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2 years ago

How does drag affect the cost of owning a car?

Physics

2 answers:

vodka [1.7K]2 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]2 years ago

4 0

Answer:

You move fast when moving only forward

Explanation:

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3 years ago

011 10.0 points

Sedbober [7]

<h2>The temperature of the air is 66.8° C</h2>

Explanation:

From the Newton's velocity of sound relationship , the velocity of sound is directly proportional to the square root of temperature .

In this case The velocity of sound = frequency x wavelength

= 798 x 0.48 = 383 m/sec

Suppose the temperature at this time = T K

Thus 383 ∝ $\sqrt{T}$ I

The velocity of sound is 329 m/s at 273 K ( given )

Thus 329 ∝ $\sqrt{273}$ II

Dividing I by II , we have

$\frac{383}{329}$ = $\sqrt{\frac{T}{273} }$

or $\frac{T}{273}$ = 1.25

and T = 339.8 K = 66.8° C

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3 years ago

75 POINTS AND BRAINLY

kramer

Answer:

B=work=power/time

Explanation:

8 0

2 years ago

Read 2 more answers

Una placa de cobre a 20°C tiene unas dimensiones de 65cm x 78 cm. Encuentra el área de la placa a 400°C; Coeficiente de dilataci

ValentinkaMS [17]

Answer:

El área de la placa es aproximadamente 5102.752 centímetros cuadrados.

Explanation:

Asumamos que el cambio dimensional como consecuencia de la temperatura es pequeña, entonces podemos estimar el área de la placa de cobre en función de la temperatura mediante la siguiente aproximación:

$A_{f} = w\cdot l \cdot [1 + 2\cdot \alpha\cdot (T_{f}-T_{o})]$ (1)

Donde:

$w$ - Ancho de la placa, en centímetros.

$l$ - Longitud de la placa, en centímetros.

$\alpha$ - Coeficiente de dilatación, en $\frac{1}{^{\circ}C}$ .

$T_{o}$ - Temperatura inicial, en grados Celsius.

$T_{f}$ - Temperatura final, en grados Celsius.

Si sabemos que $w = 65\,cm$ , $l = 78\,cm$ , $\alpha = 17\times 10^{-6}\,\frac{1}{^{\circ}C}$ , $T_{o} = 20\,^{\circ}C$ and $T_{f} = 400\,^{\circ}C$ , entonces el área de la placa a la temperatura final: