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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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2 years ago
A speeder passes a parked police car at a constant speed of 23.3 m/s. At that instant, the police car starts from rest with a un
KonstantinChe [14]

Answer:

32s

Explanation:

We must establish that by the time the police car catches up to the speeder, both have travelled a certain distance during the same amount of time. However, the police car experiences accelerated motion whereas the speeder travels at a constant velocity. Therefore we will establish two formulas for distance starting with the speeder's distance:

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and the police car distance:

x=vt+\frac{at^{2}}{2}=0+\frac{2.75\frac{m}{s^{2}} t^{2}}{2}=0.73\frac{m}{s^{2}}

Since they both travel the same distance x, we can equal both formulas and solve for t:

0 = 0.73\frac{m}{s^{2}}t^{2}-23.3\frac{m}{s} t\\\\0=t(0.73\frac{m}{s^{2}}t-23.3\frac{m}{s} )\\\\

Two solutions exist to the equation; the first one being t=0

The second solution will be:

0.73\frac{m}{s^{2}}t=23.3\frac{m}{s}\\\\t=\frac{23.3\frac{m}{s}}{0.73\frac{m}{s^{2}}}=32s

This result allows us to confirm that the police car will take 32s to catch up to the speeder

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3 years ago
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7 0
3 years ago
Kyle is flying a helicopter at 125 m/s on a heading of 325 o . If a wind is blowing at 25 m/s toward a direction of 240.0 o , wh
frosja888 [35]

Answer:

The resultant velocity of the helicopter is \vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right).

Explanation:

Physically speaking, the resulting velocity of the helicopter (\vec v_{H}), measured in meters per second, is equal to the absolute velocity of the wind (\vec v_{W}), measured in meters per second, plus the velocity of the helicopter relative to wind (\vec v_{H/W}), also call velocity at still air, measured in meters per second. That is:

\vec v_{H} = \vec v_{W}+\vec v_{H/W} (1)

In addition, vectors in rectangular form are defined by the following expression:

\vec v = \|\vec v\| \cdot (\cos \alpha, \sin \alpha) (2)

Where:

\|\vec v\| - Magnitude, measured in meters per second.

\alpha - Direction angle, measured in sexagesimal degrees.

Then, (1) is expanded by applying (2):

\vec v_{H} = \|\vec v_{W}\| \cdot (\cos \alpha_{W},\sin \alpha_{W}) +\|\vec v_{H/W}\| \cdot (\cos \alpha_{H/W},\sin \alpha_{H/W}) (3)

\vec v_{H} = \left(\|\vec v_{W}\|\cdot \cos \alpha_{W}+\|\vec v_{H/W}\|\cdot \cos \alpha_{H/W}, \|\vec v_{W}\|\cdot \sin \alpha_{W}+\|\vec v_{H/W}\|\cdot \sin \alpha_{H/W} \right)

If we know that \|\vec v_{W}\| = 25\,\frac{m}{s}, \|\vec v_{H/W}\| = 125\,\frac{m}{s}, \alpha_{W} = 240^{\circ} and \alpha_{H/W} = 325^{\circ}, then the resulting velocity of the helicopter is:

\vec v_{H} = \left(\left(25\,\frac{m}{s} \right)\cdot \cos 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \cos 325^{\circ}, \left(25\,\frac{m}{s} \right)\cdot \sin 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \sin 325^{\circ}\right)\vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right)

The resultant velocity of the helicopter is \vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right).

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3 years ago
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