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Anna007 [38]
3 years ago
3

A river barge, whose cross section is approximately rectangular, carries a load of grain. The barge is 27 ft wide and 93 ft long

. When unloaded its draft (depth of submergence) is 5 ft, and with the load of grain the draft is 7 ft. Determine: (a) the unloaded weight of the barge, and (b) the weight of the grain.

Engineering
1 answer:
Monica [59]3 years ago
8 0

Answer:

a) for equilibrium

sum of F vertical is zero

F_{vertical} =0

so that

W_{b} =F_{B}=r_{h2o} *(submerged volume)\\=(62.4lb/ft^3)(5ft*27ft*93ft)\\=783432 lb

b)

F_{vertical} =0\\W_{b}+ W_{mg} =r_{h2o} *(submerged volume)\\W_{g} =(62.4lb/ft^3)*(7ft*27ft*93ft)-783432\\W_{g} =313372lb

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How to draw the output voltage waveform rectifier
tatyana61 [14]

Answer:

Half-wave rectifier converts an AC signal into a DC signal. It's called a half-wave because it only rectify the positive part of an AC signal.

AC Signal = An electrical signal that alternates between positive and negative voltage.

DC Signal = An electrical signal that only has positive voltage.

Rectify = A fancy word for converting something.

Adding a capacitor helps the positive part of the signal stay on longer. This work because the capacitor stores energy kinda like a battery. During the negative part of the AC signal, the energy stored in the capacitor will be drained and used, then the cycle repeats.

The load resistor is just there to prevent a short circuit from happening.

7 0
3 years ago
The vertical and horizontal poles at the traffic-light assembly are erected first. Determine the additional force and moment rea
konstantin123 [22]

Answer:

1. Az=258 lb

2. My=3440 ft lb

3. ∑Mz= 0

Explanation:

∑ Fx= Ax=0

∑ Fy= Ay=0

∑ Fz= Az-86-86-86

Az=258 lb

∑Mx=86 X 31 +Mx=0

Mx=2666 ft lb

∑My=86 X 40 +My=0

My=3440 ft lb

∑Mz= 0

6 0
4 years ago
Consider an aircraft powered by a turbojet engine that has a pressure ratio of 9. The aircraft is stationary on the ground, held
77julia77 [94]

Answer:

The break force that must be applied to hold the plane stationary is 12597.4 N

Explanation:

p₁ = p₂, T₁ = T₂

\dfrac{T_{2}}{T_{1}} = \left (\dfrac{P_{2}}{P_{1}}  \right )^{\frac{K-1}{k} }

{T_{2}}{} = T_{1} \times \left (\dfrac{P_{2}}{P_{1}}  \right )^{\frac{K-1}{k} } = 280.15 \times \left (9  \right )^{\frac{1.333-1}{1.333} } = 485.03\ K

The heat supplied = \dot {m}_f × Heating value of jet fuel

The heat supplied = 0.5 kg/s × 42,700 kJ/kg = 21,350 kJ/s

The heat supplied = \dot m · c_p(T_3 - T_2)

\dot m = 20 kg/s

The heat supplied = 20*c_p(T_3 - T_2) = 21,350 kJ/s

c_p = 1.15 kJ/kg

T₃ = 21,350/(1.15*20) + 485.03 = 1413.3 K

p₂ = p₁ × p₂/p₁ = 95×9 = 855 kPa

p₃ = p₂ = 855 kPa

T₃ - T₄ = T₂ - T₁ = 485.03 - 280.15 = 204.88 K

T₄ = 1413.3 - 204.88 = 1208.42 K

\dfrac{T_5}{T_4}  = \dfrac{2}{1.333 + 1}

T₅ = 1208.42*(2/2.333) = 1035.94 K

C_j = \sqrt{\gamma \times R \times T_5} = √(1.333*287.3*1035.94) = 629.87 m/s

The total thrust = \dot m × C_j = 20*629.87 = 12597.4 N

Therefore;

The break force that must be applied to hold the plane stationary = 12597.4 N.

5 0
3 years ago
A rod that was originally 100-cm-long experiences a strain of 82%. What is the new length of the rod?
Ierofanga [76]

Answer: (b)

Explanation:

Given

Original length of the rod is L=100\ cm

Strain experienced is \epsilon=82\%=0.82

Strain is the ratio of the change in length to the original length

\Rightarrow \epsilon =\dfrac{\Delta L}{L}\\\\\Rightarrow 0.82=\dfrac{\Delta L}{100}\\\\\Rightarrow \Delta L=82\ cm

Therefore, new length is given by (Considering the load is tensile in nature)

\Rightarrow L'=\Delta L+L\\\Rightarrow L'=82+100=182\ cm

Thus, option (b) is correct.

8 0
3 years ago
Ruler game, HELPPPPP
viktelen [127]
D! :D
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3 0
3 years ago
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