Answer:
a) (Ω-m)^{-1}
b) Resistance = 121.4 Ω
Explanation:
given data:
diameter is 7.0 mm
length 57 mm
current I = 0.25 A
voltage v = 24 v
distance between the probes is 45 mm
electrical conductivity is given as
(Ω-m)^{-1}[/tex]
b)
Resistance = 121.4 Ω
Answer:
1) 63.66 ohm
2) 188.49 ohm
Explanation:
Data provided in the question:
Part 1
Capacitance, C = 5μF = 5 × 10⁻⁶ F
Frequency = 500 Hz
Now,
Impedance =
or
Impedance =
or
Impedance = 63.66 ohm
Part 2
Inductance = 60 mH = 60 × 10⁻³ H
Frequency = 500 Hz
Now,
Impedance for an inductor = 2πfL
thus,
Impedance = 2 × π × 500 × 60 × 10⁻³
= 188.49 ohm
Answer:
0.1 nm
Explanation
Potential deference of the electron is given as V =150 V
Mass of electron
Let the velocity of electron = v
Charge on the electron
plank's constant h =
According to energy conservation
Now we know that De Broglie wavelength
The classical motion for an oscillator that starts from rest at location x₀ is
x(t) = x₀ cos(ωt)
The probability that the particle is at a particular x at a particular time t
is given by ρ(x, t) = δ(x − x(t)), and we can perform the temporal average
to get the spatial density. Our natural time scale for the averaging is a half
cycle, take t = 0 → π/ ω
Thus,
ρ =
Limit is 0 to π/ω
We perform the change of variables to allow access to the δ, let y = x₀ cos(ωt) so that
ρ(x) =
Limit is x₀ to -x₀
Limit is -x₀ to x₀
This has as expected. Here the limit is -x₀ to x₀
The expectation value is 0 when the ρ(x) is symmetric, x ρ(x) is asymmetric and the limits of integration are asymmetric.
Answer:
Heat transferred from the refrigerated space = 95.93 kJ/kg
Work required = 18.45 kJ/kg
Coefficient of performance = 3.61
Quality at the beginning of the heat addition cycle = 0.37
Explanation:
From figure
is heat rejection process
is heat transferred from the refrigerated space
is high temperature = 50 °C + 273 = 323 K
is low temperature = -20 °C + 273 = 253 K
is net work of the cycle (the difference between compressor's work and turbine's work)
Coefficient of performance of a Carnot refrigerator is calculated as
From figure it can be seen that heat rejection is latent heat of vaporisation of R-12 at 50 °C. From table
From coefficient of performance definition
Energy balance gives
Vapor quality at the beginning of the heat addition cycle is calculated as (f and g refer to saturated liquid and saturated gas respectively)
From figure
Replacing with table values
Quality can be computed by other properties, for example, specific enthalpy. Rearrenging quality equation we get
By energy balance, turbine's work is
Finally, compressor's work is
