Answer:
power emitted is 1.75 W
Explanation:
given data
length l = 5 cm = 5 ×
m
diameter d = 0.074 cm = 74 ×
m
total filament emissivity = 0.300
temperature = 3068 K
to find out
power emitted
solution
we find first area that is π×d×L
area = π×d×L
area = π×74 ××5 ×
area = 1162.3892 × m²
so here power emitted is express as
power emitted = E × σ × area × (temperature)^4
put here all value
power emitted = 0.300× 5.67 × 1162.3892 × × (3068)^4
power emitted = 1.75 W
A. electrons<span> and </span>neutrons<span> B. </span>electrons<span> and </span>protons<span> C. </span>protons<span> and </span>neutrons<span> D. all particles are attracted to each other. According to atomic theory, </span>electrons<span> are usually found: A. in the </span>atomic nucleus<span> B. outside the nucleus, yet very near it because they are attracted to the </span>protons<span>.</span>
The 60 and the 5 in parallel have an equivalent resistance of 4.615 ohms. (rounded). The 10 in series makes it 14.615...
<span>The electric force is given by:
F = [ k*(q1)*(q2) ] / d^2
F = Electric force
k = Coulomb's constant
q1 = Charge of one proton
q2 = Charge of second proton
d = Distance between centers of mass
Values:
F = unknown
k = 8.98E 9 N-m^2/C^2
q1 = 1.6E-19
q2 = 1.6E-19
d = 1.0E-15 m
Insert values into F = [ k*(q1)*(q2) ] / d^2
F = [ (8.98E 9 N-m^2/C^2) * (1.6E-19) * (1.6E-19) ] / (1.0E-15 m)^2
F = </span>229.888 N
answer
the electric force of repulsion between nuclear protons is 229.888 N
