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7nadin3 [17]
3 years ago
9

A kite is an airfoil that uses the wind to produce a lift. Held in place by a string, a kite can remain aloft indefinitely. The

diamond kite is a particularly simple kite. As the wind passes under the lower surface of the diamond kite, it bends away from the surface and slows down. The air pressure under the lower surface is

Engineering
1 answer:
Anna71 [15]3 years ago
4 0

Answer:

<em>a) results in an increased pressure.</em>

<em>b) results in a pressure lower than the atmospheric pressure.</em>

<em>c) The direction of the overall force is upwards.</em>

<em>d) other forces are the weight, drag, and tension on the string.</em>

<em>e) the extra weight due to the tail adds drag and keeps the nose of the kite pointing upwards and the kite generally stable.</em>

<em></em>

Explanation:

a) As the wind passes under the lower surface of the diamond kite, it slows down. The slowing down of the wind under the lower surface results in an increased air pressure under the lower surface which now exceeds the pressure at the upper surface of the diamond kite, generating more lift upwards.

b) As the wind passes over the upper surface of the diamond kite, turbulence occurs resulting in pressure lower than the atmospheric pressure. This results in a reduce air drag on the kite.

c) The direction of the overall force exerted by the air on the kite is upwards.

d) Other force that acts on the kite in the air are; gravity force (due to the weight of the kite), drag (due to air flow around the kite), and tension force on the string from you. Once at the maximum height, the approximate net force on the kite is zero (provided the kite is stable at this height).

e) The extra weight of the kite due to the tail adds drag at the bottom of the kite, keeping the nose pointing upwards, and keeps the kite stable.

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A new 1.2-ha suburban residential development is to be drained by a storm sewer that connects to the municipal drainage system.
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6 0
3 years ago
A 1 m3 rigid tank initially contains air whose density is 1.18kg/m3. The tank is connected to a high pressure supply line throug
Elanso [62]

To solve this problem it is necessary to apply the concepts related to density in relation to mass and volume for each of the states presented.

Density can be defined as

\rho = \frac{m}{V}

Where

m = Mass

V = Volume

For state one we know that

\rho_1 = \frac{m_1}{V}

m_1 = \rho_1 V

m_1 = 1.18*1

m_1 = 1.18Kg

For state two we have to

\rho_2 = \frac{m_2}{V}

m_2 = \rho_2 V

m_1 = 7.2*1

m_1 = 7.2Kg

Therefore the total change of mass would be

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5 0
3 years ago
A 03-series cylindrical roller bearing with inner ring rotating is required for an application in which the life requirement is
-BARSIC- [3]

Answer:

\mathbf{C_{10} = 137.611 \ kN}

Explanation:

From the information given:

Life requirement = 40 kh = 40 40 \times 10^{3} \ h

Speed (N) = 520 rev/min

Reliability goal (R_D) = 0.9

Radial load (F_D) = 2600 lbf

To find C10 value by using the formula:

C_{10}=F_D\times \pmatrix \dfrac{x_D}{x_o +(\theta-x_o) \bigg(In(\dfrac{1}{R_o}) \bigg)^{\dfrac{1}{b}}} \end {pmatrix} ^{^{^{\dfrac{1}{a}}

where;

x_D = \text{bearing life in million revolution} \\  \\ x_D = \dfrac{60 \times L_h \times N}{10^6} \\ \\ x_D = \dfrac{60 \times 40 \times 10^3 \times 520}{10^6}\\ \\ x_D = 1248 \text{ million revolutions}

\text{The cyclindrical roller bearing (a)}= \dfrac{10}{3}

The Weibull parameters include:

x_o = 0.02

(\theta - x_o) = 4.439

b= 1.483

∴

Using the above formula:

C_{10}=1.4\times 2600 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{1}{\dfrac{10}{3}}}

C_{10}=3640 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{3}{10}}

C_{10} = 3640 \times \bigg[\dfrac{1248}{0.9933481582}\bigg]^{\dfrac{3}{10}}

C_{10} = 30962.449 \ lbf

Recall that:

1 kN = 225 lbf

∴

C_{10} = \dfrac{30962.449}{225}

\mathbf{C_{10} = 137.611 \ kN}

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Explanation:

7 0
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