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2 years ago

The two major parts of the optical telescope

1 answer:

5 0

The "objective" (lens or mirror) is the major major major part of
the optical telescope. It's really the only part you need in order
to make a telescope (besides something to hold the objective).
You can put a piece of film or a CCD right at the focal point of
the objective lens or mirror and capture 'images' (pictures) there.

If you want to use the telescope for looking through and seeing stuff
with your eye, then you need the other major part ... the eyepiece lens.

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PHYSICS PLEASE HELP!

Salsk061 [2.6K]

Answer:

It would take two half-lives which would be 130days

The substance has a half-life of 30 minutes. There has been two cycles that have occurred.

idk the last one

Explanation:

3 0

3 years ago

Determine the magnitude of the momentum of a ... 107-kg halfback moving eastward at 8 m/s.The halfback's momentum in kgm/s is:

Veronika [31]

Given:

The mass of the halfback is m = 107 kg

The speed of the halfback is v = 8 m/s

To find the momentum.

Explanation:

The momentum of the halfback is

$\begin{gathered} p=\text{ mv} \\ =\text{ 107}\times8 \\ =\text{ 856 kg m/s} \end{gathered}$

Thus, the momentum of the halfback is 856 kg m/s

3 0

1 year ago

A net force of 0.7 N is applied on a body. What happens to the acceleration of the body in a second trial if half of the net for

Dmitriy789 [7]

Answer:

The answer is The acceleration is double its original value.

Explanation:

<h2><u>It is because of the second trial of accelaration. Because of this, an object's acceleration doubles from its original value.</u></h2><h2><u></u></h2>

Hope this helps....

Have a nice day!!!!

6 0

2 years ago

Read 2 more answers

At a meeting of physics teacher in Montana, the teachers were asked to calculate where a flour sack would land if dropped from a

Harlamova29_29 [7]

At a distance of 469.2 m from the original point below the airplane.

Explanation:

First of all, we have to calculate the time it takes for the sack to reach the ground.

To do so, we just analyze its vertical motion, which is a free-fall motion, so we can use the suvat equation:

$s=ut+\frac{1}{2}at^2$

where, taking downward as positive direction:

s = 300 m is the vertical displacement

u = 0 is the initial vertical velocity

t is the time

$a=g=9.8 m/s^2$ is the acceleration of gravity

Solving for t, we find the it takes for the sack to reach the ground:

$t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(300)}{9.8}}=7.82 s$

Now we analyze the horizontal motion. The horizontal velocity of the pack is constant (since there are no forces along the horizontal direction) and equal to the initial speed of the airplane, so:

$v_x = 60 m/s$