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Darya [45]
2 years ago
13

The diagram shows a multi-tap transformer. Which pair of terminals would you connect to produce an output voltage of 8 V?​

Physics
1 answer:
Triss [41]2 years ago
7 0

Answer:

The pair of terminals to produce an output voltage of 8V are K and M.

Explanation:

A transformer is an electronic device that can be used to increase or decrease an output voltage. It consists of secondary and primary coils, and operates on the principle of induction.

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

where: V_{s} is the voltage in the secondary coil, V_{p} is the voltage in the primary coil, N_{s} is the number of turns in the secondary coil and N_{p} is the number of turns in the primary coil.

In the question, let us combine the number of turns in the secondary coil. So that,

  N_{s} = 15 + 5 = 20

\frac{V_{s} }{200} = \frac{20}{500}

V_{s} = \frac{4000}{500}

    = 8

V_{s} = 8V

For the output voltage to be equal to 8V, the pair of terminals required are K and M.

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In your own words, define the following terms.<br> 2. climate___________.
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A car of mass m, traveling at constant speed, rides over the top of a round hill. How do the normal force of the road on the car
dybincka [34]

Answer:

The normal force will be lower than the gravitational force acting on the car. Therefore the answer is N < mg, which is <em>option B</em>.

Explanation:

Over a round hill, the centripetal force acting toward the the radius of the hill supports the gravitational force (mg) of the car. This notion can be expressed mathematically as follows:

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Normal force = Gravitational force - centripetal force

At the foot of a round hill

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4 0
3 years ago
The barometric pressure at sea level is 30 in of mercury when that on a mountain top is 29 in. If the specific weight of air is
stealth61 [152]

To solve this problem we will apply the concepts related to pressure, depending on the product between the density of the fluid, the gravity and the depth / height at which it is located.

For mercury, density, gravity and height are defined as

\rho_m = 846lb/ft^3

g = 32.17405ft/s^2

h_1 = 1in = \frac{1}{12} ft

For the air the defined properties would be

\rho_a = 0.0075lb/ft^3

g = 32.17405ft/s^2

h_2 = ?

We have for equilibrium that

\text{Pressure change in Air}=\text{Pressure change in Mercury}

\rho_m g h_1 = \rho_a g h_2

Replacing,

(846)(32.17405)(\frac{1}{12}) = (0.0075)(32.17405)(h_2)

Rearranging to find h_2

h_2 = \frac{(846)(32.17405)(\frac{1}{12}) }{(0.0075)(32.17405)}

h = 9400ft

Therefore the elevation of the mountain top is 9400ft

7 0
2 years ago
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