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See, differentiability at a point means the existence of a unique tangent at that point.
As an example consider the graph of y = |x|
There is no unique tangent at x = 0.
Hence the function y = |x| is non differentiable at x = 0
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Sorry for the wrong solution. Didn't see the first quadrant.
Excellent solution naveen
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i = cos /2 + i sin/2 = ei/2
ii = (ei/2)i = e-/2 (As i2 = -1)
= 1/e/2
Now e lies between 2 and 3.
/2 is approximately = 1.57
Hence e/2 isn't a very large value. Infact it will be less than 6 but definitely greater than 1.
Hence its reciprocal 1/e/2 should be less than unity
And hence ii is a real number less than unity.
Please check the question. It shouldn't be a large real number.
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i = cos /2 + i sin/2 = ei/2
ii = (ei/2)i = e-/2 (As i2 = -1)
= 1/e/2
Now e lies between 2 and 3.
/2 is approximately = 1.57
Hence e/2 isn't a very large value. Infact it will be less than 6 but definitely greater than 1.
Hence its reciprocal 1/e/2 should be less than unity
And hence ii is a real number less than unity.
Please check the question. It shouldn't be a large real number.
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The spin of the electron is about itself. The spin quantum no. represents the direction of spin of the electron about itself. Then why should they crash
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I couldn't understand what you meant by "point of integration"
If you want to understand how this integration works, here it goes:
0 q (1/(CE - q)) dq
Substitute CE - q = z
then, dq = -dz
When q = 0, the z = CE
and when q = q, z = CE - q
The integral now becomes
CE CE - q(1/z)(-dz) = -CECE - q(1/z)dz = -[lnz]CECE - q = -{ ln(CE - q) - ln(q)}
= - ln((CE - q)/q)
- ln((CE - q)/q) = t/RC
((CE - q)/q) = e-t/RC
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Actually, the integration wasn't necessary in this case. I have removed it.
As far as application of integration goes,
Suppose you know acceleration a as function of time. Integrate it to get velocity. Integrate it to get position.
Suppose you know acceleration as a function of position. Then use the following
a = dv/dt = (dv/dx)(dx/dt) = (dv/dx)v [As dx/dt = v]
a = v(dv/dx)
And then integrate.
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The chair exerts a normal force on the man. By Newton's third law, the man also exerts the same amount of force on the chair. This is the force exerted by the man on the chair, given to be 450 N.
Similarly, the force that the man exerts on the rope, the rope exerts back on him. That is the tension in the rope.
Forces on the man are:
The normal N due to the chair
The weight of the man mg.
The force T that the rope exerts on him.
Then, T + N - mg = ma -----------(1)
Now, for the chair, the force are
The force T exerted by the rope supporting the chair
The normal force N exerted by the man on the chair
The weight of the chair Mg.
Then, T - N - Mg = Ma ------(2)
(1) - (2)
2N + Mg - mg = (m - M)a
Now, mg = 1000 N,  m = 100kg (As g = 10 ms-2)
Mg = 250 N  M = 25 kg
N = 450 N
Substituting these values,
a = 2 ms-2
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Net horizontal force, H = f + f 2cos 45 = 2f
Net vertical force, V = f - f 2 sin45 = 0
Hence the net force is horizontal, ie, parallel to AC.
Magnitude of resultant force, R = 2f
Let this resultant force be applied at a point x distant from C on the side BC.
Then the torque due to this force at any point must equal the sum of torques of the forces shown.
Taking torques about the point of intersection of the diagonals,
2f xsin45 = f(L/ 2) + f(L/ 2) - (f 2cos 45)((L/ 2)
x = L/2
where L is the length of the side of the square.
Hence the resultant is applied at the midpoint of the side BC
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Static friction acts till the body doesn't move. Hence to overcome the static friction 75 N force is needed.
 N = F1
 static = F1/N = 75 N/ (20 kg)(9.8 m/s2) = 0.37
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The change in momentum of the ball
 p = m(v'-v)
Where v' is the final velocity and v is the initial velocity
If  t be the time of contact, then the average force exerted is
Fav = p/t
Substitute
m = 0.15 kg
v' = 12 m/s
v = -20 m/s
t = 1/25 s
to get the answer
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The answer is A and D
The forces acting on the bob are
The tension T at an angle  with the vertical
mg downwards
q /2 horizontally, away from the sheet
For equilibrium,
Tcos = mg
Tsin = q/2
Dividing these two eauations,
tan = q/2mg ---------------(1)
Now suppose the bob is deflected by an arbitrary small angle  from the equilibrium position, away from the sheet.
Then, the equation of motion is
mgsin(+) - (q/2)cos(+) = m(dv/dt)
Expanding and applying cos  1 and sin
mgsin - (q/2)cos + [mgcos - (q/2)sin] = m(dv/dt)
Now from the relationship proved in (1), the first two terms cancel out.
[mgcos - (q/2)sin] = m(d(L )/dt)
Now, = |d(+)/dt| = - d/dt (As is an decreasing function of time). The above equation can then be written as
d2/dt2 + [(g/L)cos - (q/2mL)sin] = 0
Hence the time period is
T = 2 [L/(gcos - (q/2m)sin)]
The denominator is definitely less than g. Hence The time period is greater than
2 (L/g)
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First of all understand that the point charge is placed symmetrically wrt to all the six faces of the cube. Now, flux is related to no of lines of electric field cutting a surface. As the point charge is placed at the center of the cube, the no of lines crossing each face must be same, which means that the fulx must be same.
Suppose the flux through each face is 
By Gauss's Law
6 = q/
= q/6
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dv/dt = a
Now, at every instant before slipping the frictional force f equals the resultant of centripetal and tangential forces
f2 = (mv2/R)2 + (m(dv/dt))2
For no slipping,
f< N
(mv2/R)2 + (m(dv/dt))2 <( mg)2
(v2/R)2 + a2 <2g2
v < [(2g2 - a2 )R2]1/4
Hence the speed at which the car slips
v = [(2g2 - a2 )R2]1/4
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These problems are best solved geometrically using phasors (vectors).
The general funda is that Asin(t+ ) can be represented by Aei, that is a vector of magnitude A at an angle with +X axis.
A cos t is a vector of magnitude A along the +Y axis.
(A/2)cos(t+/4) is a vector of magnitude A/2 making an angle /4 +  /2 with X axis.
(A/4)cos(t+) is a vector of magnitude A/4 along the negative Y axis.
(A/8)cos(t+3/2) is a vector of magnitude A/8 along +X axis.
Add these vectors.
Now the answer will be in the form Asin(t+)
The resultant magnitude is the A.
The angle the resultant makes with the +X axis is the .
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