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bulgar [2K]
3 years ago
7

How could I build device that converts the blowing of wind into a form of energy a car could use?

Engineering
1 answer:
VMariaS [17]3 years ago
4 0
Wind turbines operate on a simple principle. The energy in the wind turns two or three propeller-like blades around a rotor. The rotor is connected to the main shaft, which spins a generator to create electricity
You might be interested in
std::string x[n]; int mode1count = 0; // msft int mode2count = 0; // appl int mode3count = 0; // csco int mode4count = 0; // imb
rosijanka [135]

Answer:

std::string x[n];

int mode1count = 0; // msft

int mode2count = 0; // appl

int mode3count = 0; // csco

int mode4count = 0; // imb

int i;

for(i = 0; i < n; ++i) // Set the array of strings

x[i] = modeList[i];

for(i = 0; i < n; ++i)

{

if(x[i] == "msft")

++mode1count;

else if(x[i] == "appl")

++mode2count;

else if(x[i] == "csco")

++mode3count;

else if(x[i] == "imb")

++mode4count;

}

Explanation:

It is important to ensure that there is adequate space and necessary punctuation when developing codes and programming to avoid getting error messages when running the codes. In this problem, the codes required are shown in the answer section. The codes are properly writing and arranged.

7 0
3 years ago
What is a height gage?
Elena L [17]
The answer would be -62 because 62 x 1 equals 62 so that would be the answer.
6 0
3 years ago
Find the compressibility factor Z for oxygen at 3 MPa and 160 K.
saveliy_v [14]

Answer:

Z= 0.868

Explanation:

Given that

P= 3 MPa

T = 160 K

We know that

P v= Z R T

P= Pressure

v = specific volume

R= gas constant

T = Absolute temperature

Z=  Compressibility factor

Here specific volume of gas is not given so we assume that specific volume gas

v=0.012\ m^3/kg

We know that for oxygen gas constant

R = 0.259 KJ/kg.K

Now by putting the values

P v = Z R T

3000 x 0.012 = Z x 0.259 x 160

Z= 0.868

So  Compressibility factor is 0.868.

5 0
3 years ago
A closed, 0.4-m-diameter cylindrical tank is completely filled with oil (SG 0.9)and rotates about its vertical longitudinal axis
egoroff_w [7]

Answer: p_{B} - p_{A} = 28800 Pa or 28.8 kPa

Explanation: To determine the pressure of a liquid in a rotating tank,it is used:

p = \frac{p_{fluid}.w^{2}.r^{2} }{2} - γfluid . z + c

where:

p_{fluid} is the liquid's density

w is the angular velocity

r is the radius

γfluid.z is the pressure variation due to centrifugal force.

For this question, the difference between a point on the circumference and a point on the axis will be:

p_{B} - p_{A} = \frac{p_{fluid}.w^{2}.r_{B} ^{2} }{2} - γfluid.z_{B} - (\frac{p_{fluid}.w^{2}.r_{A} ^{2} }{2} - γfluid.z_{A})

p_{B} - p_{A} = \frac{p_{fluid}.w^{2}}{2} (r_B^{2} - r_A^{2} ) - γfluid(z_{B} -z_{A})

Since there is no variation in the z-axis, z = 0 and that the density of oil is 0.9.10³kg/m³:

p_{B} - p_{A} = \frac{p_{fluid}.w^{2}}{2} (r_B^{2} - r_A^{2} )

p_{B} - p_{A} = \frac{0.9.10^3.40^2}{2}(0.2^2 - 0)

p_{B} - p_{A} = 28800

The difference in pressure between two points, one on the circumference and the other on the axis is p_{B} - p_{A} = 28800 Pa or 28.8 kPa

8 0
3 years ago
Do the coil resistances have any effect on the plots?
PolarNik [594]
Because of the skin depth effect, the current at high frequency tends to flow at very low depth from radius. Then at high frequency the effective cross section of the wire is narrower than at DC.

Fro example skin depth at 100 kHz is 0.206 mm (0.008”), a wire more thicker than AWG26 could be a waste of copper, better use a bunch of thin wire (Litz wire) to rise the Q factor.
8 0
3 years ago
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