Answer:
2.145×10^-10 V or 0.2145nV
Explanation:
From hf=eV
h= Plank's constant = 6.6×10^-34JS
f= frequency of the electromagnetic wave = 5.2×10^4 Hz
e= electronic charge= 1.6×10^-19 C
V= voltage
V= hf/e
V= 6.6×10^-34JS × 5.2×10^4 Hz/ 1.6×10^-19 C
V= 2.145×10^-10 V or 0.2145nV
Therefore the voltage created is 2.145×10^-10 V or 0.2145nV
Answer:
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Explanation:
Answer: A (
,309.8°)
B (2
, 315°)
C (
, 26.56°)
Explanation: To transform rectangular coordinates into polar coordinates use:
and 
For point A:




°
Point A is in the II quadrant, so we substract the angle for 360° since it is in degrees:

309.8°
Polar coordinates for point A is (
, 309.8°)
For point B:





°
Point B is in IV quadrant, so:

315°
Polar coordinates for point B is (
, 315°)
For point C:





26.56°
Polar coordinates for point C is (
, 26.56°)
One side of the mountain that has constant wind and rain blowing onto it, is more likely to catch what is falling than the other side leaving it dryer.