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Flura [38]
4 years ago
10

Shear strain can be expressed in units of either degrees or radians. a)True b)- False

Engineering
1 answer:
ExtremeBDS [4]4 years ago
3 0

Answer:

true

Explanation:

shear strain is define as the ratio of change in deformation to the original length perpendicular to the axes of member due to shear stress.

     ε    = deformation/original length

         

strain is a unit less quantity but shear stain is generally expressed in radians but it can also be expressed in degree.

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Which of the following statements about pitot-static systems is FALSE? a). A pitot probe measures the Total Pressure of the free
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Answer:

C

Explanation:

Pitot tube:

  Pitot tube is a device which is used to measure the velocity of flow by measuring pressure difference between the points.

As we know that stagnation pressure is the summation of dynamic and static pressure.

Stagnation pressure = Static pressure + Dynamic pressure

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3 years ago
An important fluid property is the kinematic viscosity which determines the viscous, or frictional, forces acting in a flow. The
Aleksandr [31]

Answer:- 400 K - 26.665

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Basic usage of the interpolation formula to get the values of the required answers kinematic viscosities at 400 and 600 K

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4 years ago
The Environmental Protection Agency (EPA) has standards and regulations that says that the lead level in soil cannot exceed the
DENIUS [597]

Answer:

See below

Explanation:

<u>Check One-Sample T-Interval Conditions</u>

Random Sample? √

Sample Size ≥30? √

Independent? √

Population Standard Deviation Unknown? √

<u>One-Sample T-Interval Information</u>

  • Formula --> CI=\bar{x}\pm t^*(\frac{S_x}{\sqrt{n}})
  • Sample Mean --> \bar{x}=390.25
  • Critical Value --> t^*=2.0096 (given df=n-1=50-1=49 degrees of freedom at a 95% confidence level)
  • Sample Size --> n=50
  • Sample Standard Deviation --> S_x=30.5

<u>Problem 1</u>

The critical t-value, as mentioned previously, would be t^*=2.0096, making the 95% confidence interval equal to CI=\bar{x}\pm t^*(\frac{S_x}{\sqrt{n}})=390.25\pm2.0096(\frac{30.5}{\sqrt{50}})\approx\{381.5819,398.9181\}

This interval suggests that we are 95% confident that the true mean levels of lead in soil are between 381.5819 and 398.9181 parts per million (ppm), which satisfies the EPA's regulated maximum of 400 ppm.

3 0
2 years ago
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