All categories

2 years ago

21. An AC generator has an rms output voltage of 274 V at a frequency of 57.0 Hz.

Physics

1 answer:

maks197457 [2]2 years ago

4 0

The answers are as follows;

a) the inductive reactance is 322 ohm

b) The maximum voltage is 387.5 V

c) The rms and maximum currents in the inductor are 1.2 A and 0.85 A.

<h3>What is the reactance?</h3>

The reactance is obtained from;

XL = 2πfL

XL = 2 * 3.14 * 57.0 * 0.900

XL = 322 ohm

The maximum voltage is obtained as;

Vo = Vrms * √2

Vo = 274 V * √2

Vo = 387.5 V

Io = Vo/XL

Io = 387.5 V/ 322 ohm

Io = 1.2 A

Irms = 274 V/322 ohm

Irms = 0.85 A

Learn more about inductive reactance:brainly.com/question/17129912

#SPJ1

You might be interested in

A hollow glass prism doesnot cause the dispersion of light why?<br><br>Please HELP! Its urgent..

eimsori [14]

Answer:

When white light passes through the prism ,different colours suffers deviation through different angles and light appears to be dispersed. But in a hollow prism dispersion does not take place as all the colours travel with same speed in the air inside hollow prism. Thus no angular dispersion is there.

4 0

3 years ago

Read 2 more answers

A 3.1 kg ball is thrown straight upward with a speed of 18.2 m/s. Use conservation of energy to calculate the maximum height the

spayn [35]

Answer:

h = 16.9 m

Explanation:

When a ball is thrown upward, its velocity gradually decreases, until it stops for a moment, when it reaches the maximum height, while its height increases. Thus, the law conservation of energy states in this case, that:

Kinetic Energy Lost by Ball = Potential Energy Gained by Ball

(0.5)m(Vf² - Vi²) = mgh

h = (0.5)(Vf² - Vi²)/g

where,

Vf = Final Speed of Ball = 0 m/s (Since, ball stops for a moment at highest point)

Vi = Initial Speed of Ball = 18.2 m/s

g = acceleration due to gravity = - 9.8 m/s² ( negative for upward motion)

h = maximum height the ball can reach = ?

Therefore, using values in the equation, we get:

h = (0.5)[(0 m/s)² - (18.2 m/s)²]/(-9.8 m/s²)

4 0

3 years ago

WILL GIVE BRAINLIEST

lisabon 2012 [21]

B 200N; 200N/10kg=20m/s2

4 0

4 years ago

Two ships leave a harbor at the same time. One ship travels on a bearing S 13 degrees W at 16 miles per hour. The other ship tra

Artemon [7]

Answer:

47 miles

Explanation:

From the diagrammatic representation below;we can observe the illustration of the question.

after two hours;

The first ship = 16 miles per hour × 2 hours = 32 hours

the second ship = 11 miles per hour × 2 hours = 22 hours

to solve for ΔQPR in the diagram; we have;

∠ P = 13° + (2nd quadrant) + 15°

∠ P = 13° + 90° + 15°

∠ P = 118°

To solve for the distance apart the two ship, we can see from the diagram that it is "p". So, using cosine rule; we have:

p² = q² + r² - 2qr Cos P

p² = 22² + 32² - 2(22)(32) Cos 118°

p² = 484 + 1024 - (1408) (-0.4695)

p² = 1508 + 661.056

p² = 2169.056

p = $\sqrt{2169.056}$

p = 46.5731252

p ≅ 47 miles

∴ The ships will be 47 miles far apart after 2 hours.

5 0

4 years ago

A car traveling at a velocity v can stop in a minimum distance d. What would be the car's minimum stopping distance if it were t

alexira [117]

Answer:

a. 4d.

If the car travels at a velocity of 2v, the minimum stopping distance will be 4d.

Explanation:

Hi there!

The equations of distance and velocity of the car are the following:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position of the car at time t.

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

v = velocity of the car at time t.

Let´s find the time it takes the car to stop using the equation of velocity. When the car stops, its velocity is zero. Then:

velocity = v0 + a · t v0 = v

0 = v + a · t

Solving for t:

-v/a = t

Since the acceleration is negative because the car is stopping:

v/a = t

Now replacing t = v/a in the equation of position:

x = x0 + v0 · t + 1/2 · a · t² (let´s consider x0 = 0)

x = v · (v/a) + 1/2 · (-a) (v/a)²

x = v²/a - 1/2 · v²/a

x = 1/2 v²/a

At a velocity of v, the stopping distance is 1/2 v²/a = d

Now, let´s do the same calculations with an initial velocity v0 = 2v:

Using the equation of velocity:

velocity = v0 + a · t

0 = 2v - a · t

-2v/-a = t

t = 2v/a

Replacing in the equation of position:

x1 = x0 + v0 · t + 1/2 · a · t²

x1 = 2v · (2v/a) + 1/2 · (-a) · (2v/a)²

x1 = 4v²/a - 2v²/a

x1 = 2v²/a

x1 = 4(1/2 v²/a)

x1 = 4x

x1 = 4d

If the car travels at a velocity 2v, the minimum stopping distance will be 4d.

5 0

3 years ago