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rusak2 [61]
3 years ago
6

Un transformador reductor se usa para convertir un voltaje alterno de 10000 V a 500V ¿Cuál deberá ser la razón (N1/N2) entre las

espiras si la corriente en el primario es de 1 amp. Y el transformador tiene un rendimiento del 100%. ¿Cuál es la corriente de salida?
Physics
1 answer:
Vedmedyk [2.9K]3 years ago
4 0

Answer:

(a). The value of \dfrac{N_{1}}{N_{2}} is \dfrac{1}{20}

(b). The output current is 0.05 A

Explanation:

Given that,

Secondary voltage = 10000 V

Primary voltage = 500 V

Primary current = 1 A

(a). We need to calculate the value of \dfrac{N_{1}}{N_{2}}

Put the value into the formula

\dfrac{V_{s}}{V_{p}}=\dfrac{N_{2}}{N_{1}}

\dfrac{N_{1}}{N_{2}}=\dfrac{V_{p}}{V_{s}}

Put the value into the formula

\dfrac{N_{1}}{N_{2}}=\dfrac{500}{10000}

\dfrac{N_{1}}{N_{2}}=\dfrac{1}{20}

(b). We need to calculate the output current

Using formula of current

\dfrac{I_{s}}{I_{p}}=\dfrac{N_{1}}{N_{2}}

I_{s}=I_{p}\times\dfrac{N_{1}}{N_{2}}

Put the value into the formula

I_{s}=1\times\dfrac{1}{20}

I_{s}=0.05\ A

Hence, (a). The value of \dfrac{N_{1}}{N_{2}} is \dfrac{1}{20}

(b). The output current is 0.05 A

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Ipatiy [6.2K]

The first part of the text is missing, you can find on google:

"A ball is thrown horizontally from the roof of a building 45 m. If it strikes the ground 56 m away, find the following values."

Let's now solve the different parts.

(a) 3.03 s

The time of flight can be found by analyzing the vertical motion only. The vertical displacement at time t is given by

y(t) = h -\frac{1}{2}gt^2

where

h = 45 m is the initial height

g = 9.8 m/s^2 is the acceleration of gravity

When y=0, the ball reaches the ground, so the time taken for this to happen can be found by substituting y=0 and solving for the time:

0=h-\frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(45)}{9.8}}=3.03 s

(b) 18.5 m/s

For this part, we need to analyze the horizontal motion only, which is a uniform motion at constant speed.

The horizontal position is given by

x=v_x t

where

v_x is the horizontal speed, which is constant

t is the time

At t = 3.03 s (time of flight), we know that the horizontal position is x = 56 m. By substituting these numbers and solving for vx, we find the horizontal speed:

v_x = \frac{x}{t}=\frac{56}{3.03}=18.5 m/s

The ball was thrown horizontally: this means that its initial vertical speed was zero, so 18.5 m/s was also its initial overall speed.

(c) 35.0 m/s at 58.1 degrees below the horizontal

At the impact, we know that the horizontal speed is still the same:

v_x = 18.5 m/s

we need to find the vertical velocity. This can be done by using the equation

v_y = u_y -gt

where

u_y =0 is the initial vertical velocity

g is the acceleration of gravity

t is the time

Substituting t = 3.03 s, we find the vertical velocity at the time of impact:

v_y = -(9.8)(3.03)=-29.7 m/s

So the magnitude of the velocity at the impact (so, the speed at the impact) is

v=\sqrt{v_x^2+v_y^2}=\sqrt{18.5^2+(-29.7)^2}=35.0 m/s

The angle instead can be found as:

\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{-29.7}{18.5})=-58.1^{\circ}

so, 58.1 degrees below the horizontal.

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