Answer:
I=9.6×e^{-8} A
Explanation:
The magnetic field inside the solenoid.
B=I*500*muy0/0.3=2.1×e ^-3×I.
so the total flux go through the square loop.
B×π×r^2=I×2.1×e^-3π×0.025^2
=4.11×e^-6×I
we have that
(flux)'= -U
so differentiating flux we get
so the inducted emf in the loop.
U=4.11×e^{-6}×dI/dt=4.11×e^-6×0.7=2.9×e^-6 (V)
so, I=2.9×e^{-6}÷30
I=9.6×e^{-8} A
Answer:
1. Equatorial Evergreen or Rainforest
2. Tropical forest
3. Mediterranean forest
4. Temperate broad-leaved forest
5. Warm temperate forest
Explanation:
Answer:
k = 4.21 * 10⁻³(L/(mol.s))
Explanation:
We know that
k = Ae
------------------- euqation (1)
K= rate constant;
A = frequency factor = 4.36 10^11 M⁻¹s⁻¹;
E = activation energy = 93.1kJ/mol;
R= ideal gas constant = 8.314 J/mol.K;
T= temperature = 332 K;
Put values in equation 1.
k = 4.36*10¹¹(M⁻¹s⁻¹)e![^{[(-93.1*10^3)(J/mol)]/[(8.314)(J/mol.K)(332K)}](https://tex.z-dn.net/?f=%5E%7B%5B%28-93.1%2A10%5E3%29%28J%2Fmol%29%5D%2F%5B%288.314%29%28J%2Fmol.K%29%28332K%29%7D)
k = 4.2154 * 10⁻³(M⁻¹s⁻¹)
here M =mol/L
k = 4.21 * 10⁻³((mol/L)⁻¹s⁻¹)
or
k = 4.21 * 10⁻³((L/mol)s⁻¹)
or
k = 4.21 * 10⁻³(L/(mol.s))
The amount of work done by steady flow devices varies with the particular gas volume. The kinetic energy of gas particles decreases during cooling.
When the gas is subjected to intermediate cooling during compression, the gas specific volume is reduced, which lowers the compressor's power consumption. Compression is less adiabatic and more isothermal because the compressed gas must be cooled between stages since compression produces heat. The system's thermodynamic cycle's cold sink temperature is lowered by cooling the compressor coils. By increasing the temperature difference between the heat source and the cold sink, this improves efficiency.
Learn more about thermodynamics here-
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