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7nadin3 [17]
2 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]2 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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(35-39) A student travels on a school bus in the middle of winter from home to school. The school bus temperature is 68.0° F. Th
arlik [135]

Answer:

The net energy transfer from the student's body during the 20-min ride to school is 139.164 BTU.

Explanation:

From Heat Transfer we determine that heat transfer rate due to electromagnetic radiation (\dot Q), measured in BTU per hour, is represented by this formula:

\dot Q = \epsilon\cdot A\cdot \sigma \cdot (T_{s}^{4}-T_{b}^{4}) (1)

Where:

\epsilon - Emissivity, dimensionless.

A - Surface area of the student, measured in square feet.

\sigma - Stefan-Boltzmann constant, measured in BTU per hour-square feet-quartic Rankine.

T_{s} - Temperature of the student, measured in Rankine.

T_{b} - Temperature of the bus, measured in Rankine.

If we know that \epsilon = 0.90, A = 16.188\,ft^{2}, \sigma = 1.714\times 10^{-9}\,\frac{BTU}{h\cdot ft^{2}\cdot R^{4}}, T_{s} = 554.07\,R and T_{b} = 527.67\,R, then the heat transfer rate due to electromagnetic radiation is:

\dot Q = (0.90)\cdot (16.188\,ft^{2})\cdot \left(1.714\times 10^{-9}\,\frac{BTU}{h\cdot ft^{2}\cdot R^{4}} \right)\cdot [(554.07\,R)^{4}-(527.67\,R)^{4}]

\dot Q = 417.492\,\frac{BTU}{h}

Under the consideration of steady heat transfer we find that the net energy transfer from the student's body during the 20 min-ride to school is:

Q = \dot Q \cdot \Delta t (2)

Where \Delta t is the heat transfer time, measured in hours.

If we know that \dot Q = 417.492\,\frac{BTU}{h} and \Delta t = \frac{1}{3}\,h, then the net energy transfer is:

Q = \left(417.492\,\frac{BTU}{h} \right)\cdot \left(\frac{1}{3}\,h \right)

Q = 139.164\,BTU

The net energy transfer from the student's body during the 20-min ride to school is 139.164 BTU.

7 0
2 years ago
If a hoist lifts a 4500lb load 30ft in 15s, the power delivered to the load is a) 18.00hp b) 9000hp c) 16.36hp d) None of the ab
12345 [234]

Answer:

Explanation:

load = 4500lb                   lift height= 30 ft

time =15 s

velocity=\frac{30}{15} ft/s

velocity=2 ft/s

power = force\times velocity

power={4500}\times2

power= 9000 lb ft/s

1 hp= 550 lb ft/s

power= \frac{9000}{550} =16.36 hp

5 0
3 years ago
Compare and contrast the different forms of energy that are represented below:
Kitty [74]

Explanation:

1) Wind energy is generated through a wind turbine. When wind passes through the blades of wind mill, the blades of the wind mill tend to rotate. Due to the spinning of the rotor across the turbine, the kinetic energy from the wind is converted to electrical energy

2) Incase of wind energy, the consumption gets higher when there is more wind and would be zero incase of no movement of blades.

Incase of hydroelectric power, the generation is rather stable and consumption depends on the usage of power from the consumers

Incase of gasoline generator, the generation is also stable subject to availability of gasoline and consumption again depends on the usage of power from the consumers

3) Pros of Hydroelectric power

Cost of electricity generation is less

Can produce green energy

Produce mass volume of electricity

Cons of Hydroelectric power

Requires massive initial investement

Can be installed on certain demographical area

8 0
3 years ago
Which of the following conditions is a good sign of minor
Kryger [21]

Answer:

Explanation:

d

7 0
3 years ago
A plate clutch has a single friction surface 9-in OD by 7-in ID. The coefficient of friction is 0.2 and the maximum pressure is
Talja [164]

Answer:

the torque capacity is  30316.369 lb-in

Explanation:

Given data

OD = 9 in

ID = 7 in

coefficient of friction = 0.2

maximum pressure = 1.5 in-kip = 1500 lb

To find out

the torque capacity using the uniform-pressure assumption.

Solution

We know the the torque formula for uniform pressure theory is

torque = 2/3 × \pi × coefficient of friction × maximum pressure ( R³ - r³ )    .....................................1

here R = OD/2 = 4.5 in and r = ID/2 = 3.5 in

now put all these value R, r, coefficient of friction and  maximum pressure in equation 1 and we will get here torque

torque = 2/3 × \pi × 0.2 × 1500 ( 4.5³ - 3.5³ )

so the torque =  30316.369 lb-in

3 0
3 years ago
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