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3 years ago

5

Fig. 4 shows a simply-supported beam with supports A and B. The beam is subjected to three forces, 2000 N, 4000 N and 1500 N in

the directions shown.
Calculate the vertical reaction forces at supports A and B.

1 answer:

Drupady [299]3 years ago

6 0

Answer:A

Explanation:

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Take water density and kinematic viscosity as p=1000 kg/m3 and v= 1x10^-6 m^2/s. (c) Water flows through an orifice plate with a

guapka [62]

Answer:

$K_v=12.34$

Explanation:

Given;

For orifice, loss coefficient, K₀ = 10

Diameter, D₀ = 45 mm = 0.045 m

loss coefficient of the orifice, Ko = 10

Diameter of the gate valve, Dy = 1.5D₀ = 1.5 × 0.045 m = 0.0675 m

Total head drop, Δhtotal=25 m

Discharge, Q = 10 l/s = 0.01 m³/s

Now,

the velocity of flow through orifice, Vo = Discharge / area of the orifice

or

Vo = $\frac{0.01}{\frac{\pi}{4}0.045^2}$

or

Vo = 6.28 m/s

also,

the velocity of flow through gate valve, $V_v$ = Discharge / area of the orifice

or

$V_v$ = $\frac{0.01}{\frac{\pi}{4}0.0675^2}$

or

$V_v$ = 2.79 m/s

Now,

the total head drop = head drop at orifice + head drop at gate valve

or

25 m = $K_o\frac{V_o^2}{2g}+K_v\frac{V_v^2}{2g}$

where,

$K_v$ is the loss coefficient for the gate valve

on substituting the values, we get

25 m = $10\frac{6.28^2}{2\times 9.81}+K_v\frac{2.79^2}{2\times9.81}$

or

$K_v\frac{2.79^2}{2\times9.81}$ = 4.898

or

$K_v=12.34$

3 0

4 years ago

Mining is an example of this type of business

luda_lava [24]

Answer:

Mining would go under Industry organization.

5 0

4 years ago

25 gallons of an incompressible liquid exert a force of 70 lbf at the earth’s surface. What force in lbf would 6 gallons of this

jekas [21]

Answer:

froce by 6 gallon liquid on moon surface is 2.86 lbf

Explanation:

given data:

at earth surface

volume of an incompressible liquid = Ve = 25 gallons

force by liquid = 70 lbf

on moon

volume of liquid = Vm = 6 gallons

gravitational acceleration on moon is am = 5.51 ft/s2

Due to incompressibility , the density remain constant.

mass of liquid on surface of earth $= \frac{ force}{ acceleration}$

$mass = \frac{70lbf}{32.2 ft/s2}$

mass = 2.173 pound

$density \rho = \frac{mass}{volume}$

$= \frac{2.173}{25} = 0.0869 pound/ gallon$

froce by 6 gallon liquid on moon surface is

Fm = mass * acceleration

= density* volume * am

= 0.0869 *6* 5.51

= 2.86 lbf

5 0

3 years ago

Consider a simple ideal Rankine cycle with fixed boiler and condenser pressures. If the steam is superheated to a higher tempera

Usimov [2.4K]

Answer:

Option D - the moisture content at turbine exit will decrease

Explanation:

In an ideal rankine system, the phenomenon of superheating occurs at a state where the vapor state of the fluid is heated above its saturation temperature and the phase of the fluid is changed from the vapor phase to the gaseous phase.

Now, a vapour phase has two different substances at room temperature, whereas a gas phase consists of just a single substance at a defined thermodynamic range, at standard room temperature.

At the turbine exit, since it's just a single substance in gaseous phase, it means it will have less moisture content.

Thus, the correct answer is;the moisture content at turbine exit will decrease

4 0

3 years ago

A subsurface exploration report shows that the average water content of a fine-grained soil in a proposed borrow area is 22% and

Morgarella [4.7K]

Answer:

shrinkage ratio = 1.538

Explanation:

given data

water content = 22 %

dry density γ = 82 pcf

required dry density specified γ' = 96 pcf

required to produce = 50,000 yd³ = 50000 × 27 = 1350,000 ft³

solution

we get here first volume of borrow pit that is we know that

dry density ∝ $\frac{1}{volume}$

so $\frac{\gamma d}{\gamma 'd} = \frac{v'}{v}$

$\frac{82}{96} = \frac{1350000}{v}$

v = 1580487.8 ft³

v = 58536.58 yd³

so here

shrinkage ratio will be as

shrinkage ratio = $\frac{96}{62.4}$

shrinkage ratio = 1.538

6 0

3 years ago