Answer:
A. Sedimentary
Explanation:
I took the test and got this correct. Hopefully this helps you!
Let's use the mirror equation to solve the problem:

where f is the focal length of the mirror,

the distance of the object from the mirror, and

the distance of the image from the mirror.
For a concave mirror, for the sign convention f is considered to be positive. So we can solve the equation for

by using the numbers given in the text of the problem:



Where the negative sign means that the image is virtual, so it is located behind the mirror, at 8.6 cm from the center of the mirror.
Answer:
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Explanation:
Answer:
The Magnifying power of a telescope is 
Explanation:
Radius of curvature R = 5.9 m = 590 cm
focal length of objective
= 
⇒
= 
⇒
= 295 cm
Focal length of eyepiece
= 2.7 cm
Magnifying power of a telescope is given by,



therefore the Magnifying power of a telescope is 
Answer:
I = 21.13 mA ≈ 21 mA
Explanation:
If
I₁ = 5 mA
L₁ = L₂ = L
V₁ = V₂ = V
ρ₁ = 1.68*10⁻⁸ Ohm-m
ρ₂ = 1.59*10⁻⁸ Ohm-m
D₁ = D
D₂ = 2D
S₁ = 0.25*π*D²
S₂ = 0.25*π*(2*D)² = π*D²
If we apply the equation
R = ρ*L / S
where (using Ohm's Law):
R = V / I
we have
V / I = ρ*L / S
If V and L are the same
V / L = ρ*I / S
then
(V / L)₁ = (V / L)₂ ⇒ ρ₁*I₁ / S₁ = ρ₂*I₂ / S₂
If
S₁ = 0.25*π*D² and
S₂ = 0.25*π*(2*D)² = π*D²
we have
ρ₁*I₁ / (0.25*π*D²) = ρ₂*I₂ / (π*D²)
⇒ I₂ = 4*ρ₁*I₁ / ρ₂
⇒ I₂ = 4*1.68*10⁻⁸ Ohm-m*5 mA / 1.59*10⁻⁸ Ohm-m
⇒ I₂ = 21.13 mA