Answer:
Efficent at low speeds.
Explanation:
Since water is 1,000 times more dense than air, electricity can generated from tides much more efficiently at slower speeds than wind turbines can.
Answer:
0.5 kW
Explanation:
The given parameters are;
Volume of tank = 1 m³
Pressure of air entering tank = 1 bar
Temperature of air = 27°C = 300.15 K
Temperature after heating = 477 °C = 750.15 K
V₂ = 1 m³
P₁V₁/T₁ = P₂V₂/T₂
P₁ = P₂
V₁ = T₁×V₂/T₂ = 300.15 * 1 /750.15 = 0.4 m³

For ideal gas,
= 5/2×R = 5/2*0.287 = 0.7175 kJ
PV = NKT
N = PV/(KT) = 100000×1/(750.15×1.38×10⁻²³)
N = 9.66×10²⁴
Number of moles of air = 9.66×10²⁴/(6.02×10²³) = 16.05 moles
The average mass of one mole of air = 28.8 g
Therefore, the total mass = 28.8*16.05 = 462.135 g = 0.46 kg
∴ dQ = 0.46*0.7175*(750.15 - 300.15) = 149.211 kJ
The power input required = The rate of heat transfer = 149.211/(60*5)
The power input required = 0.49737 kW ≈ 0.5 kW.
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Answer:
maximum isolator stiffness k =1764 kN-m
Explanation:
mean speed of rotation 


=65.44 rad/sec


= 0.1*(65.44)^2
F_T =428.36 N
Transmission ratio 
also
transmission ratio ![= \frac{1}{[\frac{w}{w_n}]^{2} -1}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B1%7D%7B%5B%5Cfrac%7Bw%7D%7Bw_n%7D%5D%5E%7B2%7D%20-1%7D)
![0.7 =\frac{1}{[\frac{65.44}{w_n}]^2 -1}](https://tex.z-dn.net/?f=0.7%20%3D%5Cfrac%7B1%7D%7B%5B%5Cfrac%7B65.44%7D%7Bw_n%7D%5D%5E2%20-1%7D)
SOLVING FOR Wn
Wn = 42 rad/sec

k = m*W^2_n
k = 1000*42^2 = 1764 kN-m
k =1764 kN-m
Answer:
Could ask a family member to help
Explanation: