A force that attempts to decrease the length of a structural member is___

Heat is transferred at a rate of 2 kW from a hot reservoir at 800 K to a cold reservoir at 300 K. Calculate the rate at which th

andre [41]

Answer:

4.17x10^-3 kW/K

Explanation:

Detailed explanation and calculation is shown in the image below

6 0

3 years ago

On the generalized enthalpy departure chart, the normalized enthalpy departure values seem to approach zero as the reduced press

Alekssandra [29.7K]

Answer:

Enthalpy is a function of pressure hence normalized enthalpy departure values will approach zero with reduced pressure approaching zero

Explanation:

On the generalized enthalpy departure chart, the normalized enthalpy departure values seem to approach zero as the reduced pressure PR approaches zero. this is because enthalpy is a function of pressure therefore as the Pressure is reducing towards the zero value, the gas associated with the pressure tends to behave more like an Ideal gas.

For an Ideal gas the Normalized enthalpy departure value will be approaching the zero value.

4 0

3 years ago

Describe the Range of a set of numbers and how it is calculated.

allochka39001 [22]

Answer:

The range of a set of data is the difference between the highest and lowest values in the set. To find the range, first order the data from least to greatest. Then subtract the smallest value from the largest value in the set.

Explanation:

7 0

3 years ago

Considering only (110), (1 1 0), (101), and (10 1 ) as the possible slip planes, calculate the stress at which a BCC single crys

vredina [299]

Solution :

i. Slip plane (1 1 0)

Slip direction -- [1 1 1]

Applied stress direction = ( 1 0 0 ]

τ = 50 MPa ( Here slip direction must be perpendicular to slip plane)

τ = σ cos Φ cos λ

$\cos \phi = \frac{(1,0,0) \cdot (1,1,0)}{1 \times \sqrt2}$

$=\frac{1}{\sqrt2 }$

$\cos \lambda = \frac{(1,0,0) \cdot (1,-1,1)}{1 \times \sqrt3}$

$=\frac{1}{\sqrt3 }$

τ = σ cos Φ cos λ

∴ $50= \sigma \times \frac{1}{\sqrt2} \times \frac{1}{\sqrt3}$

σ = 122.47 MPa

ii. Slip plane --- (1 1 0)

Slip direction -- [1 1 1]

$\cos \phi = \frac{(1, 0, 0) \cdot (1, -1, 0)}{1 \times \sqrt2} =\frac{1}{\sqrt2}$

$\cos \lambda = \frac{(1, 0, 0) \cdot (1, 1, -1)}{1 \times \sqrt3} =\frac{1}{\sqrt3}$

τ = σ cos Φ cos λ

∴ $50= \sigma \times \frac{1}{\sqrt2} \times \frac{1}{\sqrt3}$

σ = 122.47 MPa

iii. Slip plane --- (1 0 1)

Slip direction --- [1 1 1]

$\cos \phi = \frac{(1, 0, 0) \cdot (1, 0, 1)}{1 \times \sqrt2} =\frac{1}{\sqrt2}$

$\cos \lambda = \frac{(1, 0, 0) \cdot (1, 1, -1)}{1 \times \sqrt3} =\frac{1}{\sqrt3}$

τ = σ cos Φ cos λ

∴ $50= \sigma \times \frac{1}{\sqrt2} \times \frac{1}{\sqrt3}$

σ = 122.47 MPa

iv. Slip plane -- (1 0 1)

Slip direction ---- [1 1 1]

$\cos \phi = \frac{(1, 0, 0) \cdot (1, 0, -1)}{1 \times \sqrt2}=\frac{1}{\sqrt2}$

$\cos \lambda = \frac{(1, 0, 0) \cdot (1, -1, 1)}{1 \times \sqrt3} =\frac{1}{\sqrt3}$

τ = σ cos Φ cos λ

∴ $50= \sigma \times \frac{1}{\sqrt2} \times \frac{1}{\sqrt3}$

σ = 122.47 MPa

∴ (1, 0, -1). (1, -1, 1) = 1 + 0 - 1 = 0

3 0

2 years ago

What is land administration and cadastral survey

vaieri [72.5K]

Cadastral surveying is the sub-field of cadastre and surveying that specialises in the establishment and re-establishment of real property boundaries. ... A cadastral surveyor must apply both the spatial-measurement principles of general surveying and legal principles such as respect of neighboring titles.

8 0

2 years ago

A force that attempts to decrease the length of a structural member is____