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myrzilka [38]
2 years ago
7

The primary of an ideal transformer has 400 turns and its secondary has 200 turns. Neglecting electrical losses, if the power in

put is 200 W (average) and 240 V (rms), we can expect the output of the secondary to be
Engineering
1 answer:
Vanyuwa [196]2 years ago
6 0

Answer:

A.C. Voltage (transformers only work with A.C.) from input (primary) to output (secondary) is purely a function of the turns ratio.

Explanation:

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Select three functions of catalysts.
Korolek [52]

Answer: speed up food processing

speed up plant growth

Increase fuel efficiency

Explanation:

A catalyst simply refers to a substance that leads to an increase in the reaction rate when it's added to a substance. When the activation energy is reduced by catalysts, this.hwlpa on the speeding up of a reaction.

Therefore,the functions of catalysts include speed up food processing, speeding up plant growth and increase fuel efficiency

5 0
3 years ago
While discussing VIN numbers, Technician A says that the first digit of the VIN identifies the country where the vehicle was man
ruslelena [56]
Usually the first digit of the vin id’s the country it was built. So technician A would be correct. That’s usually how it is. Hope this helps. Please let me know if this is incorrect
4 0
3 years ago
A hole of diameter D = 0.25 m is drilled through the center of a solid block of square cross section with w = 1 m on a side. The
attashe74 [19]

Answer:

q=4.013\:\:kW\\\\T_1=278.91\:\:^{\circ}C\\\\T_2=275.82\:\:^{\circ}C

Explanation:

R_{conv,1}=(h_1\pi D_1L)^{-1}=(50*0.25*2)^{-1}=0.01273\:\:K/W\\\\R_{conv,2}=(h^2*4wL)^{-1}=(4*4*1)^{-1}=0.0625\:\:K/W\\\\R_{cond(2D)}=(Sk)^{-1}=(8.59*150)^{-1}=0.00078\:\:K/W

So, heat rate can be calculated as follows:

q=\frac{T_{\infty,1}-T_{\infty,2}}{R_{conv,1}+R_{conv,2}+R_{cond(2D)}} =\frac{330-25}{0.076} =4.013\:\:kW

Surface temperatures can be calculated as follows:

T_1=T_{\infty,1}-qR_{conv,1}=330-51.09=278.91\:\:^{\circ}C\\\\T_2=T_{\infty,2}+qR_{conv,2}=25+250.82=275.82\:\:^{\circ}C

6 0
4 years ago
While walking across campus one windy day, an engineering student speculates about using an umbrella as a "sail" to propel a bic
makvit [3.9K]

Answer:

Given data:\\While walking across campus one windy day\\Frontal area, \(A=0.3 m ^{2}\)\\Wind speed \(V=24 Km / hr\)\\The drag coefficient \(C_{D, b}=1.2\)\\The combined mass \(m=75 kg\)\\Umbrella diameter, \(D=1.22 m\)\\Velocity of wind \(V=24 \frac{ km }{ hr }\)\\The rolling resistance \(C_{R}=0.75 \%\)

Solution:

Note: Refer the diagram

Basic equation:\\'s law of motion: \(\sum F_{x}=m a_{x}\)\\Lift coefficient, \(C_{L}=\frac{F_{L}}{\frac{1}{2} \rho V^{2} A_{p}}\)\\Drag coefficient, \(C_{D}=\frac{F_{D}}{\frac{1}{2} \rho V^{2} A_{p}}\)

From force balance equation:\\\(\sum F_{x}=F_{D}-F_{R}=0\)\\But \(F_{D}=\left(C_{D, \alpha} A_{u}+C_{D, B} A_{b}\right) \frac{1}{2} \rho\left(V_{\nu}-V_{b}\right)^{2}\)\(F_{R}=C_{R} m g\)\\Area of the Umbrella \(A_{u}=\frac{\pi D_{u}^{2}}{4}\)\(A_{x}=\frac{\pi \times 1.22^{2}}{4} m ^{2}\)\(A_{v}=1.17 m ^{2}\)

Drag coefficient data for selected objects table at

Hemisphere (open end facing flow), C_{D, x}=1.42

Substituting all parameters,

\begin{aligned}&F_{R}=0.0075 \times 75 \times 9.81\\&F_{R}=5.52 N\end{aligned}

Then,

\begin{aligned}&V_{b}=V_{w}-\left[\frac{2 F_{R}}{\rho\left(C_{D, w} A_{w}+C_{D, B} A_{b}\right)}\right]^{\frac{1}{2}} \dots\\&V_{w}=24 \times 1000 \times \frac{1}{3600}\\&V_{w}=6.67 \frac{ m }{ s }\end{aligned}

And the equation becomes,

\begin{aligned}&V_{b}=6.67-\left[\frac{2 \times 5.52}{1.23(1.42 \times 1.17+1.2 \times 0.3)}\right]^{\frac{1}{2}}\\&V_{b}=6.67-2.11\\&V_{b}=4.56 \frac{ m }{ s }\end{aligned}

Thus the floyds travels at 68.3^{\circ}wind speed.

7 0
4 years ago
calculate the magnitude of the force acting on the pin at D. Pin C is fixed in DE and bears against the smooth slot in the trian
Olenka [21]

Answer:

hello your question lacks the required diagram attached below is the diagram

answer : 58.47 N

Explanation:

The magnitude of the force acting on the Pin D

Fd = \sqrt{Dx^2 + Dy^2}

    = \sqrt{16.80^2 + 56^2}

    = 58.465 N

Dx = 16.80 N

Dy = 56 N

hello attached below is the detailed solution

8 0
3 years ago
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