Answer:
a) i = -9.63 cm
, h ’= .0.24075 cm erect
b) i = 259.74 cm
,
Explanation:
For this exercise let's start by finding the focal length of the lens
1 / f = (n-1) (1 / R₁ - 1 / R₂)
1 / f = (1.70 -1)) 1 / ∞ - 1/13)
1 / f = 0.0538
f = - 18.57 cm
Now we can use the constructor equation
1 / f = 1 / o + 1 / i
1 / i = 1 / f - 1 / o
1 / i = -1 / 18.57 -1/20
1 / i = -0.1038 cm
I = -9.63 cm
For the height of the
image let's use magnification
m = h '/ h = - i / o
h ’= -h i / o
h ’= - 0.5 (-9.63) / 20
h ’= .0.24075 cm
b) we invert the lens
The focal length is
1 / f = (1.70 -1) (1/13 - 1 / int)
1 / f = 0.0538
f = 18.57 cm
1 / i = 1 / f -1 / o
1 / I = 1 / 18.57 - 1/20
1 / I = 3.85 10-3
i = 259.74 cm
h ’= - 0.5 259.74 / 20
h ’= 6.4935 cm
Vf = 0 + 3.5•8.7
= 30.45 m/s
Answer:
Explanation:
The process is isothermic, as P V = constant .
work done = 2.303 n RT log P₁ / P₂
= 2.303 x 5 / 29 x 8.3 x 303 log 2 / 1 kJ
= 300.5k J
This energy in work done by the gas will come fro heat supplied as internal energy is constant due to constant temperature.
heat supplied = 300.5k J
specific volume is volume per unit mass
v / m
pv = n RT
pv = m / M RT
v / m = RT / p M
specific volume = RT / p M
option B is correct.
