1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Travka [436]
2 years ago
13

In an a.c generator, there are rings used. What are there names and how are they adapted to reversing current.​

Physics
2 answers:
lutik1710 [3]2 years ago
7 0
  • The slip rings are used in The AC generator .
  • The slip rings maintain constant contact with the same sides of the coil which help them reversing current .

AC generator works on The Faraday's Law

Digram attached .

djyliett [7]2 years ago
6 0

Answer:

In an a.c generator, slip rings are used.

Adaptation: <em>They</em><em> </em><em>are</em><em> </em><em>not</em><em> </em><em>directly</em><em> </em><em>connected</em><em> </em><em>directly</em><em> </em><em>to</em><em> </em><em>carbon</em><em> </em><em>brushes</em><em>,</em><em> </em><em>hence</em><em> </em><em>donot</em><em> </em><em>come</em><em> </em><em>into</em><em> </em><em>contact</em><em> </em><em>when</em><em> </em><em>the</em><em> </em><em>rectangular</em><em> </em><em>coil</em><em> </em><em>rotates</em><em>.</em>

Explanation:

.

You might be interested in
If the mass of a planet is 0.231 mE and its radius is 0.528 rE, estimate the gravitational field g at the surface of the planet.
crimeas [40]

Answer:

8.1 m/s^2

Explanation:

The strength of the gravitational field at the surface of a planet is given by

g=\frac{GM}{R^2} (1)

where

G is the gravitational constant

M is the mass of the planet

R is the radius of the planet

For the Earth:

g_E = \frac{GM_E}{R_E^2}=9.8 m/s^2

For the unknown planet,

M_X = 0.231 M_E\\R_X = 0.528 R_E

Substituting into the eq.(1), we find the gravitational acceleration of planet X relative to that of the Earth:

g_X = \frac{GM_X}{R_X^2}=\frac{G(0.231M_E)}{(0.528R_E)^2}=\frac{0.231}{0.528^2}(\frac{GM_E}{R_E^2})=0.829 g_E

And substituting g = 9.8 m/s^2,

g_X = 0.829(9.8)=8.1 m/s^2

3 0
2 years ago
A loop of wire is carrying current of 2 A . The radius of the loop is 0.4 m. What is the magnetic field at a distance 0.09 m alo
HACTEHA [7]

Answer:

B=2.91\ \mu T

Explanation:

Given that,

The current in the loop, I = 2 A

The radius of the loop, r = 0.4 m

We need to find the magnetic field at a distance 0.09 m along the axis and above the center of the loop. The formula for the magnetic field at some distance is given as follows :

B=\dfrac{\mu_o}{4\pi }\dfrac{2\pi r^2 I}{(r^2+d^2)^{3/2}}

Put all the values,

B=10^{-7}\times \dfrac{2\pi \times 0.4^2 \times 2}{(0.4^2+0.09^2)^{3/2}}\\\\=2.91\times 10^{-6}\ T\\\\or\\\\B=2.91\ \mu T

So, the required magnetic field is equal to 2.91\ \mu T.

3 0
2 years ago
The starships of the Solar Federation are marked with the symbol of the Federation, a circle, whereas starships of the Denebian
Llana [10]

Complete question

The complete question is shown on the first uploaded image  

Answer:

The velocity is  v = c* \sqrt{1 -  \frac{1}{n^2} }

Explanation:

From the question we are told that

           a = nb

The length of the minor axis  of  the symbol of the Federation, a circle, seen by the observer at velocity v must be equal to the minor axis(b) of the  Empire's symbol, (an ellipse)

Now this length seen by the observer can be mathematically represented as

        h = t \sqrt{1 - \frac{v^2}{c^2} }

Here t  is the actual length of the major axis of of the  Empire's symbol, (an ellipse)

So t = a = nb

and  b is the length of the minor axis of the symbol of the Federation, (a circle) when seen by an observer at velocity v which from the question must be the length of the minor axis of the of the  Empire's symbol, (an ellipse)

 i.e    h = b

So

    b  =  nb  [\sqrt{1 - \frac{v^2}{c^2} } ]  

     [\frac{1}{n} ]^2 =  1 -  \frac{v^2}{c^2}

      v^2 =c^2 [1- \frac{1}{n^2} ]

       v^2 =c^2 [\frac{n^2 -1}{n^2} ]

        v = c* \sqrt{1 -  \frac{1}{n^2} }

     

     

6 0
3 years ago
A car starts from the state of xestIf its velocity becomes 70 km/hr in 6 minutes, i) what is the accordine acceleration of F the
Alexxandr [17]

Answer: 3.5\ km

Explanation:

Given

Car starts from the state of rest and acquires a velocity of 70\ km/hr in 6 minutes

Final velocity in m/s is v=70\approx 19.44\ m/s

Using equation of motion

v=u+at\\\Rightarrow 19.44=0+a(6\times 60)\\\Rightarrow a=0.054\ m/s^2

Distance covered in 360 s

\Rightarrow v^2-u^2=2as\\\Rightarrow 19.44^2-0=2\times 0.054\times s\\\Rightarrow s=3500.64\ m\approx 3.5\ km

4 0
3 years ago
What is the number of the lowest energy level that contains an f sublevel?. . 3. . 4. . 5. . 6
DENIUS [597]
The correct answer among all the other choices is 4. This is the number of the lowest energy level that contains an f sublevel. Thank you for posting your question. I hope that this answer helped you. Let me know if you need more help. 
7 0
2 years ago
Other questions:
  • Which element has the greatest number of valence electrons?
    12·2 answers
  • If it takes 320 N of force to push a box up a ramp, yet it would take 1,600 N of force to lift the box straight up, what is the
    9·1 answer
  • Help with this please
    8·1 answer
  • A 7.9 g bullet leaves the muzzle of a rifle with ayvaci (na25565) – Chapter 4 Basics – balantic – (APPhysPd4Balant) 2 a speed of
    7·1 answer
  • A 2500-N net force acting on a 880-kg car accelerates it at a rate of ______ m/s/s
    15·1 answer
  • How do you determine the acceleration of an object?
    11·2 answers
  • Kiko writes a roport to compare and contrast the code of plants and animals. Which of the following statements is accurato arvd
    7·1 answer
  • What statement used during the scientific method is falsifiable or framed in a way that allow other scientist to prove it false
    8·1 answer
  • A 22g bullet traveling 210 m/s penetrates a 2.0kg block of wood and emerges going 150m/s. If the block were stationary on a fric
    15·1 answer
  • Review the vocabulary associated with nuclear and wave therapies
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!