Answer:
R = 9880 yd * 3 ft/yd / 5280 ft/mi = 5.61 mi
If you do it in steps
R = 9880 yd * 3 ft/yd = 29640 ft
R = 29640 ft / 5280 ft/mi = 5.61 mi
Answer:
M g H / 2 = M g L / 2 initial potential energy of rod
I ω^2 / 2 = 1/3 M L^2 * ω^2 / 2 kinetic energy attained by rod
M g L / 2 = 1/3 M L^2 * ω^2 / 2
g = 3 L ω^2
ω = (g / (3 L))^1/2
Answer:
E= 4.35*10^6 N/C
Explanation:
Let's find the area charge density of the plate
α= 6.9*10^-6/9*10^-2 = 7.7*10^-5C/m2
Now we can calculate the electric field just of the plate
E =α/2e =7.7*10^-5/2*8.85*10^-12 = 4.35*10^6 N/C
I think that the answer to that is true hope that helps
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