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Trava [24]
2 years ago
5

A cellphone charger has a transformer with 2000 turns in the primary coil which converts 240v a.c to 6v a.c and a rectifier that

converts 6v a.c to 6v d.c. what do the letters a.c and d.c stand for?​how many turns are in the secondary coil of the transformer
Physics
1 answer:
inn [45]2 years ago
5 0

Answer:

As we are converting 220V AC into a 5V DC, first we need a step-down transformer to reduce such high voltage. Here we have used 9-0-9 1A step-down transformer, which convert 220V AC to 9V AC. In transformer there are primary and secondary coils which step up or step down the voltage according to the no of turn in the coils.

Selection of proper transformer is very important. Current rating depends upon the Current requirement of Load circuit (circuit which will use the generate DC). The voltage rating should be more than the required voltage. Means if we need 5V DC, transformer should at least have a rating of 7V, because voltage regulator IC 7805 at least need 2V more i.e. 7V to provide a 5V voltage.

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An infinitely long cylindrical insulating shell of inner radius a and outer radius b has a uniform volume charge density p. Dete
xenn [34]

Answer: The electric field is: a) r<a , E0=; b) a<r<b E=ρ (r-a)/εo;

c) r>b E=ρ b (b-a)/r*εo

Explanation: In order to solve this problem we have to use the Gaussian law in diffrengios regions.

As we know,

∫E.dr= Qinside/εo

For r<a --->Qinside=0 then E=0

for a<r<b er have

E*2π*r*L= Q inside/εo       in this case Qinside= ρ.Vol=ρ*2*π*r*(r-a)*L

E*2π*r*L =ρ*2*π*r* (r-a)*L/εo

E=ρ*(r-a)/εo

Finally for r>b

E*2π*r*L =ρ*2*π*b* (b-a)*L/εo

E=ρ*b* (b-a)*/r*εo

3 0
2 years ago
A particle initially located at the origin has an acceleration of vector a = 2.00ĵ m/s2 and an initial velocity of vector v i =
natali 33 [55]

With acceleration

\mathbf a=\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)\,\mathbf j

and initial velocity

\mathbf v(0)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i

the velocity at time <em>t</em> (b) is given by

\mathbf v(t)=\mathbf v(0)+\displaystyle\int_0^t\mathbf a\,\mathrm du

\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\displaystyle\int_0^t\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)\,\mathbf j\,\mathrm du

\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf j\bigg|_{u=0}^{u=t}

\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)t\,\mathbf j

We can get the position at time <em>t</em> (a) by integrating the velocity:

\mathbf x(t)=\mathbf x(0)+\displaystyle\int_0^t\mathbf v(u)\,\mathrm du

The particle starts at the origin, so \mathbf x(0)=\mathbf0.

\mathbf x(t)=\displaystyle\int_0^t\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf j\,\mathrm du

\mathbf x(t)=\left(\left(8.00\dfrac{\rm m}{\rm s}\right)u\,\mathbf i+\dfrac12\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u^2\,\mathbf j\right)\bigg|_{u=0}^{u=t}

\mathbf x(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)t\,\mathbf i+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)t^2\,\mathbf j

Get the coordinates at <em>t</em> = 8.00 s by evaluating \mathbf x(t) at this time:

\mathbf x(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)(8.00\,\mathrm s)\,\mathbf i+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)(8.00\,\mathrm s)^2\,\mathbf j

\mathbf x(8.00\,\mathrm s)=(64.0\,\mathrm m)\,\mathbf i+(64.0\,\mathrm m)\,\mathbf j

so the particle is located at (<em>x</em>, <em>y</em>) = (64.0, 64.0).

Get the speed at <em>t</em> = 8.00 s by evaluating \mathbf v(t) at the same time:

\mathbf v(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)(8.00\,\mathrm s)\,\mathbf j

\mathbf v(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(16.0\dfrac{\rm m}{\rm s}\right)\,\mathbf j

This is the <em>velocity</em> at <em>t</em> = 8.00 s. Get the <em>speed</em> by computing the magnitude of this vector:

\|\mathbf v(8.00\,\mathrm s)\|=\sqrt{\left(8.00\dfrac{\rm m}{\rm s}\right)^2+\left(16.0\dfrac{\rm m}{\rm s}\right)^2}=8\sqrt5\dfrac{\rm m}{\rm s}\approx17.9\dfrac{\rm m}{\rm s}

5 0
2 years ago
8. A meter reader determines that a business has used 5000 kW.h of energy in 4 months. If
ladessa [460]

Answer:

ENERGY AND COST. One kllowatt hour is 1,000 watts of power for one hour of time. ... Determine power: P = V XI ... Calculate the total kilowatt hours used. ... If the electric costs are 150 per kWh, how much does it cost to run the refrigerator in ... 8. A room was lighted with three 100-watt bulbs for 5 hours per day. If the cost of.

Explanation:

7 0
2 years ago
a telescope is oriting on a spacecraft aroudn the earth. Speed of 3.25 x 10^5 mass 7500 kg. What is the de Broglie wavelength of
Scrat [10]

Answer:

<em>2.72 x 10^-43 m</em>

<em></em>

Explanation:

mass of the telescope = 7500 kg

speed of the telescope = 3.25 x 10^5 m/s

de Broglie's  wavelength of the telescope is given as

λ = h/mv

where

λ is the wavelength of the telescope

h is the plank's constant = 6.63 × 10-34 m^2 kg/s

m is the mass of the telescope = 7500 kg

v is speed of the telescope = 3.25 x 10^5 m/s

substituting value, we have

λ = (6.63 × 10-34)/(7500 x 3.25 x 10^5)

λ = <em>2.72 x 10^-43 m</em>

8 0
3 years ago
What are the three types of friction and when does each apply
padilas [110]

Answer:static, sliding, rolling,

Explanation:

3 0
3 years ago
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