Let say the point is inside the cylinder
then as per Gauss' law we have

here q = charge inside the gaussian surface.
Now if our point is inside the cylinder then we can say that gaussian surface has charge less than total charge.
we will calculate the charge first which is given as


now using the equation of Gauss law we will have


now we will have

Now if we have a situation that the point lies outside the cylinder
we will calculate the charge first which is given as it is now the total charge of the cylinder


now using the equation of Gauss law we will have


now we will have
Answer:
(a) 3.44 x 10^-3 m^3/s
(b) 8.4 m/s
Explanation:
area of water line, A = 5.29 x 10^-3 m
number of holes, N = 15
Speed of water in line, V = 0.651 m/s
(a) Volume flow rate is given by
V = area of water line x speed of water in water line
V = 5.29 x 10^-3 x 0.651 = 3.44 x 10^-3 m^3/s
(b) area of one hole, a = 4.13 x 10^-4 m
Let v be the velocity of water in each hole
According to the equation of continuity
A x V = a x v
5.29 x 10^-3 x 0.651 = 4.1 x 10^-4 x v
v = 8.4 m/s
Answer:
5p
Explanation:
We are given that a tin has 50 electrons.
We have to find in which subshell electrons experience the lowest effective nuclear charge.
We know that the electron in outermost shell experience the lowest effective nuclear charge.
Electronic configuration is given by

Outer most sub-shell is 5p. Therefore, 5p subshell experience the lowest effective nuclear charge because the distance of 5p sub-shell is large from nucleus.
Answer: 5p
Answer:
Q=+100kj,w=-15kj,Q=100kj,w=-62kj
Explanation:
when energy is exerted into a system work done is equal to zero .hence the system does work to the surrounding.