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3 years ago

Define drag and lift forces.

Engineering

1 answer:

ra1l [238]3 years ago

3 0

Explanation:

Drag is the force which is generated parallel and also in opposition to direction of the travel for the object which is moving through the fluid.

Lift is the force which is generated perpendicular to direction of the travel for the object moving through the fluid.

Both of the two forces which are the lift and the drag force act through center of the pressure of object.

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On what kind of sectional drawing would you find the maximum length of a building?

Genrish500 [490]

Answer:

A longitudinal section

Explanation:

A longitudinal section is a section drawn along the length of an object, as opposed to a cross section, which is drawn across the width or diameter of an object.

5 0

2 years ago

A kitchen contains one section of counter that's 20 inches

Ne4ueva [31]

The number of receptacles that are needed for all of these kitchen areas are: C. Four.

<h3>What are receptacles?</h3>

Receptacles can be defined as types of sockets or series of outlets (openings) that provides a path where current can be taken in a wiring system, so as to run electrical appliances in buildings.

Based on the information provided, the number of receptacles that are needed for all of these kitchen areas are four because one would be used in each area.

Read more on receptacles here: brainly.com/question/23839796

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4 0

2 years ago

Find the total amount of heat in Q lost through a wall 10' by 18' , with R value from q. 1. Inside temperature is 70 degrees F w

marissa [1.9K]

Answer:

Just think

Explanation:

6 0

4 years ago

Read 2 more answers

A car engine with a thermal efficiency of 33% drives the air-conditioner unit (a refrigerator) besides powering the car and othe

fgiga [73]

Answer:

The rate of fuel required to drive the air conditioner $Q_h = 6.061 kW$

The flow rate of the cold air is $\r m = 0.30765 kg/s$

Explanation:

From this question we are told that

The efficiency is $\eta = 33$ % = 0.33

Temperature for the hot day is $T_h = 35^oC = 308 K \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (35+273)$

Temperature after cooling is $T_c = 5^oC = 278K$

The input power is $P_{in} = 2kW$

The rate of fuel required to drive the air conditioner can be mathematically represented as

$Q_h = \frac{P_{in}}{\eta}$

$= \frac{2}{0.33} = 6.061 kW$

From the question the air condition is assumed to be half as a Carnot refrigeration unit

This can be Mathematically interpreted in terms of COP(coefficient of performance) as

$\beta_{air} = 0.5 \beta$

where $\beta$ denotes COP and is mathematically represented as

$\beta = \frac{Q_c}{P_{in}}$

= > $Q_c = \beta P_{in}$

Where $Q_c$ is the rate of flue being burned for cold air to flow

Now if the COP of a Carnot refrigerator is having this value

$\beta_{Carnot } = \frac{T_c}{T_h - T_c}$

$= \frac{278}{308-278}$

$\beta_{Carnot} = 9.267\\$

Then

$\beta_{air} = 0.5 * 9.2667$

$= 4.6333$

Now substituting the value of $\beta$ to solve for $Q_c$

$Q_c = \beta P_{in}$

$= 4.6333 *2$

$9.2667kW$

The equation for the rate of fuel being burned for the cold air to flow

$Q_c = \r mc_p \Delta T$

Making the flow rate of the cold air

$\r m = \frac{Q_c}{c_p \Delta T}$

$= \frac{9.2667}{1.004}* (308 - 278)$

$= 0.30765 kg/s$

4 0

3 years ago

A wood pipe having an inner diameter of 3 ft. is bound together using steel hoops having a cross sectional area of 0.2 in^2. The

Minchanka [31]

Answers:

31.7 inches

Explanation:

Given:

Diameter = 3ft

Let D = Diameter

So, D = 3ft. (Convert to inches)

D = 3 * 12in = 36 inches

Coss-sectional area of the steel = 0.2in²

Gauge Pressure (P) = 4psi

Stress in Steel (σ)= 11.4ksi

Force in steel = ½ (Pressure * Projected Area)

Area (A) = 2 * Force/Pressure

Also, Area (A) = Spacing (S) * Wood Pipe Diameter

Area = Area

2*Force/Pressure = Spacing * Diameter

Substitute values I to the above expression

2 * Force / 4psi = S * 36 inches

Also

Force in steel (F) = Stress in steel (σ) × Cross-sectional area of the steel

So, F = 11.4ksi * 0.2in²

F = 11.4 * 10³psi * 0.2in²

F = 2.28 * 10³ psi.in²

So, 2 * Force / 4psi = S * 36 becomes

2 * 2.28 * 10³/4 = S * 36

S = 2 * 2.28 * 10³ / (4 * 36)

S = 4560/144

S = 31.66667 inches

S = 31.7 inches (approximated)

5 0

4 years ago