Yes I know the method through extension at equilibrium ... But I want to check out the method of COE for future purposes...
Can u please just tell me the equation that u would use and the two points that u would consider for solving through COE ?
Homework Statement
In A spring mass system , the spring stretches 2 cm from its 's frelength when a force of 10 N is applied . This spring is stretched 10 cm from it's free length , when a body of mass m = 2 kg is attached to it and released from rest at time t = 0 . Find the A) force constant...
Homework Statement
Write the equation for a particle in simple harmonic motion with amplitude a and angular frequency w considering all distances from one extreme position and time when it is at other extreme end.
Homework Equations
X = A sin (wt + ∆)
∆ = phase difference
The Attempt at a...
Homework Statement
The potential difference across resistance R carrying current I is V = IR . Now if the potential difference is measured via voltmeter of resistance r , the reading on voltmeter is V' . Prove that V' = Ir/(R+r) . For what value of r does the voltmeter measure true value...
I didn't ask for your pity mister , only solution.. also don't presume I didn't consider the non ideal voltage source with some internal resistance , I didn't add it to my answer as the question doesn't mention any internal resistance .. which is odd for the 12th level book!
Homework Statement
Figure shows a potentiometer circuit for comparison of two resistances , the balance point with standard resistor R = 10 ohm , is found to be 58.3 , while that with unknown resistance X is 68.5 cm , determine the value of X .
b) what might you do if you failed to find the...
Homework Statement
How to apply constraints in the system to get a relationship between the displacements of block of mass m and pulley of mass M.?
Homework Equations
∑T.a= 0
The Attempt at a Solution
Assuming tension in both strings to be T .
-T × a1 ( for the block) + 2T × a2 ( for the...
Homework Statement
Question 6.
Homework Equations
Time constant = RC
The Attempt at a Solution
I think answer should be 1/2 RC as Rnet = 1/2R and to convert it to single resistance form we should first find Rnet
But the answer is coming out to be RC . How?
<< Corrected Image added by...
Homework Statement
A block of mass m having charge q placed on smooth horizontal table and is connected to a wall thorough an unstretched spring of constant k . A horizontal electric field E parallel to spring is switched on. Find the ampliture of the shm by the block.
Homework Equations
kx=...
Homework Statement
In the circuit diagram shown , Xc = 100 ohm , XL = 200 ohm , R = 100 ohm , the effective current through the source is ?
Homework Equations
Z= √( R^2 + ( XL - Xc)^2)
Vrms = Irms/ Z
The Attempt at a Solution
I tried to draw the phaser diagram and calculate the relation...
But I still don't get why integration is required .. it's not like specific resistivity is changing... And shouldn't length by taken as 2R-R = R only then??
Homework Statement
1.In the WE-11 resistance is calculated by integrating and that too by taking length as dr and area as the CSA of the small cylinder ..
Shouldn't length be l and area by 2πrdr..?
I also don't understand why can't we simply use
Dl/A formula
, Where A =π((2R)^2 - R^2)
2)...
Acc to question ifthe bulb is connected to 100V supply power dissipated by it will be of 500 W
I have used that info to find resistance of the bulb which will remain constant irrespective of circuit conditions
Now if the bulb is connected to 200 V supply and the bulb still dissipates 500W that...
1. Yeah it refers to bulb
2. 10000 is for V^2 where V = 100
3. So that means that what I have calculated is basically power across R + bulb instead of only bulb? Also if the resistance R was in parallel .. then my answer would have been correct??
If the circuit has 200 V supply.
The resistance R that must be put in series with bulb so that it draws 500 w is?
2. Relevent equations
P= v^2/r
I = v/r
3. My attempt at the solution
R= V^2/P r (bulb)= 10000/500 = 20 ohm
Now for 200 v supply ..
P= 500 W
V= 200V
Rnet = 20 + R
R+20= 80
R= 60...