Answer:
a) 90 kJ
b) 230.26 kJ
Explanation:
The pressure at the first point = 10 bar —> 10 x 102 = 1020 kPa
The volume at the first point = 0.1 m^3
The pressure at the second point = 1 bar —> 1 x 102 = 102 kPa
The volume at the second point = 1 m^3
Process A.
constant volume V = C from point (1) to P = 10 bar.
Constant pressure P = C to the point (2).
Process B.
The relation of the process is PV = C
Required
For process A & B
(a) Sketch the process on P-V coordinates
(b) Evaluate the work W in kJ.
Assumption
Quasi-equilibrium process
Kinetic and potential effect can be ignored.
Solution
For process A.
V=C
There is no change in volume then
The work is defined by
║ V║limit 1--0.1
90 kJ
Process B
PV=C
By substituting with point (1) C = 10^2 x 1= 10^2
The work is defined by
║ ln(V)║limit 1--0.1
=230.26 kJ
Explanation:
The width of the central maximum is given by
W = 2 λ L / a
where W is the width of the central maximum
λ is the wavelength of the light used.
L is the distance between the aperture and screen
a is the width of the slit or aperture
So we can see that if any one quantity is varied by keeping others constant in the above formula , there would be a change in width of central maximum.
Answer:
19.53 cm
Explanation:
The computation of the height is as follows:
Here we applied the conservation of the energy formula
As we know that
P.E of the block = P.E of the spring
m g h = ( 1 ÷ 2) k x^2
where
m = 0.15
g = 9.81
k = 420
x = 0.037
So now put the values to the above formula
(0.15) (9.81) (h) = 1 ÷2 × 420 × (0.037)^2
1.4715 (h) = 0.28749
h = 0.19537 m
= 19.53 cm
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