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beks73 [17]
3 years ago
11

A disk drive plugged into a 120V outlet operates on a voltage of 12V . The transformer that powers the disk drive has 125 turns

on its primary coil.
Part A) Should the number of turns on the secondary coil be greater than or less than 125? greater than or less than
Part B) Explain
Part C) Find the number of turns on the secondary coil.
Physics
1 answer:
Serga [27]3 years ago
6 0

Answer:

(A-) Number of turns in secondary coil will be<u> less than 125.</u>

(B-) EXPLAIN- The voltage is to be lowered.<u> The voltage will be reduced from 120 to 9 volts.</u> As a result, a<u> step down transformer </u>will be used, indicating that the secondary coil's number of turns would be less than 125.

(C-)<u> The number of turns on the secondary coil is 12.5</u>

<u></u>

Explanation:

  • STEP DOWN TRANSFORMER - A step-down transformer transforms high voltage (HV) and low current from the transformer's primary side to low voltage (LV) and high current on the secondary side. A phase down transformer is the opposite of this.

In electrical networks and transmission lines, a step-down transformer can be used in a number of ways.

<u>Step down transformer formula-</u>

                  V_s=\frac{N_s\times V_p}{N_p}

We are given with the -

Primary voltage V_p = 120V

Secondary voltage Vs=12V

Number of primary turns N_p=125

  • <u>To calculate the number of turns on the secondary coil -</u>

       Since , the transformer ratio is given as -

                   \frac{N_s}{N_p} =\frac{V_s}{V_p}

So, the number of turns on the secondary coil is - N_s=N_p\times\frac{V_p}{V_s}

            Putting , the given values -:

                125\times\frac{12}{120}

         =   12.5

<u>Therefore , the number of turns on secondary coil is 12.5 .</u>

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4. I drop a pufferfish of mass 5 kg from a height of 5.5 m onto an upright spring of total length 0.5 m and spring constant 3000
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a)  0.28 m or 28 cm is the minimum  height above ground the fish reaches.

b)  at the height of 0.484 m height , the pufferfish will eventually come to rest.

c) There exists  two types of energy remain at the equilibrium point in the system. These are :

Gravitational potential energy  = 23.72J

Spring potential energy   = 0.384 J

Explanation:

Given that :

Mass of the pufferfish m =5kg

initial height of the fish h =5.5m

length of the spring l =0.5m

Spring constant K =3000N/m

a)

Assuming no energy loss to friction, what is the minimum height above the ground that the pufferfish reaches?

Lets assume that the minimum height the fish reaches is = x meters

Now by using the conservation of energy; we realize that :

Initial total energy = final total energy

Gravitational potential energy =

Gravitational potential energy' + Spring potential energy (kinetic energy is zero in both cases)

mgh = mgx + \frac{1}{2}K(l-x)^2

Replacing our given values into the above equation; we have :

(5)(9.8)(5.5) = (5)(9.5)(x) + \frac{1}{2}(3000)(0.5-x)^2

269.5 = 47.5 x + 1500(0.5 -x )²

269.5 = 47.5 x + 1500(0.25 - x²)

269.5 = 47.5 x + 375 - 1500 x²

269.5 - 375 = 47.5 x - 1500 x²

-105.5 = 47.5 x - 1500 x²

-105.5 + 1500 x² - 47.5 x = 0

1500 x² - 47.5 x - 105.5 = 0

By using quadratic equation and taking the positive value;

x = 0.28 m or 28 cm is the minimum height above ground the fish reaches.

b)

At the equilibrium position the weight of fish will be equal to the force applied by the spring thus

mg = kx

substituting  our given values ; we have:

(5)(9.8) = 3000x

x = 61.22

x = 0.016m  : so this is the compression in the spring

Now; to determine the height  the pufferfish gets to before  it eventually come to rest; we have

(0.5-0.016) m = 0.484m

therefore, at the height of 0.484 m height , the pufferfish will eventually come to rest.

c)

There exists  two types of energy remain at the equilibrium point in the system. These are :

Gravitational potential energy  = mgh' = (5)(9.8)(0.484)

= 23.72J

and spring potential energy  

=\frac{1}{2}Kx^2\\ = \frac{1}{2}(3000)(0.016)^2\\= 0.384J

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