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Either you haven't typed the time or the charge is given as charge per unit time. Please check.
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Please specify the question clearly. What is theta???
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For uniformly accelerated motion
vav = (1/2)(v + v0)
The acceleration is uniform here too
a = -g j
v0 = v0(cos i + sin j)
At highest point,
v = v0cos i
vav = (1/2)(v + v0) = (1/2)(v0cos i + sin j + v0cos i)
= (1/2)(2v0cos i +v0 sin j )
Magnitude of average velocity
vav = (1/2 [ (4v02cos2 + v02sin2 )] = (1/2) [ (4v02cos2 + v02 - v02cos2 )]
= (v0/2)[ (3cos2 + 1)]
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Consider a particle on the rim whose radius from the centre makes an acute angle of  with the vertical.
Such a point has two velocities: v horizontally due to translation and  R at a tangent due to rotation. As the case is of pure rolling, v = R. Hence the net velocity of the particle is the vector sum of v velocity horizontal and a velocity v along the tangent. You can see by drawing the figure that the angle between these two velocities is  .
Hence the magnitude of velocity of the particle is
u = (v2 + v2 + 2 v v cos) = 2v sin(/2)
The magnitude of velocity is the speed, whose integral wrt time gives distance covered.
s = 0 T u dt = 0 T 2v sin(/2) dt
where T is the time for one complete rotation.
Now = t  d = dt
s = 02 pi 2v sin(/2) / d = 2R02 pisin(/2)d = 8R
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1) The field should be gravitational.
Taking the conventional directions of XY axes, the point of projection to be the origin, and the time instant of projection to be t = 0,
v1 = v1 i where v1 = 3m/s
v2 = -v2 i where v2 = 4m/s
a = -g j
v1(t) = v1 + at = v1 i -gtj
v2(t) = v1 + at = -v2 i -gtj
Let at time instant T their velocity vectors become perpendicular
v1(T).v2(T) = 0
-v1 v2 + (gT)2 = 0
T = (v1 v2) / g
At any instant t the position vectors of the particles are
r1(t) = v1t+ (1/2)at2 = v1t i -(1/2)gt2j
r2(t) = v2t+ (1/2)at2 = -v2t i -(1/2)gt2j
Instantaneous separation between the two particles
d(t) = |r1(t) - r2(t)| = (v1 + v2)t
At the time T,
d(T) = (v1 + v2)T = (v1 + v2) (v1 v2) / g
Substitute the numerical values to get the answer.
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2)
As the particles move with constant velocities their position vectors are
r1(t) = r1 + v1t
r2(t) = r2 + v2t
At the instant of collision, say at t = T,
r1(T) = r2(T)
r1 + v1T = r2 + v2T
r1 - r2 = (v2 - v1)T
Hence r1 - r2 and (v2 - v1) are parallel vectors. As they are parallel vectors, unit vectors along them must be same.
Hence,
r1 - r2/|r1 - r2| = (v2 - v1)/|(v2 - v1)|
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The external force is mgsin30. Now, the role of friction is to prevent the slipping of the block. Friction can go upto a value of  mgcos30. Now if mgsin30 is less than this max value, why should friction apply its max strength. It will apply a value equal to mgsin30
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As we know for constant acceleration a,
 x = v0( t) + (1/2)a(t)2 x /t= v0 + (1/2)a(t)vav = = v0 + (1/2)a(t)
v = a(t)t = v/a
Substituting the value of t in the second term of the first equation
vav = v0 + (1/2)a(v/a) = v0 + (1/2)(v - v0)
vav = (1/2)(v + v0)
As can be seen this result is derived from the results for constant acceleration.
It isn't applicable to non uniform accelerations.
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Very crisp and clear solution edison
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F = x3-3x
m(dv/dt) = x3-3x
m(dv/dx)(dx/dt) = x3-3x
mv(dv/dx) = x3-3x
m0V vdv = 13 (x3-3x)dx
(1/2)V2 = 8
V2 = 16
V =  4 m/s
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You can geometrically find the anticlockwise angles the sides of the hexagon make with the +X axis. They are 0,  /3, 2 /3, ( = 3/3), 4/3, 5/3
Suppose the length of the hexagon side be L.
Then the X components are
L = L cos0
L cos/3
L cos2/3
-L = L cos = L cos3/3
L cos4/3
Lcos5/3
The vectors A1, A2, ...A6 form a closed hexagon and hence their resultant is zero.
Then the X component of their resultant must also be zero. The X component of the resultant is the sum of the X components of the individual vectors. Hence
L cos0 + L cos/3 + L cos2/3 + L cos3/3 + L cos4/3 + Lcos5/3 = 0
cos0 + cos/3 + cos2/3 + cos3/3 + cos4/3 + cos5/3 = 0
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Consider this problem
A rope of mass m and length L has 2 forces F1 and F2 applied at its end points.. find the expression for tension at any point in the rope.
Here is the solution to this problem
consider a length x of the rope from the end where F2 is acting. Let the tension at the other end be T.
Mass of this portion of the rope = (m/L)x
Acceleration of the rope = (F1 - F2)/m
So
T - F2 =[ (m/L)x] [(F1 - F2)/m]
T = F2 +[(F1 - F2)/L]x
At x = L/2
T = (F1 + F2)/2
Hence the tension at the midpoint is
T = (100 + 70)/2 = 85 N
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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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