<span>They are used to measure and map effluent and pollution discharges from factories and sewerage plants, and the movement of sand around harbours, rivers and bays. Radioactive materials used for such purposes have short half-lives and decay to background levels within days.</span>
Answer:
Vi = 0.055 m³ = 55 L
Explanation:
From first Law of Thermodynamics, we know that:
ΔQ = ΔU + W
where,
ΔQ = Heat absorbed by the system = 52.5 J
ΔU = Change in Internal Energy = -102.5 J (negative sign shows decrease in internal energy of the system)
W = Work Done in Expansion by the system = ?
Therefore,
52.5 J = - 102.5 J + W
W = 52.5 J + 102.5 J
W = 155 J
Now, the work done in a constant pressure condition is given by:
W = PΔV
W = P(Vf - Vi)
where,
P = Constant Pressure = (0.5 atm)(101325 Pa/1 atm) = 50662.5 Pa
Vf = Final Volume of System = (58 L)(0.001 m³/1 L) = 0.058 m³
Vi = Initial Volume of System = ?
Therefore,
155 J = (50662.5 Pa)(0.058 m³ - Vi)
Vi = 0.058 m³ - 155 J/50662.5 Pa
Vi = 0.058 m³ - 0.003 m³
<u>Vi = 0.055 m³ = 55 L</u>
Answer:
143 °
Explanation:
a ) If d be the distance between slits , λ be wavelength of light used and at angle θ nth dark fringe is formed then
d sinθ = ( 2n+1) λ/2
for first dark fringe
d sinθ = λ/2
d /λ = 1/ 2 sinθ
1 / 2 sin15
= 1.93
b )
For intensity of fringe at angle θ, the relation is
I = I₀ cos²θ
I / I₀ = cos²θ/2
Given I / I₀ =0. 1
0.1 = cos²θ/2
θ/2 = 71.5
θ = 143 °
<span>B. equal and in opposite directions</span>
