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3 years ago

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The voltage entering a transformer’s primary winding is 120 volts. The primary winding is wrapped around the iron core 10 times.

The secondary coil is wrapped around the core 4 times. What is the voltage leaving the transformer?

2 answers:

kolbaska11 [484]3 years ago

8 0

the answer is for the question on edge is 48..................

loris [4]3 years ago

5 0

<u>Answer</u>

48 Volts

<u>Explanation</u>
The question can be solve using the turn rule of a transformer that states;

Np/Ns = Vp/Vs
Where Np ⇒ number of turns in the primary coil.
Ns ⇒number of turns in the seconndary coil
Vp ⇒ primary voltage
Vs ⇒secondary voltage

Np/Ns = Vp/Vs

10/4 = 120/Vp

Vp = (120 × 4)/10

= 480/10
= 48 Volts

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A wheel rotating about a fixed axis with a constant angular acceleration of 2.0 rad/s2 starts from rest at t = 0. The wheel has

irga5000 [103]

Answer:

The total linear acceleration is approximately 0.246 meters per square second.

Explanation:

The total linear acceleration ( $a$ ) consist in two components, <em>radial</em> ( $a_{r}$ ) and <em>tangential</em> ( $a_{t}$ ), in meters per square second:

$a_{r} = \omega^{2}\cdot r$ (1)

$a_{t} = \alpha \cdot r$ (2)

Since both components are orthogonal to each other, the total linear acceleration is determined by Pythagorean Theorem:

$a = \sqrt{a_{r}^{2}+a_{t}^{2}}$ (3)

Where:

$r$ - Radius of the wheel, in meters.

$\omega$ - Angular speed, in radians per second.

$\alpha$ - Angular acceleration, in radians per square second.

Given that wheel accelerates uniformly, we use the following kinematic equation:

$\omega = \omega_{o}+ \alpha\cdot t$ (4)

Where:

$\omega_{o}$ - Initial angular speed, in radians per second.

$t$ - Time, in seconds.

If we know that $r = 0.1\,m$ , $\alpha = 2\,\frac{rad}{s^{2}}$ , $\omega_{o} = 0\,\frac{rad}{s}$ and $t = 0.60\,s$ , then the total linear acceleration is:

$\omega = \omega_{o}+ \alpha\cdot t$

$\omega = 1.2\,\frac{rad}{s}$

$a_{r} = \omega^{2}\cdot r$

$a_{r} = 0.144\,\frac{m}{s^{2}}$

$a_{t} = \alpha \cdot r$

$a_{t} = 0.2\,\frac{m}{s^{2}}$

$a = \sqrt{a_{r}^{2}+a_{t}^{2}}$

$a \approx 0.246\,\frac{m}{s^{2}}$

The total linear acceleration is approximately 0.246 meters per square second.

3 0

3 years ago

Convert the following to relative uncertainties <br>a) 2.70 ± 0.05cm<br>b) 12.02 ± 0.08cm

DENIUS [597]

data which is expressed in form of following way

$a = a_o + \Delta a$

here in above expression

$a_o$ = true value

$\Delta a$ = uncertainty in the value

now the relative uncertainty is given as

$\frac{\Delta a}{a_o}$

now by above formula we can say

a) 2.70 ± 0.05cm

here

True value = 2.70

uncertainty = 0.05

Relative uncertainty = $\frac{0.05}{2.70}$ = 0.0185

b) 12.02 ± 0.08cm

here

True value = 12.02

uncertainty = 0.08

Relative uncertainty = $\frac{0.08}{12.02}$ = 0.00665

4 0

3 years ago

"My distance to the center of the earth is about 4000 miles when I am on the surface. If I go to a height of 8000 miles above th

lord [1]

Given,

Distance from the surface to the center of the earth, d=4000 miles

Distance from the center to you at a height of 8000 miles, a= 8000+4000=12000 miles

The gravitational force acting on a person at the surface is equal to his weight.

From Newton's Universal Law of Gravitation, the gravitational force is

$F=\frac{G\times M\times m}{r^2}$

Where G is the gravitational constant, M is the mass of the earth, m is the mass of the object/person, r is the distance between the center of the earth and the object/person

At the surface, this force is equal to the weight of the person, W=mg

i.e.

$F_s=\frac{G\times M\times m}{d^2}=W$

On substituting the of d,

$W=\frac{\text{GMm}}{4000^2}$

At a height of 8000 miles from the surface, the gravitational force is equal to,

$F_a=\frac{GMm}{12000^2}$

On dividing the above two equations,

$\frac{F_a}{W}=\frac{4000^2^{}}{12000^2}=\frac{1}{9}$

Therefore,

$F_a=\frac{1}{9}W$

Therefore at a height of 8000 miles above the surface of the earth, the force of gravity becomes 1/9 time your weight.

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1 year ago

Calculate the orbital speed (in m/s) of a satellite that circles the Earth with a time period of 12.00 hours. The mass of the Ea

Hoochie [10]

Answer:

$v = 3869 m/s$

Explanation:

As we know that the orbital speed of the satellite is given as

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also we know that

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$T = \frac{2\pi r}{v}$

now from above equation we know that

$T = \frac{2\pi (\frac{GM}{v^2})}{v}$

$T = \frac{2\pi GM}{v^3}$

so we will have

$v = (\frac{2\pi GM}{T})^{1/3}$

now plug in all data in this equation

$v = (\frac{2\pi (6.67 \times 10^{-11})(5.97 \times 10^{24})}{12 \times 3600})^{1/3}$

$v = 3869 m/s$

3 0

3 years ago

A 120-kg object and a 420-kg object are separated by 3.00 m At what position (other than an infinitely remote one) can the 51.0-

djverab [1.8K]

Answer:

1.045 m from 120 kg

Explanation:

m1 = 120 kg

m2 = 420 kg

m = 51 kg

d = 3 m

Let m is placed at a distance y from 120 kg so that the net force on 51 kg is zero.

By use of the gravitational force

Force on m due to m1 is equal to the force on m due to m2.

$\frac{Gm_{1}m}{y^{2}}=\frac{Gm_{2}m}{\left ( d-y \right )^{2}}$

$\frac{m_{1}}{y^{2}}=\frac{m_{2}}{\left ( d-y \right )^{2}}$

$\frac{3-y}{y}=\sqrt{\frac{7}{2}}$

3 - y = 1.87 y

3 = 2.87 y

y = 1.045 m

Thus, the net force on 51 kg is zero if it is placed at a distance of 1.045 m from 120 kg.

6 0

3 years ago