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Mekhanik [1.2K]
3 years ago
6

The lab technician you recently hired tells you the following: Boss, an undisturbed sample of saturated clayey soil was brought

to me from the Mission Valley site. I measured the mass of the sample to be 600 grams. I then measured the mass of the sample after placing it in the oven for 24 hrs. I found this mass to be 200 grams. Can you help me determine the water content?" Determine the water content.
Engineering
1 answer:
harkovskaia [24]3 years ago
6 0

Answer:

The water of the saturated clayed soil is 66.67 %.

Explanation:

Given;

mass of saturated clayed soil, M_s = 600 g

mass of dry soil sample, M_d = 200 g

mass of water content, M_w = M_s - M_d = 600 g - 200 g = 400 g

The water content is determined as;

M_w(\%)  = \frac{M_s - M_d}{M_s} *100\%\\\\M_w(\%)  = \frac{600-200}{600} *100 \% \\\\M_w(\%)  = 66.67 \%

Therefore, the water of the saturated clayed soil is 66.67 %.

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Which of the following is not an electronic device ?
PolarNik [594]

Answer:

B

Explanation:

it's does not transmit any energy

6 0
1 year ago
The entire system of components that produces power and transmits it to the road is called the vehicle's _____.
IrinaK [193]

Answer:

Powertrain

Explanation:

6 0
3 years ago
How can we calculate the speed of the output gear in a simple gear train? Explain with the help of an example.
Snowcat [4.5K]

Answer:

N_3=\dfrac{T_1}{T_3}N_1

Explanation:

In the diagram there three gears in which gear 1 is input gear ,gear 2 is idle gear and gear 3 is out put gear.

Lets take

Speed\ of\ gear 1=N_1

Number\ of\ teeth\ of\ gear 1=T_1

Speed\ of\ gear 3=N_3

Number\ of\ teeth\ of\ gear 3=T_3

All external matting gears will rotates in opposite direction with respect to each other.

So the speed of gear third can be given as follows

\dfrac{T_1}{T_3}=\dfrac{N_3}{N_1}

N_3=\dfrac{T_1}{T_3}N_1

3 0
3 years ago
You are hitting a nail with a hammer (mass of hammer =1.8lb) the initial velocity of the hammer is 50 mph (73.33 ft/sec). The ti
Archy [21]

Answer:

The nail exerts a force of 573.88 Pounds on the Hammer in positive j direction.

Explanation:

Since we know that the force is the rate at which the momentum of an object changes.

Mathematically \overrightarrow{F}=\frac{\Delta \overrightarrow{p}}{\Delta t}

The momentum of any body is defines as \overrightarrow{p}=mass\times \overrightarrow{v}

In the above problem we see that the moumentum of the hammer is reduced to zero in 0.023 seconds thus the force on the hammer is calculated using the above relations as

\overrightarrow{F}=\frac{m(\overrightarrow{v_{f}}-\overrightarrow{v_{i}})}{\Delta t}

\overrightarrow{F}=\frac{m(0-(-73.33)}{0.23}=\frac{1.8\times 73.33}{0.23}=573.88Pounds

6 0
3 years ago
The size of Carvins Cove water reservoir is 3.2 billion gallons. Approximately, 11 cfs of water is continuous withdrawn from thi
Zolol [24]

Answer:

471 days

Explanation:

Capacity of Carvins Cove water reservoir = 3.2 billion gallons i.e. 3.2 x 10˄9 gallons

As,  

1 gallon = 0.133 cubic feet (cf)

Therefore,  

Capacity of Carvins Cove water reservoir in cf  = 3.2 x 10˄9 x 0.133

                                                                         = 4.28 x 10˄8

 

Applying Mass balance i.e

Accumulation = Mass In - Mass out   (Eq. 01)

Here  

Mass In = 0.5 cfs

Mass out = 11 cfs

Putting values in (Eq. 01)

Accumulation  = 0.5 - 11

                         = - 10.5 cfs

 

Negative accumulation shows that reservoir is depleting i.e. at a rate of 10.5 cubic feet per second.

Converting depletion of reservoir in cubic feet per hour = 10.5 x 3600

                                                                                       = 37,800

 

Converting depletion of reservoir in cubic feet per day = 37, 800 x 24

                                                                                         = 907,200  

 

i.e. 907,200 cubic feet volume is being depleted in days = 1 day

1 cubic feet volume is being depleted in days = 1/907,200 day

4.28 x 10˄8 cubic feet volume will deplete in days  = (4.28 x 10˄8) x                    1/907,200

                                                                                 = 471 Days.

 

Hence in case of continuous drought reservoir will last for 471 days before dry-up.

8 0
2 years ago
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