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Oxana [17]
3 years ago
6

Part A What is the correct expression of the internal torque in segment AB? A shaft is fixed at A. A clockwise distributed torqu

e with magnitude 2 kN-m/m is applied between A and B. A counterclockwise torque of 400 N-m is applied at B, and a clockwise torque of 600 N-m is applied at the free-end C. The segments AB and BC have lengths 0.8 m and 0.6 m, respectively. What is the correct expression of the internal torque in segment AB? A shaft is fixed at A. A clockwise distributed torque with magnitude 2 kN-m/m is applied between A and B. A counterclockwise torque of 400 N-m is applied at B, and a clockwise torque of 600 N-m is applied at the free-end C. The segments AB and BC have lengths 0.8 m and 0.6 m, respectively. 2000x + 1400 N-m 2000x N-m –2000x +1800 N-m –2000x – 1400 N-m 2000x – 1800 N-m –2000x – 200 N-m 2000x + 200 N-m –2000x N-m
Engineering
1 answer:
Hunter-Best [27]3 years ago
4 0

Answer:

2000x + 1400 N-m

Explanation:

The torque at a given point is determined as the perpendicular force multiply by the distance of the force from the given point. The conventional method is to take clockwise direction as positive and anticlockwise direction as negative. Therefore:

Torque at the segment AB = 2000*x + 400 = (2000x + 400) N-m

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A supercapacitor, also called an ultracapacitor, is a high-capacity capacitor with a capacitance value much higher than other capacitors, but with lower voltage limits, that bridges the gap between electrolytic capacitors and rechargeable batteries.

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1. Which of the following will cause a spark knock?
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Answer:

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3 years ago
Two common methods of improving fuel efficiency of a vehicle are to reduce the drag coefficient and the frontal area of the vehi
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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

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