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RSB [31]
3 years ago
10

A camera lens focuses on an object 75.0 cm from the lensThe image forms 3.50 cm behind the lens. What is the magnification of th

e image?
Physics
1 answer:
N76 [4]3 years ago
6 0

Answer:

7/150

Explanation:

The following data were obtained from the question:

Object distance (u) = 75cm

Image distance (v) = 3.5cm

Magnification (M) =..?

Magnification is simply defined as:

Magnification (M) = Image distance (v)/ object distance (u)

M = v /u

With the above formula, we can obtain the magnification of the image as follow:

M = v/u

M = 3.5/75

M = 7/150

Therefore, the magnification of the image is 7/150.

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Latent heat of fusion refers to which changes of state?
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The latent heat of fusion refers to the solid to liquid or liquid to solid states.

Answer: Option C

<u>Explanation: </u>

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7 0
3 years ago
We have three identical metallic spheres A, B, C. Initially sphere A is charged with charge Q, while B and C are neutral. First,
larisa [96]

Answer:

The final charges of each sphere are:   q_A = 3/8 Q , q_B = 3/8 Q ,               q_C = 3/4 Q

Explanation:

This problem asks for the final charge of each sphere, for this we must use that the charge is distributed evenly over a metal surface.

Let's start Sphere A makes contact with sphere B, whereby each one ends with half of the initial charge, at this point

                q_A = Q / 2

                q_B = Q / 2

Now sphere A touches sphere C, ending with half the charge

                q_A = ½ (Q / 2) = ¼ Q

                q_B = ¼ Q

Now the sphere A that has Q / 4 of the initial charge is put in contact with the sphere B that has Q / 2 of the initial charge, the total charge is the sum of the charge

                  q = Q / 4 + Q / 2 = ¾ Q

This is the charge distributed between the two spheres, sphere A is 3/8 Q and sphere B is 3/8 Q

                  q_A = 3/8 Q

                  q_B = 3/8 Q

The final charges of each sphere are:

                q_A = 3/8 Q

                q_B = 3/8 Q

                q_C = 3/4 Q

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