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dmitriy555 [2]
3 years ago
11

Hat do peaked roofs, convertible tops, and airplane wings have in common when air moves faster across their top surfaces? What d

o peaked roofs, convertible tops, and airplane wings have in common when air moves faster across their top surfaces?
a.The pressure underneath them is reduced.
b.The pressure underneath them is increased.
c.The pressure above them is reduced.
d.The pressure above them is increased.
Physics
2 answers:
Nutka1998 [239]3 years ago
6 0

Answer:

a.

Explanation:

The air moves faster across the top of the surface and exerts less pressure on the air below and thus due to difference in air pressure the top roof is lifted.

hence the correct option is pressure underneath them is reduced.

Mumz [18]3 years ago
4 0

Answer:

option (c)

Explanation:

As the wind blows from upper side, then the pressure at the above side is reduced and hence the pressure at the underneath of the surface is increased so that the roofs blown away.

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ExtremeBDS [4]
An impulsive force is a force that is acting only during a short time, I mean, for an instant. Impulse is a physics magnitude define by the product of the impulsive force and the time that it was acting.

Is there any mistake in my English? Please, let me know.
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A large, cylindrical water tank with diameter 3.60 m is on a platform 2.00 m above the ground. The vertical tank is open to the
zysi [14]

To solve this problem it is necessary to apply the concepts related to the geometry of a cylindrical tank and its respective definition.

The volume of a tank is given by

V = \frac{\pi d^2}{4}h

Where

d = Diameter

h = Height

Considering that there are two stages, let's define the initial and final volume as,

V_0 = \frac{\pi d^2}{4}H

V_f = \frac{\pi d^2}{4}h

We know as well by definition that

1gal = 3.785*10^{-3}m^3

Then we have for the statement that

V_f = V_0 -1gal

V_f = V_0 - 3.785*10^{-3}

Replacing the previous data

\frac{\pi d^2}{4}h = \frac{\pi d^2}{4}H- 3.785*10^{-3}

\frac{\pi (3.6)^2}{4}h = \frac{\pi (3.6)^2}{4}(2)- 3.785*10^{-3}

Solving to get h,

h = 1.99963m

Therefore the change is

\Delta h = H-h

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Therefore te change in the height of the water in the tank is 0.37mm

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3 years ago
Most people can detect frequencies as high as 20 000 Hz. Assuming the speed of sound in air is 345 m/s, determine the wavelength
Gekata [30.6K]

Wavelength = speed / frequency

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6 0
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Question 13 (1 point)
KonstantinChe [14]

Answer:

Reflective clothing. So vehicles can see them and stuff.

3 0
2 years ago
The suspension system of a 1700 kg automobile "sags" 7.7 cm when the chassis is placed on it. Also, the oscillation amplitude de
spin [16.1K]

Answer:

the spring constant k = 5.409*10^4 \ N/m

the value for the damping constant \\ \\b = 1.518 *10^3 \ kg/s

Explanation:

From Hooke's Law

F = kx\\\\k =\frac{F}{x}\\\\where \ F = mg\\\\k = \frac{mg}{x}\\\\given \ that:\\\\mass \ of \ each \ wheel = 425 \ kg\\\\x = 7.7cm = 0.077 m\\\\g = 9.8 \ m/s^2\\\\Then;\\\\k = \frac{425 \ kg * 9.8 \ m/s^2}{0.077 \ m}\\\\k = 5.409*10^4 \ N/m

Thus; the spring constant k = 5.409*10^4 \ N/m

The amplitude is decreasing 37% during one period of the motion

e^{\frac{-bT}{2m}}= \frac{37}{100}\\\\e^{\frac{-bT}{2m}}= 0.37\\\\\frac{-bT}{2m} = In(0.37)\\\\\frac{-bT}{2m} = -0.9943\\\\b = \frac{2m(0.9943)}{T}\\\\b = \frac{2m(0.9943)}{\frac{2 \pi}{\omega}}\\\\b = \frac{m(0.9943) \ ( \omega) )}{ \pi}

b = \frac{m(0.9943)(\sqrt{\frac{k}{m})}}{\pi}\\\\b = \frac{425*(0.9943)(\sqrt{\frac{5.409*10^4}{425}) }    }{3.14}\\\\b = 1518.24 \ kg/s\\\\b = 1.518 *10^3 \ kg/s

Therefore; the value for the damping constant \\ \\b = 1.518 *10^3 \ kg/s

5 0
3 years ago
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