Answer:
The surface charge density on the conductor is found to be 26.55 x 1-6-12 C/m²
Explanation:
The electric field intensity due to a thin conducting sheet is given by the following formula:
Electric Field Intensity = (Surface Charge Density)/2(Permittivity of free space)
From this formula:
Surface Charge Density = 2(Electric Field Intensity)(Permittivity of free space)
We have the following data:
Electric Field Intensity = 1.5 N/C
Permittivity of free space = 8.85 x 10^-12 C²/N.m²
Therefore,
Surface Charge Density = 2(1.5 N/C)(8.85 x 10^-12 C²/Nm²)
<u>Surface Charge Density = 26.55 x 10^-12 C/m²</u>
Hence, the surface charge density on the conducting thin sheet will be 26.55 x 10^ -12 C/m².
The force in the spring will be
.
The deflection of the beam will be
= 
<h3>What is a cantilever beam?</h3>
A rigid, horizontally extending structural member known as a cantilever is supported at only one end. Typically, it extends from a solidly affixed flat vertical surface, such as a wall.
Given that:-
- A cantilever beam AB of length L has fixed support at A and spring support at B.
- The spring behaves in a linearly elastic manner with stiffness k. If a concentrated load P is applied at B.
The spring force will be calculated as:-

Deflection will be given by:-

The spring force will be calculated by:-

The deflection of the beam will be given as:-

Therefore the force in the spring will be
..The deflection of the beam will be
= 
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Answer:
The value of heat transferred watt per foot length Q = 54.78 Watt per foot length.
Explanation:
Diameter of pipe = 2 in = 0.0508 m
Steam temperature
= 300 F = 422.04 K
Duct temperature
= 70 F = 294.26 K
Emmisivity of surface 1 = 0.79
Emmisivity of surface 2 = 0.276
Net emmisivity of both surfaces ∈ = 0.25
Stefan volazman constant
= 5.67 ×

Heat transfer per foot length is given by
Q = ∈
A (
) ------ (1)
Put all the values in equation (1) , we get
Q = 0.25 × 5.67 ×
× 3.14 × 0.0508 × 1 × (
)
Q = 54.78 Watt per foot.
This is the value of heat transferred watt per foot length.
Answer:
s= 20.4 m
Explanation:
First lets write down equations for each ball:
s=so+vo*t+1/2a_c*t^2
for ball A:
s_a=30+5*t+1/2*9.81*t^2
for ball B:
s_b=20*t-1/2*9.81*t^2
to find time deeded to pass we just put that
s_a = s_b
30+5*t-4.91*t^2=20*t-4.9*t^2
t=2 s
now we just have to put that time in any of those equations an get distance from the ground:
s = 30 + 5*2 -1/2*9.81 *2^2
s= 20.4 m