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4 years ago

The correct statement about the lift and drag on an object is:_______

Engineering

2 answers:

Lisa [10]4 years ago

8 0

Answer:

(a). the resultant force in the direction of the freestream velocity is termed the drag and the resultant force normal to the freestream velocity is termed the lift

Explanation:

When a fluid flows around the surface of an object, it exerts a force on it. This force has two components, namely lift and drag.

The component of this force that is perpendicular (normal) to the freestream velocity is known as lift, while the component of this force that is parallel or in the direction of the fluid freestream flow is known as drag.

Lift is as a result of pressure differences, while drag results from forces due to pressure distributions over the object surface, and forces due to skin friction or viscous force.

Thus, drag results from the combination of pressure and viscous forces while lift results only from the<em> pressure differences</em> (not pressure forces as was used in option D).

The only correct option left is "A"

(a). the resultant force in the direction of the freestream velocity is termed the drag and the resultant force normal to the freestream velocity is termed the lift

adoni [48]4 years ago

7 0

Answer:

Explanation:

I hope this helps! :)

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Consider a very long rectangular fin attached to a flat surface such that the temperature at the end of the fin is essentially t

kodGreya [7K]

Answer:

$\frac{T-20}{130-20}= e^{-14.28*0.05}$

And if we solve for T we got:

$T= 20 + 110e^{-14.28*0.05} = 73.86 C$

The answer for this case would be $T = 73.86 C$ at 5cm from the base of the fin.

Explanation:

Data given

For this case we have the following data given:

$h = 20 \frac{W}{m^2 K}$ represent the heat transfer coefficient.

$p$ represent the perimeter for this case and would be given by:

$p = 2*0.05m +2*0.001m= 0.102m$

$k = 200 \frac{W}{m C}$ represent the thermal conductivity

$w = 5cm =0.05 m$ represent the width

$h = 1mm =0.001m$ represent the thickness

$A= wh= 0.05m *0.001m = 0.00005 m^2$

Solution to the problem

For this case we assume that we have steady conditions, the temperature of the fins varies just in one direction, the heat transfer coefficient not changes with the time and the thermal properties of the fin not change.

We can determine the temperature if the fin at x=5 cm=0.05 m from the base with the following formula:

$\frac{T-T_{\infty}}{T_b -T_{\infty}} = e^{-mx}$

Where m is a coefficient given by:

$m = \sqrt{\frac{hp}{kA}}=\sqrt{\frac{20 W/m^2 C 0.102 m}{200 W/ mC 0.00005 m^2}}= 14.28 m^{-1}$

The value of x for this case represent the distance $x =5 cm =0.05m$

$T_b =130 C$ represent the base temperature

$T_{\infty}= 20$ represent the temperature of the sorroundings or the ambient.

If we replace we have this:

$\frac{T-20}{130-20}= e^{-14.28*0.05}$

And if we solve for T we got:

$T= 20 + 110e^{-14.28*0.05} = 73.86 C$

The answer for this case would be $T = 73.86 C$ at 5cm from the base of the fin.

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3 years ago

A discrete-time LTI system H has input x[n] and output y[n] related by the linear constant coefficient difference equation y[n]

Molodets [167]

Answer:

See explaination and attachment

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

Have the solution as an attachment.

4 0

3 years ago

Which of the following is an example of an observation?

Novay_Z [31]

The monkey is brown!

6 0

3 years ago

Read 2 more answers

When -iron is subjected to an atmosphere of hydrogen gas, the concentration of hydrogen in the iron, CH (in weight percent), is

djyliett [7]

Answer:

See attachments for step by step explanation towards getting answer.

Explanation:

Given that;

College Engineering 10+5 pts

When -iron is subjected to an atmosphere of hydrogen gas, the concentration of hydrogen in the iron, CH (in weight percent), is a function of hydrogen pressure, (in MPa), and absolute pH2temperature (T) according to(5.14)Furthermore, the values of D0 and Qd for this diffusion system are 1.4  10-7 m2/s and 13,400 J/mol, respectively. Consider a thin iron membrane 1 mm thick that is at 250C. Compute the diffusion flux through this membrane if the hydrogen pressure on one side of the membrane is 0.15 MPa (1.48 atm), and on the other side 7.5 MPa (74 atm).

See attachlent for complete solving.

6 0

4 years ago

Find the dy/dx using implicit differentiation

irina [24]

Answer:

dy/dx = (1 − 2x + 8y) / (4 + 3x − 12y)

Explanation:

d/dx (x − 4y) = d/dx (e^(2x + 3y − 1))

1 − 4 dy/dx = e^(2x + 3y − 1) (2 + 3 dy/dx)

Since x − 4y = e^(2x + 3y − 1):

1 − 4 dy/dx = (x − 4y) (2 + 3 dy/dx)

1 − 4 dy/dx = 2 (x − 4y) + 3 (x − 4y) dy/dx

1 − 4 dy/dx = 2x − 8y + (3x − 12y) dy/dx

1 − 2x + 8y = (4 + 3x − 12y) dy/dx

dy/dx = (1 − 2x + 8y) / (4 + 3x − 12y)

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3 years ago