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kumpel [21]
3 years ago
11

A stargazer has an astronomical telescope with an objective whose focal length is 195 cm and an eyepiece whose focal length is 1

.08 cm. He wants to increase the angular magnification of a galaxy under view by replacing the telescope's eyepiece. Once the eyepiece is replaced, the barrel of the telescope must be adjusted to bring the galaxy back into focus. If the barrel can only be shortened by 0.56 cm from its current length, what is the best angular magnification the stargazer will be able to achieve
Physics
1 answer:
Colt1911 [192]3 years ago
6 0

Answer:

The best angular magnification is

-375

Explanation:

Fe = Fo - L

From the question, (Fo) = 1.08 cm while (L) = 0.56cm

Thus, Fe = 1.08 - 0.56 = 0.52cm

Now, let's apply the angular magnification equation ;

M = - Di/Do

Where Di is the distance from the lens to the image and Do is the distance of the object to the lens.

In this question, Di is represented by f which is the distance from the lens to the image and it is f= 195cm

Also,Do is represented by Fe which is 0.52cm and it is the distance of the object to the lens.

Thus m = - f/fe = - 195/0.52 = - 375

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A research vessel is mapping the bottom of the ocean using sonar. It emits a short sound pulse called "ping" downward. The frequ
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Answer:

d = 4180.3m

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Explanation:

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an echo travels a distance equivalent to 2d, that is to and fro after it reflects from the obstacle.

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7 0
3 years ago
(I) In a ballistic pendulum experiment, projectile 1 results in a maximum height h of the pendulum equal to 2.6 cm. A second pro
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Answer:

The second projectile was 1.41 times faster than the first.

Explanation:

In the ballistic pendulum experiment, the speed (v) of the projectile is given by:  

v = \frac{m + M}{m} \cdot \sqrt{2gh}

<em>where m: is the mass of the projectile, M: is the mass of the pendulum, g: is the gravitational constant and h: is the maximum height of the pendulum.   </em>

To know how many times faster was the second projectile than the first, we need to take the ratio for the velocities for the projectiles 2 and 1:    

\frac{v_{2}}{v_{1}} = \frac{\frac{m_{2} + M}{m_{2}} \cdot \sqrt{2gh_{2}}}{\frac{m_{1} + M}{m_{1}} \cdot \sqrt{2gh_{1}}}           (1)

<em>where m₁ and m₂ are the masses of the projectiles 1 and 2, respectively, and h₁ and h₂ are the maximum height reached by the pendulum by the projectiles 1 and 2, respectively.  </em>

Since the projectile 1 has the same mass that the projectile 2, we can simplify equation (1):

\frac{v_{2}}{v_{1}} = \frac{\sqrt{h_{2}}}{\sqrt{h_{1}}}  

\frac{v_{2}}{v_{1}} = \frac{\sqrt{5.2 cm}}{\sqrt{2.6 cm}}

\frac{v_{2}}{v_{1}} = 1.41  

Therefore, the second projectile was 1.41 times faster than the first.  

I hope it helps you!

8 0
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1. GPE - 40 * 2 * 10 = 800j
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