To control whether an object is solid or incorporeal (things can pass through it) you would use the:
1 answer:
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The electric potential energy of the charge is equal to the potential at the location of the charge, V, times the charge, q:
The potential is given by the magnitude of the electric field, E, times the distance, d:
So we have
(1)
However, the electric field is equal to the electrical force F divided by the charge q:
Therefore (1) becomes
And if we use the data of the problem, we can calculate the electrical potential energy of the charge:
The potential is given by the magnitude of the electric field, E, times the distance, d:
So we have
However, the electric field is equal to the electrical force F divided by the charge q:
Therefore (1) becomes
And if we use the data of the problem, we can calculate the electrical potential energy of the charge:
Answer:
Explanation:
given,
J = 50 kg-m²
frequency, f = 20 Hz
time ,t = 5 s
we know,
angular velocity = 2 π f
ω = 2 π x 20
ω = 125.66 rad/s
now, angular acceleration calculation
α = 25.13 rad/s²
Torque given to the flywheel.
Torque of the given flywheel is equal to
Answer:
a)<em> 2000 W/m² </em><em>; </em>b) 636.94 W/m<em>².sr ; </em><em>c) </em>0.5
Explanation:
a)
The formula for calculation of total emissive power is:
Total emissive power = E = E'<em>λdλ</em>
<em> </em>= (0)d<em>λ + </em>
(100)d<em>λ + </em>
(200)d<em>λ + </em>
(100)d<em>λ </em>
(0)d<em>λ</em>
<em>where a = 5; b = 10; c = 15; d = 20; e = 25</em>
<em> = 0 +100(10-5) + 200(15-10) +100(20-15) + 0</em>
<em> = 2000 W/m²</em>
b)
The formula for total intensity of radiation is:
I = E/π = 200/3.14 = 636.94 W/m<em>².sr </em>
<em>c)</em>
Fo submissive power leaving the surface in range π/4 ≤θ≤π/2
[E(π/4 ≤θ≤π/2)]/E = Icosθsinθ dθdΦdλ
where f = infinity, g=2π, h=π/4, i=π/2
By simplifying, we get
= (-1/2)[cos(2π/2)-cos(2π/2)]
= -0.5(-1-0)
=0.5