Answer:
R = 1.2295 10⁵ m
Explanation:
After reading your problem they give us the diameter of the lens d = 4.50 cm = 0.0450 m, therefore if we use the Rayleigh criterion for the resolution in the diffraction phenomenon, we have that the minimum separation occurs in the first minimum of diffraction of one of the bodies m = 1 coincides with the central maximum of the other body
θ = 1.22 λ / D
where the constant 1.22 leaves the resolution in polar coordinates and D is the lens aperture
how angles are measured in radians
θ = y / R
where y is the separation of the two bodies (bulbs) y = 2 m and R the distance from the bulbs to the lens
R = 
let's calculate
R = 
R = 1.2295 10⁵ m
We have all the charges for q1, q2, and q3.
Since k = 8.988x10^2, and N=m^2/c^2
F(1) = F (2on1) + F (3on1)
F(2on1) = k |q1 q2| / r(the distance between the two)^2
k^ | 3x10^-6 x -5 x 10^-6 | / (.2m)^2
F(2on1) = 3.37 N
Since F1 is 7N,
F(1) = F (2on1) + F (3on1)
7N = 3.37 N + F (3on1)
Since it wil be going in the negative direction,
-7N = 3.37 N + F (3on1)
F(3on1) = -10.37N
F(3on1) = k |q1 q3| / r(the distance between the two)^2
r^2 x F(3on1) = k |q1 q3|
r = sqrt of k |q1 q3| / F(3on1)
= .144 m (distance between q1 and q3)
0 - .144m
So it's located in -.144m
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Explanation:
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Answer:
6.746 ft/s^2
Explanation:
v(t)=50
v(0)=27
t=5/3600 = 1/720 hours
v(t)-v(0)= a(t-0)
50-27= a(1/720)
a= 23*720= 16560 mi/h^2
16560mi/h^2 * 5280/3600^2 (ft/s^2) =6.746 ft/s^2
