Which of the following is the last step in creating a budget?

5. The pin support at A allows _______. Select the one that applies. (a) displacement in the x direction (b) rotation about its

soldier1979 [14.2K]

Answer: Diagram associated with your question is attached below

5) B

6) C

Explanation:

5) The pin support at A allows ; Rotation about its central axis

This is because pin supports does not allow the translation of its structural member in any direction i.e. y or x but only rotation about its axis

6) The support at B does not allow displacement in y direction

This is because roller support allows displacement only in the direction that they are situated and in this case it is the x - direction

7 0

2 years ago

In engineering, economic cost is a decision-making tangible factor. Group of answer choices True False

salantis [7]

Economic cost is a rescission making tangible factor true

8 0

2 years ago

Read 2 more answers

Two Carnot engines operate in series such that the heat rejected from one is the heat input to the other. The heat transfer from

kykrilka [37]

Answer:

Given:

high temperature reservoir $T_{H} =1000k$

low temperature reservoir $T_{L} =400k$

thermal efficiency $n_{1}= n_{2}$

The engines are said to operate on Carnot cycle which is totally reversible.

To find the intermediate temperature between the two engines, The thermal efficiency of the first heat engine can be defined as

$n_{1} =1-\frac{T}{T_{H} }$

The thermal efficiency of second heat engine can be written as

$n_{2} =1-\frac{T_{L} }{T}$

The temperature of intermediate reservoir can be defined as

$1-\frac{T}{T_{H} } =1-\frac{T_{L} }{T} \\T^2=T_{L} T_{H} \\T=\sqrt{T_{L} T_{H} }\\T=\sqrt{400*1000} =632k$

8 0

2 years ago

A wastewater treatment plant has two primary clarifiers, each 20m in diameter with a 2-m side-water depth. the effluent weirs ar

jasenka [17]

Answer:

overflow rate 20.53 m^3/d/m^2

Detention time 2.34 hr

weir loading 114.06 m^3/d/m

Explanation:

calculation for single clarifier

$sewag\ flow Q = \frac{12900}{2} = 6450 m^2/d$

$surface\ area =\frac{pi}{4}\times diameter ^2 = \frac{pi}{4}\times 20^2$

$surface area = 314.16 m^2$

volume of tank $V = A\times side\ water\ depth$

$=314.16\times 2 = 628.32m^3$

$Length\ of\ weir = \pi \times diameter of weir$

$= \pi \times 18 = 56.549 m$

overflow rate = $v_o = \frac{flow}{surface\ area} = \frac{6450}{314.16} = 20.53 m^3/d/m^2$

Detention time $t_d = \frac{volume}{flow} = \frac{628.32}{6450} \times 24 = 2.34 hr$

weir loading $= \frac{flow}{weir\ length} = \frac{6450}{56.549} = 114.06 m^3/d/m$

6 0

2 years ago

A high molecular weight hydrocarbon gas A is fed continuously into a heated mixed flow reactor (0.1liter) where it is thermally

dimulka [17.4K]

Answer:

Space velocity = 30 hr⁻¹

Explanation:

Space velocity for reactors express how much reactor volume of feed or reactants can be treated per unit time. For example, a space velocity of 3 hr⁻¹ means the reactor can process 3 times its volume per hour.

It is given mathematically as

Space velocity = (volumetric flow rate of the reactants)/(the reactor volume)

Volumetric flowrate of the reeactants

= (molar flow rate)/(concentration)

Molar flowrate of the reactants = 300 millimol/hr

Concentration of the reactants = 100 millimol/liter

Volumetric flowrate of the reactants = (300/100) = 3 liters/hr

Reactor volume = 0.1 liter

Space velocity = (3/0.1) = 30 /hr = 30 hr⁻¹

Hope this Helps!!!

5 0

3 years ago