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Amanda [17]
3 years ago
14

A 9.0-V battery (with nonzero resistance) and switch are connected in series across the primary coil of a transformer. The secon

dary coil is connected to a light bulb that operates on 120 V. Determine the ratio of the secondary to primary turns needed for the bells transformer. Determine the ratio of the secondary to primary turns needed for the bells transformer. Ns/Np=?
Physics
1 answer:
qwelly [4]3 years ago
3 0

Answer:

N₂ / N₁ = 13.3

Explanation:

A transformer is a system that induces a voltage in the secondary due to the variation of voltage in the primary, the ratio of voltages is determined by the expression

           ΔV₂ = N₂ /N₁  ΔV₁

where ΔV₂ and ΔV₁ are the voltage in the secondary and primary respectively and N is the number of windings on each side.

In this case, they indicate that the primary voltage is 9.0 V and the secondary voltage is 120 V

therefore we calculate the winding ratio

         ΔV₂ /ΔV₁ = N₂ / N₁

         N₂ / N₁ = 120/9

         N₂ / N₁ = 13.3

s good clarify that in transformers the voltage must be alternating (AC)

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Which of the following is the most important factor in determining climate
olga_2 [115]

Answer:

temperature

The two most important factors in the climate of an area are temperature and precipitation. The yearly average temperature of the area is obviously important, but the yearly range in temperature is also important.

Explanation:

7 0
3 years ago
An insulating cup contains 200 grams of water at 25 ∘C. Some ice cubes at 0 ∘C is placed in the water. The system comes to equil
Nataly [62]

Answer:

The amount of ice added in gram is 32.77g

Explanation:

This problem bothers on the heat capacity of materials

Given data

Mass of water Mw= 200g

Temperature of water θw= 25°c

Temperature of ice θice= 0°c

Equilibrium Temperature θe= 12°c

Mass of ice Mi=???

The specific heat of ice Ci= 2090 J/(kg ∘C)

specific heat of water Cw = 4186 J/(kg ∘C)

latent heat of the ice to water transition Li= 3.33 x10^5 J/kg

heat heat loss by water = heat gained by ice

N/B let us understand something, heat gained by ice is in two phases

Heat require to melt ice at 0°C to water at 0°C

And the heat required to take water from 0°C to equilibrium temperature

Hence

MwCwΔθ=MiLi +MiCiΔθ

Substituting our data we have

200*4186*(25-12)=Mi*3.3x10^5+

Mi*2090(12-0)

837200*13=Mi*3.3x10^5+Mi*2090

10883600=332090Mi

Mi=10883600/332090

Mi= 32.77g

4 0
4 years ago
Water enters a baseboard radiator at 180 °F and at a flow rate of 2.0 gpm. Assuming the radiator releases heat into the room at
beks73 [17]

Answer:

Temperature of water leaving the radiator = 160°F

Explanation:

Heat released = (ṁcΔT)

Heat released = 20000 btu/hr = 5861.42 W

ṁ = mass flowrate = density × volumetric flow rate

Volumetric flowrate = 2 gallons/min = 0.000126 m³/s; density of water = 1000 kg/m³

ṁ = 1000 × 0.000126 = 0.126 kg/s

c = specific heat capacity for water = 4200 J/kg.K

H = ṁcΔT = 5861.42

ΔT = 5861.42/(0.126 × 4200) = 11.08 K = 11.08°C

And in change in temperature terms,

10°C= 18°F

11.08°C = 11.08 × 18/10 = 20°F

ΔT = T₁ - T₂

20 = 180 - T₂

T₂ = 160°F

8 0
3 years ago
You have two resistors with the same cross sectional area and resistivity. Resistor A has length L1 and resistor B has length L2
oee [108]

Answer:

Explanation:

Given

Resistor A has length L_1

and Resistor B has Length L_2

and Resistance is given by

R=\frac{\rho L}{A}

Considering \rhoand A to be constant thus

R_2>R_1 because L_2>L_1

(a)When they are connected in series

As the current in series is same and power is i^2R

therefore P_2>P_1 as R is greater for second resistor

(b)if they are connected in Parallel

In Parallel connection Voltage is same

P=\frac{V^2}{R}

resistance of 2 is greater than 1 thus Power delivered by 1 is greater than 2

8 0
3 years ago
What item can be arranged to transform electrical energy to mechanical energy
meriva

Answer:

battery iron wire

Explanation:

7 0
3 years ago
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