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3 years ago

5

A transformer has 150 turns in the primary coil and 350 turns in its secondary coil. If the primary coil has a voltage of 200 vo

lts, how many volts will the secondary coil have?
242 volts
288
353
467

1 answer:

Ghella [55]3 years ago

4 0

Answer:

467 volts

Explanation:

Vs/Vp = Ns/Np

Vs = Ns/Np × Vp

Vs = 350/150 × 200 = 7/3 × 200

Vs = 467 volts

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Indica qué es una propiedad específica de la materia. Además explica por qué son útiles las propiedades específicas de la materi

Katyanochek1 [597]

Answer:

Check Explanation

Comprobar explicación

Explanation:

English Translation

Indicate what a specific property of matter is. Also explain why the specific properties of matter are useful compared to the general ones.

Solution

The specific properties of matter are properties that describes the intensive properties of the system. They are properties that do not depend on or change with the extent or size of the system. They are usually obtained by dividing the generalised properties or extensive properties by the extent or size of matter to make them independent of size/extent/Mass.

Examples of specific properties include specific heat capacity, specific volume etc. They usually have units of general units/Mass units.

The specific properties of matter are more important than the general ones because

- They help in general comparisons of the properties of different materials. They are used to rank, classify and compare properties of different materials.

- They are used in reference table/data to easily record easily accessible properties of matter. It helps to record standards that are general and independent of sizes/extents/Mass, thereby keeping the reference table/data/chart precise and concise.

- They provide us with values that are easy to memorize and remember, unlike trying to cram the different properties of different masses/sizes of matter.

In Spanish/En español

Las propiedades específicas de la materia son propiedades que describen las propiedades intensivas del sistema. Son propiedades que no dependen ni cambian con la extensión o el tamaño del sistema. Por lo general, se obtienen dividiendo las propiedades generalizadas o las propiedades extensivas por la extensión o el tamaño de la materia para hacerlas independientes del tamaño / extensión / masa.

Los ejemplos de propiedades específicas incluyen capacidad calorífica específica, volumen específico, etc. Usualmente tienen unidades de unidades generales / unidades de masa.

Las propiedades específicas de la materia son más importantes que las generales porque

- Ayudan en las comparaciones generales de las propiedades de diferentes materiales. Se utilizan para clasificar, clasificar y comparar propiedades de diferentes materiales.

- Se utilizan en la tabla / datos de referencia para registrar fácilmente propiedades de materia fácilmente accesibles. Ayuda a registrar estándares que son generales e independientes de tamaños / extensiones / masa, manteniendo así la tabla / datos / tabla de referencia precisa y concisa.

- Nos proporcionan valores que son fáciles de memorizar y recordar, a diferencia de tratar de agrupar las diferentes propiedades de diferentes masas / tamaños de materia.

Hope this Helps!!!

¡¡¡Espero que esto ayude!!!

7 0

3 years ago

An object of unknown mass is initially at rest and dropped from a height h. It reaches the ground with a velocity v1 . The same

Hatshy [7]

Answer:

$v_2=\sqrt{2}v_1$

Explanation:

The velocity v₁ can be calculated with the kinematic formula:

$v_1^{2} =v_0^{2} +2gh$

Since the object is initially at rest, v₁ becomes:

$v_1=\sqrt{2gh}$

Where g is the acceleration due to gravity. Now, the velocity v₂ can be calculated with the same formula, but now the initial velocity is v₁:

$v_2^{2}=v_1^{2} +2gh$

Substituting v₁ in this expression and solving for v₂, we get:

$v_2^{2}=(\sqrt{2gh} )^{2} +2gh=4gh\\\\\implies v_2=\sqrt{4gh}=2\sqrt{gh}$

Now, dividing v₂ over v₁, we get the expression:

$\frac{v_2}{v_1}=\frac{2\sqrt{gh} }{\sqrt{2gh}}=\sqrt{2}\\ \\\implies v_2=\sqrt{2}v_1$

It means that v₂ is √2 times v₁.

4 0

4 years ago

Suppose the ball is thrown from the same height as in the PRACTICE IT problem at an angle of 32.0°below the horizontal. If it st

scoray [572]

The figure of the problem is missing: find in attachment.

(a) 1.64 s

The ball follows a projectile motion path. The horizontal displacement is given by

$x(t) = v_0 cos \theta t$

where

$v_0$ is the initial speed

t is the time

$\theta=32.0^{\circ}$ is the angle below the horizontal

We can rewrite this equation as

$t=\frac{x(t)}{v_0 cos \theta}$ (1)

The vertical displacement instead is given by

$y(t) = -v_0 sin \theta t - \frac{1}{2}gt^2$ (2)

where

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

Substituting (1) into (2),

$y(t) = -x(t) tan \theta - \frac{1}{2}gt^2$

We know that for t = time of flight, the horizontal displacement is

$x(t) =50.8 m$

We also know that the vertical displacement is

$y(t) = -45 m$

Substituting everything into the equation, we can find the time of flight:

$\frac{1}{2}gt^2=-y -x tan \theta\\t=\sqrt{\frac{2(-y-xtan \theta)}{g}}=\sqrt{\frac{2(-(-45)-50.8 tan 32.0^{\circ})}{9.8}}=1.64 s$

(b) 36.5 m/s

We can now find the initial speed directly by using the equation for the horizontal displacement:

$x(t) = v_0 cos \theta t$

where we have

x = 50.8 m

$\theta=32.0^{\circ}$

Substituting the time of flight,

t = 1.64 s

We find:

$v_0 = \frac{x}{t cos \theta}=\frac{50.8}{(1.64)(cos 32.0^{\circ})}=36.5 m/s$

(c) 47.1 m/s at 48.8 degrees below the horizontal

As the ball follows a projectile motion, its horizontal velocity does not change, so its value remains equal to

$v_x = v_0 cos \theta = (36.5)(cos 32.0^{\circ})=31.0 m/s$

The initial vertical velocity is instead

$u_y = -v_0 sin \theta = -(36.5)(sin 32.0^{\circ})=-19.3 m/s$

And it changes according to the equation

$v_y = u_y -gt$

So at t = 1.64 s (when the ball hits the ground),

$v_y = -19.3 - (9.8)(1.64)=-35.4 m/s$

So the impact speed is:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(31.0)^2+(-35.4)^2}=47.1 m/s$

While the direction is:

$\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{-35.4}{31.0})=-48.8^{\circ}$

8 0

3 years ago

How did this formula come about? Please show me the formula derivation process

kirill [66]

Answer:

so simple it is a square formula

6 0

3 years ago

A capacitor is charges with 9.6 nC and has a 120 V potential difference between its terminals. Compute

IRINA_888 [86]

Explanation:

Q = CV

where C = capacitance

V = potential difference

Solving for C,

C = Q/V = (9.6×10^-9)(120 v)

= 1.15 microFarads

4 0

3 years ago