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3 years ago

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Question 4 of 10

1 answer:

Alexus [3.1K]3 years ago

7 0

Answer:

22.3 kg•m/s

Explanation:

Apex;)

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In your own words explain efficiency

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Efficiency is completing a task as fast a possible the the least amount of effort.

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The air that flows over the top part of an airplane's wing moves faster than the air that flows across the bottom. This faster m

kaheart [24]

As an airplane moves through the air, its wings cause changes in the speed and pressure of the air moving past them. These changes result in the upward force called lift.

The Bernoulli principle states that an increase in the speed of a fluid occurs simultaneously with a decrease in the pressure exerted by the fluid.

A wing is shaped and tilted so the air moving over it moves faster than the air moving under it. As air speeds up, its pressure goes down. So the faster-moving air above exerts less pressure on the wing than the slower-moving air below. The result is an upward push on the wing—lift!

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3 years ago

Two parallel disks of diameter D 5 0.6 m separated by L 5 0.4 m are located directly on top of each other. Both disks are black

oksian1 [2.3K]

To solve this problem it is necessary to apply the concepts related to the Stefan-Boltzmann law which establishes that a black body emits thermal radiation with a total hemispheric emissive power (W / m²) proportional to the fourth power of its temperature.

Heat flow is obtained as follows:

$Q = FA\sigma\Delta T^4$

Where,

F =View Factor

A = Cross sectional Area

$\sigma =$ Stefan-Boltzmann constant

T= Temperature

Our values are given as

D = 0.6m

$L = 0.4m\\T_1 = 450K\\T_2 = 450K\\T_3 = 300K$

The view factor between two coaxial parallel disks would be

$\frac{L}{r_1} = \frac{0.4}{0.3}= 1.33$

$\frac{r_2}{L} = \frac{0.3}{0.4} = 0.75$

Then the view factor between base to top surface of the cylinder becomes $F_{12} = 0.26$ . From the summation rule

$F_{13} = 1-0.26$

$F_{13} = 0.74$

Then the net rate of radiation heat transfer from the disks to the environment is calculated as

$\dot{Q_3} = \dot{Q_{13}}+\dot{Q_{23}}$

$\dot{Q_3} = 2\dot{Q_{13}}$

$\dot{Q_3} = 2F_{13}A_1 \sigma (T_1^4-T_3^4)$

$\dot{Q_3} = 2(0.74)(\pi*0.3^2)(5.67*10^{-8})(450^4-300^4)$

$\dot{Q_3} = 780.76W$

Therefore the rate heat radiation is 780.76W

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3 years ago

Julie is cycling at a speed of 3.4 meters/second. If the combined mass of the bicycle and Julie is 30 kilograms, what is the kin

nexus9112 [7]

Answer:

A

Explanation:

KE = 1/2 mv^2

=1/2(30kg)( 3.4 m/s)^2

=173.4 joules

=1.7×10^2 joules

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What is the connection between the x- and y-motions of a projectile?

Sunny_sXe [5.5K]

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