Ask question

Ask question

All categories

Ksenya-84 [330]

3 years ago

10

2. Which part of a refracting telescope forms the image?

1 answer:

Bezzdna [24]3 years ago

4 0

<em> In a refracting telescope </em><u><em>convex lens</em></u><em> forms the image. </em>

<u>Answer:</u> <em>c. Convex mirror</em>

<u>Explanation:</u>

Telescope is an instrument used for magnification of distant objects. The convex lenses objective and eyepiece are the two parts of a refracting telescope.

Objective has a greater focal length when compared with the eyepiece. Image of a distant object is formed at the second focal point of the objective. This image is magnified by the eyepiece.

The objective and eyepiece lenses can only produce an inverted image since they both are convex lenses. The function of producing a final erect image is performed by a pair of inverting lenses.

You might be interested in

I need help plz!!! Please help...

RSB [31]

Answer: 60

Explanation:

4 0

2 years ago

No links please, links never work, this is science not physics

stiks02 [169]

Answer:

b and c are the answers

A is an opinion, D is a superstitious belief because they haven’t found Jesus obviously

7 0

3 years ago

If i have a kinematic equation vf^2=vi^2-2*a(xf-xi), how can i solve for xi step by step

azamat

Answer:

Explanation:

$v_f^2 = v_i^2-2a(x_f-x_i)$

Subtract both sides by $v_i^2$ :

$- v_i^2+v_f^2 = -2a(x_f-x_i)$

Divide both sides by -2*a:

$\frac{v_i^2 - v_f^2}{2a} =x_f-x_i$

Add both sides by $x_i$ :

$x_i+\frac{v_i^2 - v_f^2}{2a} =x_f$

Subtract both sides by $\frac{v_i^2 - v_f^2}{2a}$ :

$x_i=x_f-\frac{v_i^2 - v_f^2}{2a}$

8 0

3 years ago

How much time will it take for a truck to travel 20 meters if it is traveling at 4 m/s?

ss7ja [257]

Answer:

<h2>The answer is 5 s</h2>

Explanation:

The time taken can be found by using the formula

$t = \frac{d}{v} \\$

d is the distance

v is the velocity

From the question we have

$t = \frac{20}{4} \\$

We have the final answer as

<h3>5 s</h3>

Hope this helps you

4 0

2 years ago

Read 2 more answers

A dragster starts from rest and travels 1/4 mi in 6.80 s with constant acceleration. What is its velocity when it crosses the fi

Ahat [919]

<h2>Its velocity when it crosses the finish line is 117.65 m/s</h2>

Explanation:

We have equation of motion s = ut + 0.5 at²

Initial velocity, u = 0 m/s

Acceleration, a = ?

Time, t = 6.8 s

Displacement, s = 1/4 mi = 400 meters

Substituting

s = ut + 0.5 at²

400 = 0 x 6.8 + 0.5 x a x 6.8²

a = 17.30 m/s²

Now we have equation of motion v = u + at

Initial velocity, u = 0 m/s

Final velocity, v = ?

Time, t = 6.8 s

Acceleration, a = 17.30 m/s²

Substituting

v = u + at

v = 0 + 17.30 x 6.8

v = 117.65 m/s

Its velocity when it crosses the finish line is 117.65 m/s

6 0

3 years ago