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Hoochie [10]
3 years ago
9

How does drag affect the cost of owning a car?

Physics
2 answers:
vodka [1.7K]3 years ago
5 0

Answer:

You’ve probably noticed over the years that car design has become more streamlined. In other words, most consumer vehicles don’t look very boxy anymore. Some of this is due to aesthetics, but much of it is meant to decrease a vehicle’s drag coefficient. A cube has a high drag coefficient, whereas a teardrop has a low one.

By decreasing the drag coefficient, car makers are helping vehicles “slip” through air more easily. That reduces the amount of fuel needed to move the vehicle, and the difference shows up in your wallet and in the environment.

Explanation:

~Ban~

Elena L [17]3 years ago
4 0

Answer:

You move fast when moving only forward

Explanation:

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We will have that the diagram will be the following:

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When cattle are processed, what is the most likely use for the animals' skin?
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How does force and acceleration work together to create motion?
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5 0
3 years ago
A steel plate weighing 180 lb with center of gravity at point G is supported by a roller at point A, a bar DE, and a horizontal
melamori03 [73]

Answer:

Therefore, the force supported  by the hydraulic cylinder is = -70.01 lb

Thus, the force supported by the bar = -233.84 lb

The reaction force supported by the reaction of the roller = 341.31 lb

Explanation:

The attached diagrams is shown in the file below:

From there; taking moments about point A

\sum M__A }=0

- 40 (180) - (80) F_E Cos 37° + 55  

-7200 + F_E (-80 Cos 37° + 55 sin 37° ) = 0

= - 7200 - 30.79  F_E

- 30.79  F_E = 7200

F_E = -\frac{7200}{30.79}

F_E  = -233.84 lb

Thus, the force supported by the bar = -233.84 lb

Taking the equilibrium of forces in the vertical direction

\sum f_y = 0

F_E sin 37 + F_A cos 20 - F_G = 0

- 233.84 sin 37 + F_A cos 20 - 180 = 0

F_A cos 20  = 320.73

F_A = \frac{320.73}{cos 20}

F_A = 341.31 lb

The reaction force supported by the reaction of the roller = 341.31 lb

Taking the equilibrium of forces on the horizontal direction.

\sum f_y = 0

F_E cos 37 - F_{CB} + F_A sin 20° = 0

-233.84 cos 37 - F_{CB} + 341.31 sin 20° = 0  

-186.75 -  F_{CB} + 116.74 = 0

-  F_{CB} -70.01 = 0

-  F_{CB} = 70.01

F_{CB} = - 70.01 lb

Therefore, the force supported  by the hydraulic cylinder is = -70.01 lb

7 0
4 years ago
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