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DochEvi [55]
3 years ago
12

The distribution circuit of a residential power line is operated at 2000 V rms. This voltage must be reduced to 240 V rms for us

e within the residences. If the secondary side of the transformer has 475 turns, how many turns are in the primary?
Physics
1 answer:
forsale [732]3 years ago
7 0

Answer:

N_p\approx3958

Explanation:

In an ideal transformer the relationship between the voltages is proportional to the ratio between the number of turns of the windings. Thus:

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

Where:

V_p=Voltage\hspace{3}in\hspace{3}primary\hspace{3}coil

V_s=Voltage\hspace{3}in\hspace{3}secondary\hspace{3}coil

N_p=Turns\hspace{3}on\hspace{3}primary\hspace{3}coil

N_s=Turns\hspace{3}on\hspace{3}secondary\hspace{3}coil

So, solving for N_p

N_p=N_s*\frac{V_p}{V_s} =475*\frac{2000}{240} =3958.333333\approx3958

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Glare appears on a computer screen when light from the surroundings reflects off of the screen’s surface. Some computer screens
hoa [83]

Answer:

The correct answer is D.

D:The surface of the coating allows light from the room to pass through but blocks the light from the screen.

Explanation:

Glare is produced on a computer screen when light from some external source reflects on the screen.

Anti-glare coating do not absorb light to reduce glare but they actually reduce glare by encouraging the light from the room to pass through the screen so that less light is reflected. Polarized lenses absorbs light to reduce glare, not anti-glare coating.

7 0
3 years ago
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You build a grandfather clock, whose timing is based on a pendulum. You measure its period to be 2s on Earth. You then travel wi
Elenna [48]

Answer:

\frac{g_{2}}{g_{1}} = \frac{1}{4}

Explanation:

The period of the simple pendulum is:

T = 2\pi\cdot \sqrt{\frac{l}{g} }

Where:

l - Cord length, in m.

g - Gravity constant, in \frac{m}{s^{2}}.

Given that the same pendulum is test on each planet, the following relation is formed:

T_{1}^{2}\cdot g_{1} = T_{2}^{2}\cdot g_{2}

The ratio of the gravitational constant on planet CornTeen to the gravitational constant on planet Earth is:

\frac{g_{2}}{g_{1}} = \left(\frac{T_{1}}{T_{2}} \right)^{2}

\frac{g_{2}}{g_{1}} = \left(\frac{2\,s}{4\,s} \right)^{2}

\frac{g_{2}}{g_{1}} = \frac{1}{4}

5 0
3 years ago
A world-class sprinter can burst out of the blocks to essentially top speed (of about 11.5 m/s) in the first 15.0 m of the race.
wolverine [178]
Given:
u = 0, initial speed (sprinter starts from rest)
v = 11.5 m/s, final speed
s = 15 m, distance traveled to attain final speed.

Let
a =  average acceleration,
t = time taken to attain final speed.

Then
v² = u² + 2as
or
(11.5 m/s)² = 2*(a m/s²)*(15 m)
a = 11.5²/(2*15) = 4.408 m/s²

Also
v = u +a t
or
(11.5 m/s) = (4.408 m/s²)*(t s)
t = 11.5/4.408 = 2.609 s

Answer:
The average acceleration is 4.41 m/s² (nearest hundredth).
The time required is 2.61 s (nearest hundredth).
8 0
3 years ago
If 3,600 J of work is done in 3.0 s, what is the power?
Firlakuza [10]

Answer:

1,200 watts

Explanation:

1 watt = 1 Joule (J) of work / second

So, 3600 Joules of work / 3 seconds is:

3600 J / 3 seconds = 1,200 watts

8 0
3 years ago
A centrifugal pump is operating at a flow rate of 1 m3/s and a head of 20 m. If the specific weight of water is 9800 N/m3 and th
tamaranim1 [39]

Answer:

<em>The power required by the pump is nearly 230.588 kW</em>

Explanation:

Flow rate of the pump Q = 1 m^3/s

the head flow H = 20 m

specific weight of water γ = 9800 N/m^3

efficiency of the pump η = 85%

First note that specific gravity of water is the product of the density of water and acceleration due to gravity.

γ = ρg

where ρ is density. For water its value is 1000 kg/m^3

g is the acceleration due to gravity = 9.81 m/s^2

The power to lift this water at this rate will be gotten from the equation

P = ρgQH

but ρg = γ

therefore,

P = γQH

imputing values, we'll have

P = 9800 x 1 x 20 = 196000 W

But the centrifugal pump that will be used will only be able to lift this amount of water after the efficiency factor has been considered. The power of pump needed must be greater than this power.

we can say that

196000 W is 85% of the power of the pump power needed, therefore

196000 = 85% of P_{p}

where P_{p} is the power of the pump needed

85% = 0.85

196000 = 0.85P_{p}

P_{p} = 196000/0.85 = 230588.24 W

<em>Pump power = 230.588 kW</em>

3 0
3 years ago
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