A particle of mass m is projected with velocity v and angle theta . Find its angular momentum
Answers
if the point of projection is taken as origen then
coordinates of maximum height (H,R/2) or r =Hi + (R/2)j in vector form
R is horizontal range and H is maximum height ....
at maximum height there is only horizontal component of velocity no vertical component exists ...
so , V= vcos@i ......
linear momentam = mvcos@i ( @ is the angle of projection)
we know that angular momentam(L) about point of projection = r*p (* is cross product)
L =(Hi + (R/2)j)*(vi)
L =Rv/2 (-K)
L=v3 /2g (-k) ( R =v2 /g at @=45 degree)
We know that according to Ohm’s law
V = IR
where
V= potential difference
I= Current
R = Resistance
We can also modify the equation as
I=V/R ——- (i)
Now given that the potential difference across the two ends of the component decreases to half
∴ So let the new potential difference be Vʹ=V/2
Resistance remains constant across the electrical component
So the new current is drawn through the electrical component is Iʹ = Vʹ/R
= (V/2)/R {Substituting Vʹ=V/2 in the above equation}
= (1/2) (V/R)
= (1/2) I = I/2
Therefore, the amount of current flowing through the electrical component is reduced by half.