Consider a simple ideal Rankine cycle with fixed turbine inlet conditions. What is the effect of lowering the condenser pressure
1 answer:
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Answer:
Magnification, m = 3
Explanation:
It is given that,
Focal length of the lens, f = 15 cm
Object distance, u = -10 cm
Lens formula :
v is image distance
Magnification,
So, the magnification of the lens is 3.
Answer:
<em>radius of the loop = 7.9 mm</em>
<em>number of turns N ≅ 399 turns</em>
Explanation:
length of wire L= 2 m
field strength B = 3 mT = 0.003 T
current I = 12 A
recall that field strength B = μnI
where n is the turn per unit length
vacuum permeability μ = = 1.256 x 10^-6 T-m/A
imputing values, we have
0.003 = 1.256 x 10^−6 x n x 12
0.003 = 1.507 x 10^-5 x n
n = 199.07 turns per unit length
for a length of 2 m,
number of loop N = 2 x 199.07 = 398.14 ≅ <em>399 turns</em>
since there are approximately 399 turns formed by the 2 m length of wire, it means that each loop is formed by 2/399 = 0.005 m of the wire.
this length is also equal to the circumference of each loop
the circumference of each loop =
0.005 = 2 x 3.142 x r
r = 0.005/6.284 = = 0.0079 m =<em> 7.9 mm</em>
Answer:
V = 2.5 J/C
Explanation:
<u><em>Given:</em></u>
Energy = E = 20 J
Charge = Q = 8 C
<u><em>Required:</em></u>
Potential Difference = V = ?
<u><em>Formula:</em></u>
V =
<u><em>Solution:</em></u>
V = 20/8
V = 2.5 J/C
To find the number of electrons transferred, we should divide the total charge acquired by the rod 
by the charge of a single electron (
), and we find:
