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Lostsunrise [7]
4 years ago
10

Question 6: Which of the following is NOT true if you are changing into a

Engineering
1 answer:
baherus [9]4 years ago
6 0

Answer:

You should activate your signals 100 feet before you turn, becuase of that you should not wait until you are turning to turn on your signals. If you do not have working lights use your hands. So the answer that is inccorect is B.

You should activate your signals when you get into the turn lane is <u>NOT true</u>

Because you are supposed to single beforhand not while you're in the lane.

Im taking my drivers ed too. :)

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Why doesn't any one give as many points as i do for random questions on brainly
erastova [34]

good question i like to give about 20. i think people just are cheapskates for most of their time on this planet. didnt need the 50 though but thanks.

4 0
2 years ago
Read 2 more answers
Water is pumped from one large reservoir to another at a higher elevation. If the flow rate is 2.5 ft3 /s and the pump delivers
4vir4ik [10]

Answer:

2132hp  ed enregia

e

Explanation:

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6 0
3 years ago
Please help will give brainliest please answer all 3
larisa [96]

Answer: For #1 I'm going to go with A because that has to do with biology

For #2 I'm going to go with B oceans because that has to do with plant life (and life in general).

For #3 I'll say marine/maritime engineer (you can just say marine)

Hope it helps!

8 0
4 years ago
True/False
sweet [91]

Answer:

false jdbebheuwowjwjsisidhhdd

7 0
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
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
3 years ago
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