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liq [111]
4 years ago
12

The linkage is made of three pin-connected A-36 steel members, each having a cross-sectional area of 0.75 in2, Determine the mag

nitude of the force P needed to displace point B 0.10 in. downward.
Engineering
1 answer:
algol134 years ago
8 0

Answer:

0.029N

Explanation:

Pressure = displacement × density of steel × acceleration due to gravity

displacement = 0.1in = 0.1×0.0254m = 0.00254m, density of steel = 7900kg/m^3, acceleration due to gravity = 9.8m/s^2

Pressure = 0.00254×7900×9.8 = 196.65N/m^2

Force(P) = pressure × area

Area of 1 pin-connected A-36 steel member = 0.75in^2 = 4.84×10^-6m^2

Area of 3 pin-connected A-36 steel member = 3×4.84×10^-6 = 1.452×10^-5m^2

Force(P) = 196.65 × 1.4652×10^-5 = 0.0029N

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Two common methods of improving fuel efficiency of a vehicle are to reduce the drag coefficient and the frontal area of the vehi
qaws [65]

Answer:

\Delta V = 209.151\,L, \Delta C = 217.517\,USD

Explanation:

The drag force is equal to:

F_{D} = C_{D}\cdot \frac{1}{2}\cdot \rho_{air}\cdot v^{2}\cdot A

Where C_{D} is the drag coefficient and A is the frontal area, respectively. The work loss due to drag forces is:

W = F_{D}\cdot \Delta s

The reduction on amount of fuel is associated with the reduction in work loss:

\Delta W = (F_{D,1} - F_{D,2})\cdot \Delta s

Where F_{D,1} and F_{D,2} are the original and the reduced frontal areas, respectively.

\Delta W = C_{D}\cdot \frac{1}{2}\cdot \rho_{air}\cdot v^{2}\cdot (A_{1}-A_{2})\cdot \Delta s

The change is work loss in a year is:

\Delta W = (0.3)\cdot \left(\frac{1}{2}\right)\cdot (1.20\,\frac{kg}{m^{3}})\cdot (27.778\,\frac{m}{s})^{2}\cdot [(1.85\,m)\cdot (1.75\,m) - (1.50\,m)\cdot (1.75\,m)]\cdot (25\times 10^{6}\,m)

\Delta W = 2.043\times 10^{9}\,J

\Delta W = 2.043\times 10^{6}\,kJ

The change in chemical energy from gasoline is:

\Delta E = \frac{\Delta W}{\eta}

\Delta E = \frac{2.043\times 10^{6}\,kJ}{0.3}

\Delta E = 6.81\times 10^{6}\,kJ

The changes in gasoline consumption is:

\Delta m = \frac{\Delta E}{L_{c}}

\Delta m = \frac{6.81\times 10^{6}\,kJ}{44000\,\frac{kJ}{kg} }

\Delta m = 154.772\,kg

\Delta V = \frac{154.772\,kg}{0.74\,\frac{kg}{L} }

\Delta V = 209.151\,L

Lastly, the money saved is:

\Delta C = \left(\frac{154.772\,kg}{0.74\,\frac{kg}{L} }\right)\cdot (1.04\,\frac{USD}{L} )

\Delta C = 217.517\,USD

4 0
4 years ago
How do I get my son to do his work?
valentinak56 [21]

Answer:

Explanation:

Reward him if he does his homework/work

If he doesnt do his homework, take things that he loves off of him. and tell him if he does his homework/work he will get them things back

8 0
3 years ago
A sensor produces a signal with amplitude 15 mV. A voltage amplifier must amplify the signal such that the amplitude of the outp
Nady [450]

Answer:

42.50 dB

Explanation:

Determine the minimum voltage gain

amplitude of input signal ( Vi ) = 15 mV

amplitude of output signal ( Vo) = 2 V

Vo = 2 v

therefore ; minimum gain = Vo / Vi =  2 / ( 15 * 10^-3 )

                                                         = 133.33

Minimum gain in DB = 20 log ( 133.33 )

                                  = 42.498 ≈ 42.50 dB

8 0
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sveta [45]

Answer:

I don't know ask my dad he would

Explanation:

but I can't ask him because he went to get milk and forgot to come back

8 0
3 years ago
Why is sssniperwolf so fine
RSB [31]
Bc she’s just a baddie
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3 years ago
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