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fomenos
3 years ago
9

The wet density of a sand was found to be 1.9 Mg/m3 and the field water content was 10%. In the laboratory, the density of solid

s was found to be 2.66 Mg/m3, and the maximum and minimum void ratios were 0.62 and 0.44, respectively.
a. What is the field relative density?

b. How much will a 3 m thick stratum of this sand settle if the sand is densified to a relative density of 65%?

Engineering
1 answer:
Nookie1986 [14]3 years ago
3 0

Answer:

a) 44.4%

b) 72 mm

Explanation:

See attached pictures.

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A(n) _____________ is used commonly in open split-phase motors to disconnect the start winding from the electrical circuit when
Alenkasestr [34]

A <u>centrifugal switch</u> is used commonly in open split-phase motors to disconnect the start winding from the electrical circuit when the motor reaches approximately 75% of its rated speed.

Hope that helps!

7 0
2 years ago
Working with which of these systems requires a technician that has been certified in an EPA-approved course?
makvit [3.9K]

EPA Regulations provides a certified course for the technicians involved in the Air-conditioning system.

Answer: Option (b)

<u>Explanation:</u>

The EPA regulation has implemented an act called the "Clean Air Act" under the "section of 609".

This act provides some basic requirements for EPA Regulation such as follows;

  • Refrigerant: This unit must be approved by EPA Regulations before being implemented into the atmosphere.
  • Servicing: This system provides a certified course for technicians in service and also approve them with proper refrigerant equipment.
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6 0
3 years ago
Air modeled as an ideal gas enters a combustion chamber at 20 lbf/in.2
motikmotik

Answer:

The answer is "112.97 \ \frac{ft}{s}"

Explanation:

Air flowing into thep_1 = 20 \ \frac{lbf}{in^2}

Flow rate of the mass m  = 230.556 \frac{lbm}{s}

inlet temperature T_1 = 700^{\circ} F

PipelineA= 5 \times 4 \ ft

Its air is modelled as an ideal gas Apply the ideum gas rule to the air to calcule the basic volume v:

\to \bar{R} = 1545 \ ft \frac{lbf}{lbmol ^{\circ} R}\\\\ \to M= 28.97 \frac{lb}{\bmol}\\\\ \to pv=RT \\\\\to v= \frac{\frac{\bar{R}}{M}T}{p}

      = \frac{\frac{1545}{28.97}(70^{\circ}F+459.67)}{20} \times \frac{1}{144}\\\\=9.8 \frac{ft3}{lb}

V= \frac{mv}{A}

   = \frac{230.556 \frac{lbm}{s} \times 9.8 \frac{ft^3}{lb}}{5 \times 4 \ ft^2}\\\\= 112.97 \frac{ft}{s}

8 0
3 years ago
The state of plane strain on an element is:
balu736 [363]

Answer:

a. ε₁=-0.000317

   ε₂=0.000017

θ₁= -13.28° and  θ₂=76.72°  

b. maximum in-plane shear strain =3.335 *10^-4

Associated average normal strain ε(avg) =150 *10^-6

θ = 31.71 or -58.29

Explanation:

\epsilon _{1,2} =\frac{\epsilon_x + \epsilon_y}{2}  \pm \sqrt{(\frac{\epsilon_x + \epsilon_y}{2} )^2 + (\frac{\gamma_xy}{2})^2} \\\\\epsilon _{1,2} =\frac{-300 \times 10^{-6} + 0}{2}  \pm \sqrt{(\frac{-300 \times 10^{-6}+ 0}{2}) ^2 + (\frac{150 \times 10^-6}{2})^2}\\\\\epsilon _{1,2} = -150 \times 10^{-6}  \pm 1.67 \times 10^{-4}

ε₁=-0.000317

ε₂=0.000017

To determine the orientation of ε₁ and ε₂

tan 2 \theta_p = \frac{\gamma_xy}{\epsilon_x - \epsilon_y} \\\\tan 2 \theta_p = \frac{150 \times 10^{-6}}{-300 \times 10^{-6}-\ 0}\\\\tan 2 \theta_p = -0.5

θ= -13.28° and  76.72°

To determine the direction of ε₁ and ε₂

\epsilon _{x' }=\frac{\epsilon_x + \epsilon_y}{2}  + \frac{\epsilon_x -\epsilon_y}{2} cos2\theta  + \frac{\gamma_xy}{2}sin2\theta \\\\\epsilon _{x'} =\frac{-300 \times 10^{-6}+ \ 0}{2}  + \frac{-300 \times 10^{-6} -\ 0}{2} cos(-26.56)  + \frac{150 \times 10^{-6}}{2}sin(-26.56)\\\\

=-0.000284 -0.0000335 = -0.000317 =ε₁

Therefore θ₁= -13.28° and  θ₂=76.72°  

b. maximum in-plane shear strain

\gamma_{max \ in \ plane} =2\sqrt{(\frac{\epsilon_x + \epsilon_y}{2} )^2 + (\frac{\gamma_xy}{2})^2} \\\\\gamma_{max \ in \ plane} = 2\sqrt{(\frac{-300 *10^{-6} + 0}{2} )^2 + (\frac{150 *10^{-6}}{2})^2}

=3.335 *10^-4

\epsilon_{avg} =(\frac{\epsilon_x + \epsilon_y}{2} )

ε(avg) =150 *10^-6

orientation of γmax

tan 2 \theta_s = \frac{-(\epsilon_x - \epsilon_y)}{\gamma_xy} \\\\tan 2 \theta_s = \frac{-(-300*10^{-6} - 0)}{150*10^{-6}}

θ = 31.71 or -58.29

To determine the direction of γmax

\gamma _{x'y' }=  - \frac{\epsilon_x -\epsilon_y}{2} sin2\theta  + \frac{\gamma_xy}{2}cos2\theta \\\\\gamma _{x'y' }=  - \frac{-300*10^{-6} - \ 0}{2} sin(63.42)  + \frac{150*10^{-6}}{2}cos(63.42)

= 1.67 *10^-4

4 0
3 years ago
______ are an idication that your vehicle may be developing a cooling system problem.
iris [78.8K]

Answer:

The temperature gauge showing that the vehicle has been running warmer or has recently began to have issues from overheating is  an idication that your vehicle may be developing a cooling system problem.

Explanation:

8 0
3 years ago
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