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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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Galileo’s pendulum theory stated that the time taken to swing through one complete cycle of a pendulum depends on what?
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Explanation:

7 0
3 years ago
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Minchanka [31]

Answer:

4 seconds

Explanation:

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3 years ago
A straight wire, labeled as Wire A, lies horizontally on a tabletop and is oriented to run north-south. A conventional current o
Semmy [17]

Answer:

b. If Wire B carries a northward conventional current and lies to the left (west) of wire A, then it will experience an attractive force to the right (towards Wire A).

d. If Wire B carries a southward conventional current and lies to the left (west) of wire A, then it will experience a repulsive force to the left (away from Wire A).

Explanation:

Two parallel conductors experience attractive force when the current flowing in the conductors are in the same direction.

Also two parallel conductors experience repulsive force when the current flowing in the conductors are in opposite direction.

Therefore, b and d are the correct options.

b. If Wire B carries a northward conventional current and lies to the left (west) of wire A, then it will experience an attractive force to the right (towards Wire A).

d. If Wire B carries a southward conventional current and lies to the left (west) of wire A, then it will experience a repulsive force to the left (away from Wire A).

3 0
2 years ago
The greatest height reported for a jump into an airbag is 99.4 m by stuntman Dan Koko. In 1948 he jumped from rest from the top
vaieri [72.5K]

Answer:

v = 44,16 m/s

Explanation:

We will fixate our reference in the starting point from where Dan jumped of, at the top of the Casino. Therefore, the displacement made when dan reached the airbag would be of y= -99,4 m viewed from our reference. We describe the motion of dan with the equation:

v_y^2 =v_0^2 +2ay

Dan jumped from the rest, that means that the initial velocity v_0=0, therefore:

 v_y^2 =2ay \rightarrow v_y = \sqrt{2ay}

Since Dan is moving in the negative axis regarding our reference point, we take the negative root of the equation.

v_y=-√(2*(-9,81 m/s^2 )*(-99,4 m) )=44,1613 m/s  v_y =- \sqrt{2*(-9,81 m/s^2)*(-99,4 m)} = 44,1613 m

So, if we don’t take the air resistance into account, Dan would have achieved an velocity of 44,16 m/s when he reached the airbag.

I hope everything was clear with my explanation. If you need anything else, let me know. Have a great day :D

7 0
3 years ago
What is the net displacement of the particle between 0 seconds and 80seconds?
Ira Lisetskai [31]

Answer:

Option (A)

Explanation:

Displacement of a particle on a velocity time graph is represented by the area between the line representing velocity and x-axis (time).

Displacement of a particle from t = 0 o t = 40 seconds = Area of ΔAOB

Area of triangle AOB = \frac{1}{2}(\text{Base})(\text{height})

                                   = \frac{1}{2}(40)(4)

                                   = 80 m

Similarly, displacement of the particle from t = 40 to t = 80 seconds = Area of ΔBCD

Area of ΔBCD = \frac{1}{2}(40)(4)

                        = 80 m

Total displacement of the particle from t = 0 to t = 80 seconds,

= 80 + 80

= 160 m

Option (A) will be the answer.

7 0
3 years ago
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