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grandymaker [24]
4 years ago
7

In using the drag coefficient care needs to be taken to use the correct area when determining the drag force. What is a typical

example of the appropriate area to use?
Engineering
1 answer:
stealth61 [152]4 years ago
6 0

Answer:

Explanation:

We know that Drag forceF_D

  F_D=\dfrac{1}{2}C_D\rho AV^2

Where

             C_D is the drag force constant.

                 A is the projected area.

                V is the velocity.

                ρ is the density of fluid.

Form the above expression of drag force we can say that drag force depends on the area .So We should need to take care of correct are before finding drag force on body.

Example:

 When we place our hand out of the window in a moving car ,we feel a force in the opposite direction and feel like some one trying to pull our hand .This pulling force is nothing but it is drag force.

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3. Which of the following statements is false?
MrRa [10]

Answer: c.An accumulator is not used in a system with a receiver/dryer

Explanation:

In a refrigeration system, a condenser is used to transfer heat and this occurs from the refrigerant to the air or water.

Then, the refrigerant then condenses to liquid when the hear has been transferred.

We should note that the condenser is normally mounted in front of the radiator. The receiver/dryer is a storage tank for the liquid refrigerant from the condenser.

The statement that an accumulator is not used in a system with a receiver/dryer is not true. This is because, the accumulator gives protection to the compressor which helps to prevent the failure of the compressor.

Therefore, the answer is C.

7 0
3 years ago
Ayuda!<br> es para mañana
ExtremeBDS [4]
No se ve la foto :( puedes a ser la pregunta otra ves?
8 0
3 years ago
If copper (which has a melting point of 1085°C) homogeneously nucleates at 849°C, calculate the critical radius given values of
____ [38]

Answer:

The critical radius is -1.30 nm

Explanation:

Temperature for homogenous nucleation of copper, T_{H} = 849^{0} C = 849 + 273 = 1122 K

Melting point of copper, T_{cu} = 1085^{0} C = 1085 + 273 = 1358 K

Latent heat of fusion, H_{f} = -1.77 * 10^{9} J/m^{3}

Surface free energy, \gamma = 0.200 J/m^{2}

Critical radius, r = ?

The formula for the critical radius is given by:

r = \frac{2 \gamma T_{cu} }{H_{f}(T_{cu} - T_{H})  }

r = \frac{2 * 0.2*1358 }{(-1.77 * 10^{9}) (1358 - 1122)  }

r = \frac{543.2 }{(-1.77 * 10^{9}) 236}\\r = -1.30 * 10^{-9} m\\r = -1.30 nm

The critical radius is -1.30 nm

8 0
4 years ago
Question 11 (1 point)
kirill115 [55]

Answer:

  False

Explanation:

Bella counts products in finished goods inventory and she counts kits in various stages of manufacturing.

4 0
3 years ago
3. Given the Taylor series expansion for sin x = x − x3/ 3! + x5/5! − x7/7! . Write a program that takes in an arbitrary integer
katen-ka-za [31]

Answer:

clc

clear

x = input('type value of angle in degrees:\n');

x = x*pi/180; %convverting fron degree to radian

sin_x = x; %as first term of taylor series is x

E = 1; %just giving a value of error greater than desired error

n = 0;

while E > 0.000001

previous = sin_x;

n = n+1;

sin_x = sin_x + ((-1)^n)*(x^(2*n+1))/factorial(2*n+1);

E = abs(sin_x - previous); %calculating error

end

a = sprintf('sin(x) = %1.6f',sin_x);

disp(a)

Explanation:

8 0
4 years ago
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