Answer:
The water level will drop by about 1.24 cm in 1 day.
Explanation:
Here Mass flux of water vapour is given as

where
is the mass flux of the water which is to be calculated.
- D is diffusion coefficient which is given as

- l is the thickness of the film which is 0.15 cm thick.
is given as

In this
is the saturated water pressure, which is look up from the saturated water property at 20°C and 0.5 saturation given as 2.34 Pa
is the air pressure which is given as 0.5 times of 
- R is the universal gas constant as

- T is the temperature in Kelvin scale which is

By substituting values in the equation

Converting
into 
As 1 mole of water 18
so

Putting this in the equation of mass flux equation gives

For calculation of water level drop in a day, converting mass flux as

So the water level will drop by about 1.24 cm in 1 day.
Answer:
Force is repulsive hence direction of force is away from wire
Explanation:
The first thing will be to draw a figure showing the condition,
Lets takeI attractive force as +ve and repulsive force as - ve and thereafter calculating net force on outer left wire due to other wires, net force comes out to be - ve which tells us that force is repulsive, hence direction of force is away from wire as shown in figure in the attachment.
Answer:
A.)1.52cm
B.)1.18cm
Explanation:
angular speed of 120 rev/min.
cross sectional area=0.14cm²
mass=12kg
F=120±12ω²r
=120±12(120×2π/60)^2 ×0.50
=828N or 1068N
To calculate the elongation of the wire for lowest and highest point
δ=F/A
= 1068/0.5
δ=2136MPa
'E' which is the modulus of elasticity for alluminium is 70000MPa
δ=ξl=φl/E =2136×50/70000=1.52cm
δ=F/A=828/0.5
=1656MPa
δ=ξl=φl/E
=1656×50/70000=1.18cm

Answer:

Explanation:
Given that,
The radius of sphere, r = 5 cm = 0.05 m
Net charge carries, q = 7.5 µC = 7.5 × 10⁻⁶ C
We need to find the surface charge density on the sphere. Net charge per unit area is called the surface charge density. So,

So, the surface charge density on the sphere is
.
Answer:
a)W=8.333lbf.ft
b)W=0.0107 Btu.
Explanation:
<u>Complete question</u>
The force F required to compress a spring a distance x is given by F– F0 = kx where k is the spring constant and F0 is the preload. Determine the work required to compress a spring whose spring constant is k= 200 lbf/in a distance of one inch starting from its free length where F0 = 0 lbf. Express your answer in both lbf-ft and Btu.
Solution
Preload = F₀=0 lbf
Spring constant k= 200 lbf/in
Initial length of spring x₁=0
Final length of spring x₂= 1 in
At any point, the force during deflection of a spring is given by;
F= F₀× kx where F₀ initial force, k is spring constant and x is the deflection from original point of the spring.

Change to lbf.ft by dividing the value by 12 because 1ft=12 in
100/12 = 8.333 lbf.ft
work required to compress the spring, W=8.333lbf.ft
The work required to compress the spring in Btu will be;
1 Btu= 778 lbf.ft
?= 8.333 lbf.ft----------------cross multiply
(8.333*1)/ 778 =0.0107 Btu.