To solve this
problem, we should remember that:
Energy = Force x Distance
Since we are talking about charges, therefore we make use
of Coulumb’s law for the electrical force between the two charges:
F = k q1 q2 / d^2
Where,
k = Coulumb’s constant = 9 x 10^9 N m^2/ c^2
q = charge
d = distance between the charges
Plugging back into the energy equation:
E = (k q1 q2 / d^2) * d
E = k q1 q2 / d
Solving for E using the given values:
E = (9 x 10^9 N m^2/ c^2) (3.4 E -6 c) (6.6 E -6 c) /
0.10 m
<span>E = 2.02 N m = 2.02 J</span>
Answer:
ω2 = 216.47 rad/s
Explanation:
given data
radius r1 = 460 mm
radius r2 = 46 mm
ω = 32k rad/s
solution
we know here that power generated by roller that is
power = T. ω ..............1
power = F × r × ω
and this force of roller on cylinder is equal and opposite force apply by roller
so power transfer equal in every cylinder so
( F × r1 × ω1) ÷ 2 = ( F × r2 × ω2 ) ÷ 2 ................2
so
ω2 =
ω2 = 216.47
Explanation:
When Michelson-Morley apparatus is turned through
then position of two mirrors will be changed. The resultant path difference will be as follows.

Formula for change in fringe shift is as follows.
n = 

v = 
According to the given data change in fringe is n = 1. The data is Michelson and Morley experiment is as follows.
l = 11 m
c =
m/s
Hence, putting the given values into the above formula as follows.
v = 
= 
= 
Thus, we can conclude that velocity deduced is
.
Answer:
x2 = 0.99
Explanation:
from superheated water table
at pressure p1 = 0.6MPa and temperature 200 degree celcius
h1 = 2850.6 kJ/kg
From energy equation we have following relation



![2850.6 + [\frac{50^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}] = h2 +[ \frac{600^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}]](https://tex.z-dn.net/?f=2850.6%20%2B%20%5B%5Cfrac%7B50%5E2%7D%7B2%7D%20%2A%20%5Cfrac%7B1%20kJ%2Fkg%7D%7B1000%20m%5E2%2FS%5E2%7D%5D%20%3D%20h2%20%2B%5B%20%5Cfrac%7B600%5E2%7D%7B2%7D%20%2A%20%5Cfrac%7B1%20kJ%2Fkg%7D%7B1000%20m%5E2%2FS%5E2%7D%5D)
h2 = 2671.85 kJ/kg
from superheated water table
at pressure p2 = 0.15MPa
specific enthalpy of fluid hf = 467.13 kJ/kg
enthalpy change hfg = 2226.0 kJ/kg
specific enthalpy of the saturated gas hg = 2693.1 kJ/kg
as it can be seen from above value hf>h2>hg, so phase 2 is two phase region. so we have
quality of steam x2
h2 = hf + x2(hfg)
2671.85 = 467.13 +x2*2226.0
x2 = 0.99