Answer:
efficiency of solar panel is 18%
Explanation:
Energy received by the sunlight is given as
![Power = intensity \times area](https://tex.z-dn.net/?f=Power%20%3D%20intensity%20%5Ctimes%20area)
here we know the dimensions of the plate as
![L = 0.57 m](https://tex.z-dn.net/?f=L%20%3D%200.57%20m)
![W = 1.3 m](https://tex.z-dn.net/?f=W%20%3D%201.3%20m)
now we have
![Area = (0.57)(1.3) m^2](https://tex.z-dn.net/?f=Area%20%3D%20%280.57%29%281.3%29%20m%5E2)
![A = 0.741 m^2](https://tex.z-dn.net/?f=A%20%3D%200.741%20m%5E2)
now the power received by the sun light is given as
![P_1 = 0.741(300) = 222.3 W](https://tex.z-dn.net/?f=P_1%20%3D%200.741%28300%29%20%3D%20222.3%20W)
now the output power due to solar panel is given as
![P_{out} = V i](https://tex.z-dn.net/?f=P_%7Bout%7D%20%3D%20V%20i)
![P_{out} = (6.6 A)(6 V)](https://tex.z-dn.net/?f=P_%7Bout%7D%20%3D%20%286.6%20A%29%286%20V%29)
![P_{out} = 39.6 Watt](https://tex.z-dn.net/?f=P_%7Bout%7D%20%3D%2039.6%20Watt)
now the efficiency is given as
![\eta = \frac{P_{out}}{P_{in}}](https://tex.z-dn.net/?f=%5Ceta%20%3D%20%5Cfrac%7BP_%7Bout%7D%7D%7BP_%7Bin%7D%7D)
![\eta = \frac{39.6}{222.3}](https://tex.z-dn.net/?f=%5Ceta%20%3D%20%5Cfrac%7B39.6%7D%7B222.3%7D)
![\eta = 0.18](https://tex.z-dn.net/?f=%5Ceta%20%3D%200.18)
Answer:
(a) The final angular speed is 12.05 rad/s
(b) The time taken to turn 5.5 revolutions is 5.74 s
Explanation:
Given;
number of revolutions, θ = 5.5 revolutions
acceleration of the wheel, α = 20 rpm/s
number of revolutions in radian is given as;
θ = 5.5 x 2π = 34.562 rad
angular acceleration in rad/s² is given as;
![\alpha = \frac{20 \ rev}{min} *\frac{1}{s} *(\frac{2\pi \ rad}{1 \ rev } *\frac{1 \ min}{60 \ s}) \\\\\alpha = 2.1 \ rad/s^2](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B20%20%5C%20rev%7D%7Bmin%7D%20%2A%5Cfrac%7B1%7D%7Bs%7D%20%2A%28%5Cfrac%7B2%5Cpi%20%5C%20rad%7D%7B1%20%5C%20rev%20%7D%20%2A%5Cfrac%7B1%20%5C%20min%7D%7B60%20%5C%20s%7D%29%20%5C%5C%5C%5C%5Calpha%20%3D%202.1%20%5C%20rad%2Fs%5E2)
(a)
The final angular speed is given as;
![\omega _f^2 = \omega_i ^2 + 2\alpha \theta\\\\\omega _f^2 = 0 + 2\alpha \theta\\\\\omega _f^2 = 2\alpha \theta\\\\\omega _f = \sqrt{2\alpha \theta}\\\\ \omega _f = \sqrt{2(2.1) (34.562)}\\\\ \omega _f = 12.05 \ rad/s](https://tex.z-dn.net/?f=%5Comega%20_f%5E2%20%3D%20%5Comega_i%20%5E2%20%2B%202%5Calpha%20%5Ctheta%5C%5C%5C%5C%5Comega%20_f%5E2%20%3D%200%20%2B%20%202%5Calpha%20%5Ctheta%5C%5C%5C%5C%5Comega%20_f%5E2%20%3D%20%20%202%5Calpha%20%5Ctheta%5C%5C%5C%5C%5Comega%20_f%20%3D%20%5Csqrt%7B2%5Calpha%20%5Ctheta%7D%5C%5C%5C%5C%20%5Comega%20_f%20%20%3D%20%5Csqrt%7B2%282.1%29%20%2834.562%29%7D%5C%5C%5C%5C%20%5Comega%20_f%20%3D%2012.05%20%5C%20rad%2Fs)
(b) the time taken to turn 5.5 revolutions is given as
![\omega _f = \omega _i + \alpha t\\\\12.05 = 0 + 2.1t\\\\t = \frac{12.05}{2.1} \\\\t = 5.74 \ s](https://tex.z-dn.net/?f=%5Comega%20_f%20%3D%20%5Comega%20_i%20%2B%20%5Calpha%20t%5C%5C%5C%5C12.05%20%3D%200%20%2B%202.1t%5C%5C%5C%5Ct%20%3D%20%5Cfrac%7B12.05%7D%7B2.1%7D%20%5C%5C%5C%5Ct%20%3D%205.74%20%5C%20s)
Answer:
Explanation:
1 g is 9.8 m/s^2 the problem wants the results in km/h so we'll fix that really quick.
9.8 m/s^2 (1 km/1000m)(60 sec/1 min)^2(60 min/1 hour)^2 = 127008 km/hour^2
Now, I'm assuming the ship is starting from rest, and hopefully you know your physics equations. We are going to use vf = vi + at. Everything is just given, or we can assume, so I'll just solve.
vf = vi + at
vf = 0 + 127008 km/hour^2 * 24 hours
vf = 3,048,192 km/hour
If there's anything that doesn't make sense let me know.
d. fishing
The other options all directly affect algal bloom production because they affect the nutrients in the water, and an overabundance of certain nutrients in the water is what causes algal bloom.