All categories

3 years ago

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

Physics

1 answer:

Bezzdna [24]3 years ago

4 0

In a refracting telescope convex lens forms the image.

Answer: c. Convex mirror

Explanation:

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

A 120 resistor a 60 ohm resistor and a 40 ohm resistor are connected in parallel to a 120 volt power source. what is the current

larisa [96]

Answer:

6 A

Explanation:

First of all, we need to calculate the equivalent resistance of the circuit. The three resistors are connected in parallel, so their equivalent resistance is given by:

$\frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}=\frac{1}{120 \Omega}+\frac{1}{60 \Omega}+\frac{1}{40 \Omega}=\frac{3+2+1}{120 \Omega}=\frac{6}{120 \Omega}\\R_T = \frac{120}{6} \Omega$

And now we can use Ohm's law to find the current in the circuit:

$V=R_T II=\frac{V}{R_T}=\frac{120 V}{\frac{120}{6}\Omega}=6 A$

6 0

3 years ago

A cat dozes on a stationary merry-go-round, at a radius of 5.5 m from the center of the ride. Then the operator turns on the rid

bixtya [17]

Answer:

$u_{s}=0.56$

Explanation:

For the cat to stay in place on the merry go round without sliding the magnitude of maximum static friction must be equal to magnitude of centripetal force

$F_{s.max}=\frac{mv^2}{r} \\$

Where the r is the radius of merry-go-round and v is the tangential speed

but

$F_{s.max}=u_{s}F_{N}=u_{s}mg$

So we have

$u_{s}mg=\frac{mv^2}{r}\\ u_{s}=\frac{v^2}{gr}\\ Where\\v=\frac{2\pi R}{T} \\So\\u_{s}=\frac{(\frac{2\pi R}{T} )^2}{gr} \\u_{s}=\frac{4\pi^2 r}{gT^2}$

Substitute the given values

$u_{s}=\frac{4\pi^2 5.5m}{(9.8m/s^2)(6.3s)^2} \\u_{s}=0.56$

6 0

3 years ago

Which type of macromolecules a cell brake down food?

kobusy [5.1K]

Answer:

Hope this helps you find the answer

Explanation:

The proteins, lipids, and polysaccharides that make up most of the food we eat must be broken down into smaller molecules before our cells can use them—either as a source of energy or as building blocks for other molecules.

8 0

3 years ago

g A 2.3 kg block is attached to the spring, and it is released from rest 0.7 m from the spring's equilibrium position. Neglectin

DedPeter [7]

Answer:

3.71 m/s

Explanation:

From the law of conservation of linear momentum, since we are neglecting minor energy losses due to friction then we can express it as $mgh=0.5mv^{2}$ since all the potential energy is transformed to kinetic energy

Making v the subject of the formula then $v=\sqrt {2gh}$ and here m is the mass of the block, g is acceleration due to gravity, h is the height. Substituting 0.7 m for h and 9.81 for g then we obtain that $v=\sqrt {2\times 9.81\times 0.7}=3.705941176 m/s\approx 3.71 m/s$