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3241004551 [841]
4 years ago
13

Compare the light gathering power of a 1 meter diameter telescope to that of the human eye ,which has a diameter of roughly 2.5

centimeters. How many times more light can the telescope gather than the human eye?
Physics
1 answer:
lesantik [10]4 years ago
7 0

Answer:

The telescope can gather light 1600 times more than the human eyes can!

Explanation:

The light gathering ability of an optical element is directly proportional to its area of opening.

So, in comparing the light gathering abilities for two objects, it is just the ratio of their area of opening.

Let the diameter of the telescope be D = 1 m

And the diameter of the human eyes be d = 2.5 cm = 0.025 m

Light gathering ability of the telescope compared to the eyes = D² ÷ d²

= (D²/d²) = (1²/0.025²) = 1600 times.

The telescope can gather light 1600 times more than the human eyes can!

Hope this Helps!!!

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A basketball player shoots toward a basket 5.3 m away and 3.0 m above the floor. If the ball is released 1.9 m above the floor a
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Answer:

Vi = 8.28 m/s

Explanation:

This problem is related to the projectile motion.

As we know there are two components of motion associated with this, the horizontal component and vertical component.

The horizontal distance covered by the ball is

Vx*t = x

Vx*t = 5.3

Vx = 5.3/t  eq. 1

Also we know that

Vx = Vicos(60)

Vx = Vi*0.5  eq. 2

equate eq. 1 and eq. 2

5.3/t = Vi*0.5

5.3/0.5 = Vi*t

Vi*t = 10.6  eq. 3

The vertical distance is

Vy = y1 + Vyi*t - 0.5gt²

also we know that

Vyi = Visin(60)

Vyi = Vi*0.866

It is given that V1 = 1.9 m and and Vy = 3 m is the vertical distance

3 = 1.9 + Vi*0.866*t - 0.5gt²

3 = 1.9 + Vi*0.866*t - 0.5(9.8)t²

3 = 1.9 + 0.866(Vi*t) - 0.5(9.8)t²

3 = 1.9 + 0.866(Vi*t) - 0.5(9.8)t²

1.1 = 0.866(Vi*t) - 4.9t²

0.866(Vi*t) = 4.9t² + 1.1

substitute Vi*t = 10.6 in above equation

0.866(10.6) = 4.9t² + 1.1

9.18 = 4.9t² + 1.1

4.9t² = 8.08

t² = 8.08/4.9

t² = 1.648

t = 1.28 sec

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Vi*t = 10.6

Vi = 10.6/t

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The traffic on the freeway is moving at a constant speed of 24 m/sm/s. What distance does the traffic travel while the car is mo
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Incomplete question as there is so much information is missing.The complete question is here

A car sits on an entrance ramp to a freeway, waiting for a break in the traffic. Then the driver accelerates with constant acceleration along the ramp and onto the freeway. The car starts from rest, moves in a straight line, and has a speed of 24 m/s (54 mi/h) when it reaches the end of the 120-m-long ramp. The traffic on the freeway is moving at a constant speed of 24 m/s. What distance does the traffic travel while the car is moving the length of the ramp?

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