Answer:
1.6 m/s^2
Explanation:
Hello!
To calculate the acceleration we must know the electric field. The electric field and the potential are related by:
If the particle starts at 2.3m, the electric field is:
E = 36.869 V/m = 36.869 N/C
So, the force on the particle is:
F = q E = 2.3×10^−6 C * 36.869 N/C = 8.48 x 10^-5 N
And its acceleration is :
a = F/m = 8.48 x 10^-5 N / 5.4×10−5 kg = 1.57 m/s^2
Rounded to two significant figures:
1.6 m/s^2
Answer:
(A) therefore the image is
(B) therefore the image is
Explanation:
height of the insect (h) = 5.25 mm = 0.525 cm
distance of the insect (s) = 25 cm
radius of curvature of the flat left surface (R1) = ∞
radius of curvature of the right surface (R2) = -12.5 cm (because it is a planoconvex lens with the radius in the direction of the incident rays)
index of refraction (n) = 1.7
(A) we can find the location of the image by applying the formula below
where
f =
f = 17.9 cm
now that we have the focal length we can apply
=\frac{1}{s'}
=\frac{1}{s'}
s' = \frac{447.5}{7.1}[/tex] = 63 cm to the right of the lens
magnification = where y' is the height of the image, therefore
y' = x 0.525 = -13.22 cm
therefore the image is
(B) if the lens is reversed, the radius of curvatures would be interchanged
radius of curvature of the flat left surface (R1) = ∞
radius of curvature of the right surface (R2) = 12.5 cm
we can find the location of the image by applying the formula below
where
f =
f = 17.9 cm
now that we have the focal length we can apply
=\frac{1}{s'}
=\frac{1}{s'}
s' = \frac{447.5}{7.1}[/tex] = 63 cm to the right of the lens
magnification = where y' is the height of the image, therefore
y' = x 0.525 = -13.22 cm
therefore the image is