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Grace [21]
3 years ago
5

A typical wall outlet voltage in the United States is 120 volts. Personal MP3 players require much smaller voltages, typically 4

87.0 mV 487.0 mV . If the number of turns in the primary coil is 2464 2464 , calculate the number of turns on the secondary coil of the adapter transformer.
Physics
2 answers:
alexgriva [62]3 years ago
5 0

Answer:

Number of turns on the secondary coil of the adapter transformer is 10.

Explanation:

For a transformer,

    \frac{V_{s} }{V_{p} } = \frac{N_{s} }{N_{p} }

where V_{s} is the voltage induced in the secondary coil

           V_{p} is the voltage in the primary coil

          N_{s} is the number of turns of secondary coil

         N_{p} is the number of turns of primary coil

From the given question,

    \frac{487*10^{-3} }{120} = \frac{N_{s} }{2464}

⇒    N_{s} = \frac{2462*487*10^{-3} }{120}

            = 9.999733

  ∴   N_{s} = 10 turns

Scrat [10]3 years ago
5 0

Answer:

607,145 turns

Explanation:

Output voltage, that is secondary voltage,Es = 120 volts

Input voltage, that is primary voltage, Ep = 487/1000 = 0.487 volts

Number of turns in secondary = Ns

Number of turns in primary, Np = 2464

∴ Es/Ep = Ns/Np

Ns = Es * Np/Ep = 1`20 X 2464/0.487 =  607,145 turns ( Step up transformer)

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Answer:

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A student has a mass (including clothes and shoes) of 65.0 kg. She drinks a 12 oz. can of soda, with a nutritional energy conten
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She can climb 0.92 m without losing weight.

<u>Explanation</u>:

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1 cal = 4184 J

140 cal = 585760 J

Energy = 585760 J,  m = 65.0 kg = 65000 g,   Efficiency = 20 %

                                 GPE = mgh

where m represents the mass

          g represents the gravity,

           h represents the height.

                             585760 = 65000 \times 9.8 \times h

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7 0
3 years ago
An airplane accelerates at 16 m/s² from rest. If it lifts off from an aircraft carrier after 6.0 s, how far did it go down the r
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2 years ago
A point charge q is located at the center of a spherical shell of radius a that has a charge −q uniformly distributed on its sur
muminat

Answer:

a) E = 0

b) E =  \dfrac{k_e \cdot q}{ r^2 }

Explanation:

The electric field for all points outside the spherical shell is given as follows;

a) \phi_E = \oint E \cdot  dA =  \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}

From which we have;

E \cdot  A =  \dfrac{{\Sigma Q}}{\varepsilon _{0}} = \dfrac{+q + (-q)}{\varepsilon _{0}}  = \dfrac{0}{\varepsilon _{0}} = 0

E = 0/A = 0

E = 0

b) \phi_E = \oint E \cdot  dA =  \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}

E \cdot  A  = \dfrac{+q }{\varepsilon _{0}}

E  = \dfrac{+q }{\varepsilon _{0} \cdot A} = \dfrac{+q }{\varepsilon _{0} \cdot 4 \cdot \pi \cdot r^2}

By Gauss theorem, we have;

E\oint dS =  \dfrac{q}{\varepsilon _{0}}

Therefore, we get;

E \cdot (4 \cdot \pi \cdot r^2) =  \dfrac{q}{\varepsilon _{0}}

The electrical field outside the spherical shell

E =  \dfrac{q}{\varepsilon _{0} \cdot (4 \cdot \pi \cdot r^2) }= \dfrac{q}{4 \cdot \pi \cdot \varepsilon _{0} \cdot r^2 }=  \dfrac{q}{(4 \cdot \pi \cdot \varepsilon _{0} )\cdot r^2 }

k_e=  \dfrac{1}{(4 \cdot \pi \cdot \varepsilon _{0} ) }

Therefore, we have;

E =  \dfrac{k_e \cdot q}{ r^2 }

5 0
2 years ago
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