75

Moisture /water content, w =   \frac{w1-w2}{w2} = \frac{63.96-50.75}{50.75} = 0.26 = 26%

Void ratio =  water content X specific gravity of solid

                  = 0.26 X 2.80 =0.728

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A strain gage is mounted at an angle of 30° with respect to the longitudinal axis of the cylindrical pressure. The pressure vess
GuDViN [60]

Answer:

1790 μrad.

Explanation:

Young's modulus, E is given as 10000 ksi,

μ is given as 0.33,

Inside diameter, d = 54 in,

Thickness, t = 1 in,

Pressure, p = 794 psi = 0.794 ksi

To determine shear strain, longitudinal strain and circumferential strain will be evaluated,

Longitudinal strain, eL = (pd/4tE)(1 - 2μ)

eL = (0.794 x 54)(1 - 0.66)/(4 x 1 x 10000)

eL = 3.64 x 10-⁴ radians

Circumferential strain , eH = (pd/4tE)(2-μ)

eH = (0.794 x 54)(2 - 0.33)/(4 x 1 x 10000)

eH = 1.79 x 10-³ radians

The maximum shear strain is 1790 μrad.

4 0
3 years ago
Consider two Carnot heat engines operating in series. The first engine receives heat from the reservoir at 1400 K and rejects th
Aleksandr-060686 [28]

Answer:

The temperature T= 648.07k

Explanation:

T1=input temperature of the first heat engine =1400k

T=output temperature of the first heat engine and input temperature of the second heat engine= unknown

T3=output temperature of the second heat engine=300k

but carnot efficiency of heat engine =1 - \frac{Tl}{Th} \\

where Th =temperature at which the heat enters the engine

Tl is the  temperature of the environment

since both engines have the same thermal capacities <em>n_{th} </em> therefore n_{th} =n_{th1} =n_{th2}\\n_{th }=1-\frac{T1}{T}=1-\frac{T}{T3}\\ \\= 1-\frac{1400}{T}=1-\frac{T}{300}\\

We have now that

\frac{-1400}{T}+\frac{T}{300}=0\\

multiplying through by T

-1400 + \frac{T^{2} }{300}=0\\

multiplying through by 300

-420000+ T^{2} =0\\T^2 =420000\\\sqrt{T2}=\sqrt{420000}  \\T=648.07k

The temperature T= 648.07k

5 0
3 years ago
An 1800-W toaster, a 1400-W electric frying pan, and a 75-W lamp are plugged into the same outlet in a 15-A, 120-V circuit. The
Mila [183]

Answer:

a) Current drawn by the toaster = 15A

Current drawn by the electric frying pan = 11.67A

Current drawn by the lamp = 0.625A

b) This combination will blow the 15A fuse as the total current requirement for this setup exceeds the 15A rating of the fuse.

Explanation:

a) For parallel connection, there exists, the same voltage and different currents across all the devices.

Voltage cross each of the 3 devices = outlet voltage of 120V

From their respective power rating, current drawn by each device will be calculated.

P = IV

For the toaster, P = 1800 W, V = 120V

I = 1800/120 = 15A

For the electric frying pan, P = 1400 W, V = 120 V

I = 1400/120 = 11.67 A

For the lamp, P = 75 W, V = 120V

I = 75/120 = 0.625 A

b) Total current in a parallel connection setup = Sum total of all the currents.

Total current drawn by all 3 devices = 15 + 11.67 + 0.625 = 27.295A = 27.3 A

This total current requirement surpasses the 15A current rating of the fuse, therefore, this combination will blow the fuse.

Hope this Helps!!!

6 0
3 years ago
Calculate the rate at which body heat is conducted through the clothing of a skier in a steady- state process, given the followi
olga2289 [7]

Answer:

230.4W

Explanation:

Heat transfer by conduction consists of the transport of energy in the form of heat through solids, in this case a jacket.

the equation is as follows

Q=\frac{KA(T2-T1)}{L} \\

Where

Q=heat

k=conductivity=0.04

A=Area=1.8m^2

T2=33C

T1=1C

L=thickness=1cm=0.01mQ=\frac{(0.04)(1.8m^2)(33-1)}{0.01m}

Q=230.4W

the skier loses heat at the rate of 230.4W

4 0
3 years ago
Which of the following factors does not promote safety in the shop?
sergey [27]
“Thinking about pleasant things to pass the time” would not promote safety in the shop because it would be taking the focus away from important tasks, which in turn decreases safety.
6 0
3 years ago
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