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KatRina [158]
3 years ago
8

An ideal transformer has 75 turns in the primary coil and n turns in the secondary coil. a 120 v rms 60hz ac voltage source is c

onnected to the primary coil. a 10 ω resistor is connected to the secondary coil, forming a secondary circuit. the average power dissipated in the secondary circuit is 160 w. what is the number ns of turns in the secondary coil?
Physics
1 answer:
denpristay [2]3 years ago
4 0

think you messed up the symbol for resistor as resistors are measured in ohms where the symbol used for ohms is Greek omega

solving for average power in secondary coil:

average power =(current rms)^2*resistance⇒with a little algebra:

current rms=(√average power/resistance)

current rms=√160W/10Ω

current rms=4amps.

average power is also equal to current rms*voltage rms

with some algebra we can solve for voltage in the secondary wire:

voltage rms= average power/ current rms

voltage rms= 160W/4A

voltage rms=40Volts

now that we have voltage in the soecondary we can solve for the amount of turns in the secondary: Voltage secondary/voltage primary=number of turns in secondary/ number of turns in primary. using some algerbra we can solve for number of turns in secondary: (Voltage secondary/voltage primary)*number of turns in primary=number of turns in secondary

(40V/120V)*75turns=number of turns in secondary

number of turns in secondary=25turns

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dem82 [27]

Answer:

Frequency, f=2.30\times 10^{18}\ Hz

Explanation:

Given that,

The wavelength of the x-rays, \lambda=0.13\ nm=0.13\times 10^{-9}\ m

We need to find the frequency of an x-ray. All electromagnetic wave travel with a speed of light. It is given by the formula as :

c=f\lambda

f is the frequency

f=\dfrac{c}{\lambda}\\\\f=\dfrac{3\times 10^8}{0.13\times 10^{-9}}\\\\f=2.30\times 10^{18}\ Hz

So, the frequency of an x-ray is 2.30\times 10^{18}\ Hz. Hence, this is the required solution.

5 0
3 years ago
1. Given a list of atomic model descriptions:
laiz [17]

Answer: A: electron shells outside a central nucleus

B: hard, indivisible sphere

C: mostly empty space

Which list of atomic model descriptions represents

the order of historical development from the earliest

to most recent?

Explanation:

3

8 0
3 years ago
Two large, parallel, nonconducting sheets of positive charge face each other. What is at points (a) to the left of the sheets, (
alexira [117]

Answer:

a)The electric Field will be zero at the point between the sheets

b)E_1=\dfrac{\sigma}{\epsilon_0}

c)E_2=\dfrac{\sigma}{\epsilon_0}

Explanation:

Let \sigma be the surface charge density of the of the non conducting parallel sheet.Let consider a Gaussian surface in the form of of cylinder such that its cross-sectional is A . Then there will be flux only due to cross sectional area as the curved sectional is perpendicular to the the electric field  so the Electric Flux due to it is zero.

Now using Gauss law we have, E be the electric Field at the distance r from the sheet then

E\times 2A=\dfrac{\sigma A}{\epsilon_0}\\E=\dfrac{\sigma}{2\epsilon_0}

The Field will be away from the sheet and perpendicular to it.

a) The Electric Field between them

E_1=\dfrac{\sigma}{2\epsilon_0}-\dfrac{\sigma}{2\epsilon_0}\\=0

b)The Electric Field to the right of the sheets

E_1=\dfrac{\sigma}{2\epsilon_0}+\dfrac{\sigma}{2\epsilon_0}\\=\dfrac{\sigma}{\epsilon_0}

c)The Electric Field to the left of the sheets

E_2=\dfrac{\sigma}{2\epsilon_0}+\dfrac{\sigma}{2\epsilon_0}\\=\dfrac{\sigma}{\epsilon_0}

3 0
3 years ago
Consider the space between a point charge and the surface of a neutral spherical conducting shell. If the charge sits at the cen
Furkat [3]

Answer:

True

Explanation:

If a thin, spherical, conducting shell carries a negative charge, We expect the excess electrons to mutually repel one another, and, thereby, become uniformly distributed over the surface of the shell. The electric field-lines produced outside such a charge distribution point towards the surface of the conductor, and end on the excess electrons. Moreover, the field-lines are normal to the surface of the conductor. This must be the case, otherwise the electric field would have a component parallel to the conducting surface. Since the excess electrons are free to move through the conductor, any parallel component of the field would cause a redistribution of the charges on the shell. This process will only cease when the parallel component has been reduced to zero over the whole surface of the shell

According to Gauss law

∅ = EA =-Q/∈₀

Where ∅  is the electric flux through the gaussian surface and E is the electric field strength

If the gaussian surface encloses no charge, since all of the charge lies on the shell, so it follows from Gauss' law, and symmetry, that the electric field inside the shell is zero. In fact, the electric field inside any closed hollow conductor is zero

8 0
3 years ago
*NEED ANSWER STAT*
tresset_1 [31]
Becomes a +1 ion for this
6 0
3 years ago
Read 2 more answers
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