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tatuchka [14]
2 years ago
14

Two converging lenses are separated by 27.0 cm. The focal length of each lens is 8.90 cm. An object is placed 33.0 cm to the lef

t of the lens that is on the left. Determine the final image distance relative to the lens on the right.
Physics
1 answer:
Ymorist [56]2 years ago
7 0

Answer:

1/i + 1/o = 1/f     thin lens equation

i = 33 * 8.9 / (33 - 8.9) = 12.2 cm  to right of first lens

27 - 12.2 = 14.8 cm to left of second lens

i = 14.8 * 8.9 / (14.8 - 8.9) = 22,3 cm to right of second lens

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2 years ago
The area of the velocity time graph gives displacement of the body true or false?
mote1985 [20]

The area of the velocity time graph gives displacement of the body true or false?

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3 years ago
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A mother pushes a baby stroller 10 meters by applying 40 newtons of force. how much work was done?
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3 years ago
A skydiver of mass 80.0 kg jumps from a slow-moving aircraft and reaches a terminal speed of 50.0 m/s. (a) What is her accelerat
kirill [66]

Answer:

6.22²

Explanation:

Given that

Mass of the skydiver, m = 80 kg

Terminal speed of the skydiver, v(f) = 50 m/s

Speed of the skydiver, v(i) = 30 m/s

Acceleration of the skydiver, a = ?

To solve this, we use the formula

W - k v² = ma, where

W = weight of the skydiver

k = constant

v = speed of the skydiver

m = mass of the skydiver

So, if we substitute the values into it we have

W = mg = 80 * 9.8 = 784 N

784 - k 50² = 80 *0

784 - 2500k = 0

784 = 2500k

k = 0.3136

Now, we use this value of k to find the needed acceleration using the same formula at a speed of 30 m/s

784 - 0.3136 * 30² = 80 * a

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