Question

A particle of mass m is projected with velocity v and angle theta . Find its angular momentum

Answer 1

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)

Answer 2

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.