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3 years ago

6

Consider a step-down transformer with 15 turns in the primary and 6 turns in the secondary windings. Calculate the input impedan

ce of this transformer loaded to an 8 Ω resistor, assuming that series resistances of the windings are negligible.

1 answer:

AURORKA [14]3 years ago

8 0

Answer:

Input impedance of this transformer is 50 ohms.

Explanation:

Given that,

Number of turns in the primary coil, $N_p=15$

Number of turns in the secondary coil, $N_s=6$

Output impedance of the transformer, $V_o=8\ \Omega$

The number of turns and the impedance ratio in the step down transformer is given by :

$\dfrac{15}{6}=\sqrt{\dfrac{Z_p}{Z_s}}\\\\\dfrac{15}{6}=\sqrt{\dfrac{Z_p}{8}}\\\\Z_p=50\ \Omega$

So, the input impedance of this transformer is 50 ohms. Hence, this is the required solution.

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(a) If the coefficient of kinetic friction between tires and dry pavement is 0.80, what is the shortest distance in which you ca

Colt1911 [192]

a) 52.5 m

b) 16.0 m/s

Explanation:

a)

The motion of a car slowed down by friction is a uniformly accelerated motion, so we can use the following suvat equation:

$v^2-u^2=2as$

where

v = 0 is the final velocity (the car comes to a stop)

u = 28.7 m/s is the initial velocity of the car

a is the acceleration

s is the stopping distance

For a car acted upon the force of friction, the acceleration is given by the ratio between the force of friction and the mass of the car, so:

$a=\frac{-\mu mg}{m}=-\mu g$

where:

$\mu=0.80$ is the coefficient of friction

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

Substituting and solving for s, we find:

$s=\frac{v^2-u^2}{-2\mu g}=\frac{0-(28.7)^2}{-2(0.80)(9.8)}=52.5 m$

b)

In this case, the car is moving on a wet road. Therefore, the coefficient of kinetic friction is

$\mu=0.25$

Here we want the stopping distance of the car to remain the same as part a), so

$s=52.5 m$

We can use again the same suvat equation:

$v^2-u^2=2as$

And since the final velocity is zero

u = 0

We can find the initial velocity of the car:

$v=\sqrt{2as}=\sqrt{2\mu gs}=\sqrt{2(0.25)(9.8)(52.5)}=16.0 m/s$

7 0

3 years ago

A ball is spun around in circular motion such that it completes 50 rotations in 25 s. What is the frequency of its rotation? 2.

Elis [28]

Answer:

1. $f = 2 Hz$

2. $f = 0.011 Hz$

3. $f = 0.067 Hz$

4. $t = 4 s$

Explanation:

1. The frequency of rotation is given by:

$f = \frac{\omega}{2\pi}$

Where:

ω: is the angular speed = 50 rotations (revolutions) in 25 s.

We need to convert the units of ω.

$\omega = \frac{50 rev}{25 s}*\frac{2\pi rad}{1 rev} = 4\pi rad/s$

Now, the frequency is:

$f = \frac{4\pi rad/s}{2\pi} = 2 Hz$

2. The frequency is:

We know:

5 laps = 5 revolutions

t: time = 450 s

$f = \frac{\omega}{2\pi} = \frac{\frac{5 rev}{450 s}*\frac{2\pi rad}{1 rev}}{2\pi} = 0.011 Hz$

3. The frequency of the pendulum is:

$f = \frac{\omega}{2\pi} = \frac{\frac{1 rev}{15 s}*\frac{2\pi rad}{1 rev}}{2\pi} = 0.067 Hz$

4. We have:

θ: number of revolutions = 48 rev

f = 12 Hz

t =?

The time can be calculated as follows:

$f = \frac{\omega}{2\pi} = \frac{\theta}{2\pi t}$

$t = \frac{\theta}{2\pi f} = \frac{48 rev*\frac{2\pi rad}{1 rev}}{2\pi*12 Hz} = 4 s$

I hope it helps you!

4 0

3 years ago

We beat a blanket with a stick to remove dust particles why

lutik1710 [3]

Answer:

removing the dust particles cleans the blanket

6 0

3 years ago

Poverty and backwardness are also considered as the cause of rapid population growth. Why????

alexandr402 [8]

because poor and people of back places may not have enough knowledge about contraceptive devices and may not have enough money to buy it.That's why poverty and backwardness are considered to be a cause of population growth

6 0

3 years ago

You are driving your car along a country road at a speed of 27.0 m/s. as you come over the crest of a hill, you notice a farm tr

Bezzdna [24]

speed of the car = 27 m/s

speed of truck ahead = 10 m/s

relative speed of car with respect to truck

$v_r = 27 - 10 = 17 m/s$

relative deceleration of car

$a_r = -7 m/s^2$

now the distance before they stop with respect to each other is given by

$v_f^2 - v_i^2 = 2 a d$

$0 - 17^2 = 2 *(-7)*d$

$d = 20.6 m$

so it will come at the same speed of truck after 20.6 m distance and hence it will not hit the truck as the distance of the truck is 25 m from car

Part b)

Distance traveled by car before it stops is given by

$v_f^2 - v_i^2 = 2 a s$

$0^2 - 27^2 = 2 * (-7)* s$

$s = 52.1 m$

so it will stop after it will cover total 52.1 m distance

Part c)

time taken by the car to stop

$v_f - v_i = at$

$0 - 27 = (-7) * t$

$t = 3.86 s$

now the distance covered by truck in same time

$d = 3.86 * 10 = 38.6 m$

now after the car will stop its distance from the truck is

$D = 25 + 38.6 - 52.1 = 11.5 m$

<em>so the distance between them is 11.5 m</em>

6 0

3 years ago