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Synchronization ofalternatormeansconnecting an alternator into gridin parallel with many other alternators, that is in a live system of constant voltage and constant frequency. Many alternators and loads are connected into a grid, and all the alternators in grid are having same output voltage and frequency (whatever may be the power). It is also said that thealternatoris connected to infinite bus-bar.A stationary alternator is never connected to live bus-bars, because it will result in short circuit in the stator winding (since there is no generated emf yet). Beforeconnecting an alternator into grid, following conditions must be satisfied:

Equal voltage: The terminal voltage of incoming alternator must be equal to the bus-bar voltage.

Similar frequency: The frequency of generated voltage must be equal to the frequency of the bus-bar voltage.

Phase sequence: The phase sequence of the three phases of alternator must be similar to that of the grid or bus-bars.

Phase angle: The phase angle between the generated voltage and the voltage of grid must be zero.

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Joule's lawstates the amount ofheatproduction in a conductor is : Directly proportional to the square of electric current flowing through it. Is directly proportional to the resistance of the conductor. Directly proportional to the timeforwhich electric current flows through the conductor.

Coulomb's law states that: The magnitude of the electrostatic force of attraction or repulsion between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them.

Electro-osmosis refers to the movement of liquid in a porous material due to an applied electric field.Electro-osmosis is a very effective instrument when treating heterogeneous, silt and clay-rich soil.

The phenomenon of electro-osmosis is very useful in chemical separation techniques and buffered solutions.

Electro-osmosis can be used for organics removal. It reduces the need for specialized electrodes.

the symbol is of a rheostat or variable resistanceHeating Effect Demonstrate with an electric coil in beaker of water plus thermometer.

Magnetic Effect* Demonstrate with an electromagnet

Chemical Effect

tq so much

Electromechanical energy conversion is a reversible process except for the losses in the system. The term "reversible" implies that the energy can be transferred back and forth between the electrical and the mechanical systems. ... This is essentially the principle of operation of all electric motors.

because in generating stations the power is created with the help of turbines and those turbines are basically mechanical equipment and that mechanical power is converted into electrical power and that power is being transmitted to the other cities all other areas where it is required

Option (B) is correct.

WorkingPrinciple ofElectrical Fuse. The primary use of anelectric fuseis to protectelectricalequipment from excessive current and to prevent short circuits or mismatched loads. ... However, when an excessive amount of current flows through thefusewire, the heating effect of current causes thefusewire to melt

thx

The electric fuse is a piece of wire having a verylowmelting point and high resistance.

An electric fuse is used as a safety device for the protection of electric circuits and appliances due to short-circuiting or overloading of the electric circuits.

The primary use of an electric fuse is to protect electrical equipment from excessive current and to prevent short circuits or mismatched loads. Electrical fuses play the role of miniature circuit breakers. Apart from protecting equipment, they are also used as safety measures to prevent any safety hazards to humans.

Working:Under normal conditions, the fuse wire is a part of the circuitry, contributing to a complete loop for charges to flow through it. However, when an excessive amount of current flows through the fuse wire, the heating effect ofcurrentcauses the fuse wire to melt. This is because the fuse wire is chosen such that it has a low melting point. This causes the loop to break thereby stopping the flow of charges in the circuit.

Distance covered in nth second is given bys = u + 1/2 a (2n - 1)

Here,u = 0,a = 2 m/s^2n = 5

Therefore,Distance covered in 5th second= 0 + 1/2 * 2(2*5 - 1)= 10 - 1= 9 m

(a) is correct option

the distance travelled by

the distance travelled by the object in 10 th second is 38.5 m

d is right answer bro

KE=pv/2now is p=kethenp=pv/2henceV=2m/s

calculate velocity of a particle when its momentum is numerically equal to its k.E (a) 2 m/s (b) zero (c) 4 m/s (d) 1m/s

chutiya hai kya skti gautam

ok

given s = 2.5t²

so, distance travelled by particle after 5secs is = 2.5*(5)² = 62.5m/s²

now, average speed = total distance/time = 62.5/5 = 12.5 m/s

and instantaneous speed at 5 sec is given by

V = ds/dt = 5t = 5*5 = 25m/s.....

but distance m/s² why?

The instantaneous speed and the average speed of a body are equal when the body has uniform motion, i.e. when the acceleration is zero. In other words no external unbalanced force is acting on the body.

how we multiplied with 5 in B part ??

velocity is 5×t ... so, when t = 5 sec then v. = 5×5 = 25m/s

but it is ask to find instantaneous speed at t= 5

that's why only we put t = 5 , for example if it was asked

velocity at t = 0 , then v = 5×0 = 0Velocity at t = 1 , then v = 5×1 = 5velocity at t = 2 , then v = 5×2 = 10velocity at t = 3 , then v = 5×3 = 15

velocity will change every second like this ↑↑

Option c) is correct.

center of mass of ring and cylinder will be at it's geometrical center

Please find below the solution to the asked query:

Here, the given relation is,

P=(αβ)e−βtHere, in exponent, the term should be unit less. Therefore,

[β]=[M⁰L⁰T⁻¹]

And, the ratio of alpha and beta should have the dimensions of power. Therefore,

[α/β]=[ML²T⁻³]⇒[α]=[β][ML²T⁻³]⇒[α]=[M⁰L⁰T⁻¹][ML²T⁻³]⇒[α]=[ML²T⁻⁴]

Delivering power implies doing work and changing energy, since that is how power is defined. For a pendulum, the only force doing work (and thus changing the pendulum's energy) is gravity. The tension force that holds the pendulum on its circular path is a centripetal force, which is always perpendicular -- at each moment -- to the direction of travel, and thus does no work.

To get the power precisely at the moment when the pendulum, released from rest at 60 degrees (I will presume this is 60 degrees with respect to the vertical, since I am not told otherwise), is at 30 degrees from the vertical, we can use the formula:

P = Fv

where

P = power

F = force doing work

v = (linear) speed of object

If the bob is released from rest, we can calculate the speed v at the 30-degree point using a conservation of (mechanical) energy argument. There would also be a way to calculate v using simple harmonic equations... but that 60-degree release angle is a little large for the pendulum's motion to be truly simple harmonic. That is slightly over 1 radian, and we need the angles to be no larger, I would say, than a few tenths of a radian to have the pendulum's motion be more nearly simple harmonic.

So let's calculate v using (mechanical) energy conservation. It turns out that, when a pendulum bob is held so that the pendulum makes an angleθ with the vertical, the vertical level (h) of the bob above its level at the lowest point in the swing is

h = L(1 - cosθ)

where L is the length of the pendulum. Thus, for a 1-m pendulum, whenθ = 60 degrees, we have

h60= (1 m)(1 - cos(60°)) = 1(1 - 0.5) = 0.5 m above lowest point in the swing

Whenθ = 30°, we have

h30= (1 m)(1 - cos(30°)) = 0.134 m

So, when the bob is released from rest at 60 degrees, it has (mechanical) energy:

ME60= KE60+ GPE60

KE60= kinetic energy at the start = 0 (since it is released from rest)

GPE60= gravtiational potential energy at 60 degrees (using the lowest level in the swing as h = 0)

= mgh60= mg(0.5 m)

ME30= KE30+ GPE30= (1/2)mv2+ mgh30= (1/2)mv2+mg(0.134 m)

Since gravity is the only force doing work, (mechanical) energy is conserved. Thus:

ME60= ME30

mg(0.5 m) = (1/2)mv2+mg(0.134 m)

Notice that the mass cancels out of the equation when these two mechanical energy values are set equal. We get:

0.5g = v2/2 + 0.134g

From there, you can solve for v, the speed of the bob when the pendulum is at 30 degrees.

Now, we need to determine the force of gravity on the bob at that point. Actually, we only need the component of gravity in the bob's direction of motion (tangent to the circular path it is on), because this is the only component of gravity doing work on the bob (and thus delivering power to it).

In a very similar derivation to the one that calculates the component of gravity down an inclined plane of angleθ, it turns out the component of gravity in the bob's direction of motion is:

Fg-tangential= mgsinθ

This is tricky to describe without a diagram, but see if you can demonstrate it with a drawing.

Then:

P = Fv = Fg-tangential(v)

using the values calculated above. This will be the power delivered to the bob at that instant.

To find the distance covered put in t=6sechencex=(6)^3-6(6)^2+9(6)+2x=216-216+54+2x=56m

Electric Iron, electric heater.

thanks

1 M eV= 10^6 eVHence, option (B) is correct.

1

2

The rate of change of velocity momentum of the body propotional to applied net force

distance travelled in n seconds:

S1= un + 1/2a(n^2)

distance travelled in (n-1) seconds:

S2 = u(n-1) + 1/2a((n-1)^2)

distance travelled in nth second:

S1-S2

= u + 1/2a(2n-1)

Answer:A)20N

Pressure is force/area =F/A

app settings chay hai

Pressure is equal to force/Area=F/Awhere A is area and F is force.

Here these are as follows

Check yourself and others for injuries. ...

Check water, gas, and electriclinesfor damage. ...

Turnon theradio. ...

Stay out of damaged buildings.

Be careful around brokenglassand debris. ...

Be careful ofchimneys(they may fall on you).

Stay away from beaches. ...

Stay away from damaged areas

i) Electric Fuse: An electric fuse is connected in series it protects the circuit from overloading and prevents it from short circuiting.

(ii) Proper earthing of all electric circuit in which any leakage of current in an electric appliance is transferred to the ground and people using the appliance do not get the shock.

thanks

option 1st is correct. because when charge connected in series its charge will not divide.

Groundwater rechargeor deep drainage or deep percolation is a hydrologic process where water moves downward from surface water togroundwater.Rechargeis the primary method through which water enters an aquifer.

This is an interesting question and tests your concepts about the following topics :1. Electric Flux2. Solid AnglesSee picture below: