Answer:
I = 27.65A < 40.59°
PowerFactor = 0.76
Explanation:
Current on the heating load is:
I1 = 30KW / 4KV = 7.5A < 0°
Current on the inductive load:
I2 = (150KVA*0.6) /4KV = 22.5A with an angle of acos(0.6)=53.1°
The sum of both currents is:
It = I1 + I2 = 7.5<0° + 22.5<53.1° = 27.65<40.59°
Now, the power factor will be:
pf = cos (40.59°) = 0.76
Answer:
if you stretch a spring with k = 2, with a force of 4N, the extension will be 2m. the work done by us here is 4x2=8J. in other words, the energy transferred to the spring is 8J. but, the stored energy in the spring equals 1/2x2x2^2=4J (which is half of the work done by us in stretching it).
The wavelength emitted is indirectly proportional to the difference in the change in the energy level. For the wavelength 278 nm the change in energy level is significantly high. Further change in energy level is indicated by 454nm light but the difference in energy level for this wavelength to be emitted is not greater than the previous one. There is a possibility that these subsystems have now very low energy which should result in wavelengths ranging from 700 to 900 nm. There is another possibility that there is some metastable subsystems in the system which may cause LASER emission.
Answer:
The actual elevation angle is 12.87 degrees
Explanation:
In the attachment you can clearly see the situation. The angle of elevation as seen for the scuba diver is shown in magenta, we conclude that 
.
Using Snell's Law we can write:
,
Let's approximate the index of refraction of the air (medium 1 in the picture) to 1.
We thus have:
![\implies\theta_1=\arcsin[n_2\sin(\theta_2)]=\arcsin[1.333\sin(47)]\approx 77.13](https://tex.z-dn.net/?f=%5Cimplies%5Ctheta_1%3D%5Carcsin%5Bn_2%5Csin%28%5Ctheta_2%29%5D%3D%5Carcsin%5B1.333%5Csin%2847%29%5D%5Capprox%2077.13)
. Calling 
the actual angle of elevation, we get from the picture that 
