Too easy:
100Ω/4 sec = 200Ω/t
t= 8 sec
and physics and physics and physics and physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physics
Answer:
Efficiency = 52%
Explanation:
Given:
First stage
heat absorbed, Q₁ at temperature T₁ = 500 K
Heat released, Q₂ at temperature T₂ = 430 K
and the work done is W₁
Second stage
Heat released, Q₂ at temperature T₂ = 430 K
Heat released, Q₃ at temperature T₃ = 240 K
and the work done is W₂
Total work done, W = W₁ + W₂
Now,
The efficiency is given as:
or
Work done = change in heat
thus,
W₁ = Q₁ - Q₂
W₂ = Q₂ - Q₃
Thus,
or
or
also,
or
thus,
thus,
or
or
Efficiency = 52%
Answer:
d = 52 μm
Explanation:
given,
wavelength of the light source (λ)= 550 nm
distance to form interference pattern(D) = 1.5 m
y = 1.6 cm = 0.016 m
width of the slits = ?
now, using displacement formula
for the first maxima, m = 1
d = 5.2 x 10⁻⁶ m
d = 52 μm
hence, the width of her slits is equal to d = 52 μm
