To solve this problem we will apply the concept of magnification, which is given as the relationship between the focal length of the eyepieces and the focal length of the objective. This relationship can be expressed mathematically as,
Here,
= Magnification
= Focal length eyepieces
= Focal length of the Objective
Rearranging to find the focal length of the objective
Replacing with our values
Therefore the focal length of th eobjective lenses is 27.75cm
Answer
A thin atmosphere does not supply much oxygen, and the heat from the sun would evaporate it, because mercury is close to the sun.
Answer:
r = 6.5*10^-3 m
Explanation:
I'm assuming you meant to ask the diameters of the disk, if so, here's it
Given
Quantity of charge on electron, Q = 1.4*10^9
Electric field strength, e = 1.9*10^5
q = Q * 1.6*10^-19
q = 2.24*10^-10
E = q/ε(0)A, making A the subject of formula, we have
A = q / [E * ε(0)], where
ε(0) = 8.85*10^-12
A = 2.24*10^-10 / (1.9*10^5 * 8.85*10^-12)
A = 2.24*10^-10 / 1.6815*10^-6
A = 1.33*10^-4 m²
Remember A = πr²
1.33*10^-4 = 3.142 * r²
r² = 1.33*10^-4 / 3.142
r² = 4.23*10^-5
r = 6.5*10^-3 m
D is the answer
Velocity maybe negative or positive
while speed is always positive
Answer:
real, and then virtual
Explanation:
A converging lens is known as convex lens. This lens is called converging lens because it converges all light rays incident on the lens and parallel to the principal axis at the focus.
The nature of image formed by objects placed in front of this lens as mostly REAL IMAGES. The image formed becomes virtual only when the object is almost in close contact with the lens.
Based on the explanation, it can be deduced that an object placed far from a convex lens forms real images but as we move closer to the lens (almost touching the lens), the image formed overtime tends to be virtual.
