Four 10F capacitors are connected in series, calculate the equivalent capacitance.
a) 1.5F
b) 2.5F
c) 3.5F
d) 0.5F

Answer

A, B, C, D

Explanation

As seen from the truck the ball moves vertically upward comes back.
Time taken time taken by truck to cover 58.8 m.
`:. Time = S/v = 58.8/14.7 = 4 sec`

Hencer `v= 14.7 m/s for truck`
`u=? , v=0= -9.8 m/s^2 `
(going upward)
`t= 4/2 = 2sec`
`:. v= u+at`
`rarr 0=u+9.8x2`
`rarr = 19.6 m/s`
(vertical upward velocity)
b. From road, it seems to be projectivel motion.
tota time of flight = 4 sec
Horizontal range covered in this time
`= 58.8 m=X`
`:. x= u cos thetas t`
rarr u cos theta t = 14.7`
Taking vertical component of veocity into consideration.
`y= (0^2-(19.6)^2)/(2xx(-9.8))`
`= 19.6m ` {from (a)}
`:. y= u sin theta t- 1/2 gt^2`
`rarr 19.6 = u sin theta (2)- 1/2 (9.8) 2^2`
`rarr 2u sin theta=19.6xx2`
`rarr 2u sin theta = 19.6xx 2`
`rarr u sin theta = 19.6`
Dividing equation (2) by equatiion 1 we get
`(u sin theta)/(u cos theta) = tan theta = 19.6/14.7 = 1.333`
`rarr theta tan^-1 (1.333)=53^0`
Again, ` u cos theta = 14.7`
`rarr u = 14.7/(cos 53^0)= 24.42 m/s`
The speed of ball is 24.42 m/s at an angle `53^0` with horizontal as seen from the road.