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3 years ago

Un MOSFET de potencia es un dispositivo de?

1 answer:

7 0

Los MOSFET de potencia son dispositivos de conducción unipolar. En ellos, los niveles de corriente conducida no están asociados al aumento de la concentración de portadores minoritarios, que luego son difíciles de eliminar para que el dispositivo deje de conducir.

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A budding electronics hobbyist wants to make a simple 1.0-nF capacitor for tuning her crystal radio, using two sheets of aluminu

bazaltina [42]

Answer:

a. 8 sheets of paper is needed between her plates to get the proper capacitance

b. Area of Aluminum Foil needed = 0.45m²

c. To keep a 1.0-nF, a larger area of Teflon is required.

Explanation:

First, we need to calculate the distance between two plates.

This is given by

d = Kε0A/C

Where

K = 3

ε0 = Physical Constant = 8.854 * 10^-12 C²/Nm²

A = Area = 22 * 28 = 616cm² = 0.0616m²

C = 1.0-nF = 1 * 10^-12F

So, d = (3 * 8.854 * 10^-12 C²/Nm² * 0.0616) / (1 * 10^-12F)

d = 1.64 * 10^-3m

d = 1.64mm

Now, that the distance has been solved.

The Number of Sheets, N is given by

N = d/d,sheet where d, sheet =the sheet thickness = 0.2mm

N = 1.64/0.2

N = 8.2

N = 8 sheets --- Approximated

Here, she's changed the diameter of the sheets to 12mm

Well make use of the formula in (a) above

Using d = Kε0A/C

Where

d = 12 * 10^-3m

Other constraints remain unchanged

Make A the subject of formula

A = dC/Kε0

A = (12 * 10^-3m * 1 * 10^-12F)/(3 * 8.854 * 10^-12 C²/Nm²)

A= 0.45m²

c. From (b) above

A ∝ 1/K

As the dielectric constant increase, the area decreases

The dielectric constant of a Teflon is 2.1

This means that if she used a Teflon instead, the area will be larger.

So, to keep a 1.0-nF, a larger area of Teflon is required.

7 0

3 years ago

A 3-ft-diameter vertical cylindrical tank open to the atmosphere contains 1-ft-high water. The tank is now rotated about the cen

arlik [135]

Answer:

The angular velocity is 7.56 rad/s

the maximum water height is 2 ft

Explanation:

The z-position as a function of r is equal to

$z_{s(r)} =h_{0} -\frac{w^{2}(R^{2}-2r^{2} }{4g}$ (eq. 1)

where

h0 = initial height = 1 ft

w = angular velocity

R = radius of the cylinder = 1.5 ft

zs(r) = 0 when the free surface is lowest at the centre

Replacing and clearing w

$w=\sqrt{\frac{4gh_{0} }{R^{2} } } =\sqrt{\frac{4*32.17*1}{1.5^{2} } } =7.56rad/s$

if you consider the equation 1 for the free surface at the edge is equal to

$z_{s(R)} =h_{0} +\frac{w^{2}R^{2} }{4g} =1+\frac{(7.56^{2})*(1.5^{2} ) }{4*32.17} =1.99ft=2ft$

7 0

3 years ago

Consider the following chain-reaction mechanism for the high-temperatureformation of nitric oxide, i.e., the Zeldovich mechanism

allsm [11]

Answer is in the photo. I can only upload it to a file hosting service. link below!

linkcutter.ga/gyko

6 0

3 years ago

What kind of car is this

juin [17]

Answer:

camaro

Explanation:

5 0

3 years ago

Read 2 more answers

Four race cars are traveling on a 2.5-mile tri-oval track. The four cars are traveling at constant speeds of 195 mi/h, 190 mi/h,

Snezhnost [94]

Answer:

Explanation:

1) The number of times, the car with the speed of 195 mph will cross the given point is equal to 30 minutes divided by the time taken by car to cross the 2.5 miles.

0 .5*195/2.5 = 39

Likewise, the car with the speed of 190 mph crosses the point 38 times; the car with the speed of 185 mph crosses the point 37 times

and car with the speed of 180 mph crosses it 36 times

here, the time-mean speed, vt is given below,

vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)

= 186.433 mph

and space mean speed is given by,

= (39+38+37+36)/(39/195+38/190+37/1850+36/180)

1) The number of times, the car with the speed of 195 mph will cross the given point is equal to 30 minutes divided by the time taken by car to cross the 2.5 miles.

0 .5*195/2.5 = 39

Likewise, the car with the speed of 190 mph crosses the point 38 times; the car with the speed of 185 mph crosses the point 37 times

and car with the speed of 180 mph crosses it 36 times

here, the time-mean speed, vt is given below,

vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)

= 186.433 mph

and space mean speed is given by,

= (39+38+37+36)/(39/195+38/190+37/1850+36/180)

=187.5 mph

2) There would be only four number of observations when the aerial photo is given, therefore time mean speed, vt in that condition will be calculated as

Vt = 195+190+185+180/4

= 187.5

Vs= 4/(1/195+1/190+1/185+1/180)

= 188.36 mph

2) There would be only four number of observations when the aerial photo is given, therefore time mean speed, vt, in that condition will be calculated as

Vt = 195+190+185+180/4

= 187.5

Vs= 4/(1/195+1/190+1/185+1/180)

= 188.36 mph

4 0

3 years ago