Answer:
462 nm
Explanation:
Given: width of the slit, d = 5.6 × 10⁻⁴ m
Distance of the screen, D = 4.0 m
Fringe width, β = 3.3 mm = 3.3 × 10⁻³ m
First dark fringe means n =1
Wavelength of the light, λ = ?
When all forces on an object are balanced, the object is in .. What?
2 answers:
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Answer:
Thermal Efficiency, η = . . . . . . . . . . . . . . . . Eqn 1
where W₀ = Work Output = Q₁ - Q₀ =82500KW . . . . . . . . . . . . . . . Eqn 2
Q₁ = Heat Supplied/Input = mC(ΔT₁)
Q₁ = Heat Rejected/Output = mC(ΔT₀) . . . . . . . . . . . . . . . . . . . . . . . . Eqn 3
Note: From Carnot's theorem, for any engine working between these two temperatures (T₀/T₁), The maximum attainable efficiency is the Carnot efficiency given as follows;
Therefore, η = 1 - = 1 -
Remember, T₁ = 475K and T₀ = 308K
η = 1 - (308/475) = 1 - 0.648 = 0.352
Hence, the maximum efficiency at which this plant can operate = 35%
2. To determine the minimum amount of rejected heat that must be removed from the condenser every twenty-four hours.
Remember from Eqn 1, Q₁ = W/η,
Therefore, Q₁= 82500/0.35
Q₁=235,714KW,
So, from Eqn 2, Q₀ = 235714 - 82500
Q₀ = 153214KW (KJ/s) (Released Heat)
In t =24 hours, we can then use this to determine the minimum amount of heat rejected qₓ (KiloJoule), = Q₀t (Remember, you have to convert the time, t, unit to seconds)
= 153214 x t (KiloJoule)
qₓ = 153214 x 24 x 3600 (KiloJoule)
qₓ = 13238 MegaJoule
<h3>Therefore, the minimum amount of rejected heat that must be removed from the condenser every twenty-four hours is 13238 MJoule</h3>Answer:
A 8.0 cm diameter horizontal pipe gradually narrows to 5.0 cm. The the water flows through this pipe at certain rate, the gauge pressure in these two sections is 31.0 kPa and 24.0 kPa, respectively. What is the volume of rate of flow?
The flow rate is 3.1175×10⁻³ m³/s
Explanation:
To solve the question we rely on Bernoulli's principle as follows
thus where the pipe is horizontal we have
z₁ = z₂ hence the above equattion becomes
since the flow rate is constant then
Q = v₁A₁ = v₂A₂
Where is the area of the two sections given by A₁ = π·D₁²÷4 and
A₂ = π·D₂²÷4
Thereffore A₁ = π·0.08²÷4 = 5.02×10⁻³ m²
and A₂ = π·0.05²÷4 = 1.96×10⁻³ m²
v₁ = v₂A₂/A₁ =0.391×v₂
The given pressures are P₁ = 31.0 kPa and P₂ = 24.0 pKa and
ρ = 1000 kg/m³
Plugging the values into the above equation we get
31.0 kPa +0.5× 1000 kg/m³× (0.391×v₂)² = 24.0 pKa +0.5×1000 kg/m³×v₂²
= 31000+76.3·v₂² =24000+500·v₂²
or 423.706·v₂² = 7000
v₂² = 7000/423.706 = 16.52 or v₂ = 4.065 m/s and v₁ 0.391×4.065 = 1.59 m/s
The flow rate = v₂A₂ = 1.59×1.96×10⁻³ = 3.1175×10⁻³ m³/s