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
Answer:
1.25kg
Explanation:
Simply multiply volume and density together
Answer:
m = 14*26 = 364
Explanation:
overall magnification is given as m

mo magnification of objective lens
me magnification of EYE lens
where mo is given as

and me as

d is distant of distinct vision = 25.0 cm for normal eye
fe = focal length of eye piece
focal length of objective lense is 0.140 cm
we know that








m = 14*26 = 364
I think its B or D, most likely D.