All categories

4 years ago

Which collects light in a reflecting telescope?

Physics

2 answers:

Temka [501]4 years ago

8 0

Answer: Option a, concave lens.

Explanation: The object that collects the light in a reflecting telescope is "kinda" concave lens that collects the light and directs it to a mirror, that is where the observer can see. So the correct answer is option a.

Alecsey [184]4 years ago

4 0

Concave lens. These are used in making the objectives of reflection telescopes

You might be interested in

Convert <img src="https://tex.z-dn.net/?f=%5Cfrac%7B%280.779mg%29%28min%29%7D%7BL%7D" id="TexFormula1" title="\frac{(0.779mg)(mi

Orlov [11]

The number converted is $0.0467 \frac{(kg)(s)}{m^3}$

Explanation:

In order to convert from the original units to the final units, we have to keep in mind the following conversion factors:

$1 kg = 1000 g = 10^6 mg$

$1 min = 60 s$

$1 m^3 = 1000 L$

The original unit that we have is

$\frac{mg\cdot min}{L}$

Therefore, it can be rewritten as:

$=\frac{mg \frac{1}{10^6 mg/kg}\cdot min\cdot 60 s/min}{L\frac{1}{1000L/m^3}}=0.06 \frac{(kg)(s)}{m^3}$

Therefore, since the initial number was 0.779, the final value is

$0.779\cdot 0.06 \frac{(kg)(s)}{m^3}=0.0467 \frac{(kg)(s)}{m^3}$

#LearnwithBrainly

5 0

4 years ago

An air-conditioner with refrigerant-134a as the working fluid is used to keep a room at 238C by rejecting the waste heat to the

tatiyna

Answer:

Explanation:

<h3>Please refer to the attachment below for explanation.</h3>

7 0

3 years ago

A rising pendulum bob gains _____ energy.

puteri [66]

Answer:

B. Kinetic energy.............

4 0

3 years ago

Read 2 more answers

The water from a fire hose follows a path described by y equals 3.0 plus 0.8 x minus 0.40 x squared (units are in meters). If

ivanzaharov [21]

Explanation:

It is given that, the water from a fire hose follows a path described by equation :

$y=3+0.8x-0.4x^2$ ........(1)

The x component of constant velocity, $v_x=5\ m/s$

We need to find the resultant velocity at the point (2,3).

Let $\dfrac{dx}{dt}=v_x$ and $\dfrac{dy}{dt}=v_y$

Differentiating equation (1) wrt t as,

$\dfrac{dy}{dt}=0.8\times \dfrac{dx}{dt}-0.8x\times \dfrac{dx}{dt}$

$v_y=0.8\times v_x-0.8x\times v_x$

$v_y=0.8v_x(1-x)$

When x = 2 and $v_x=5\ m/s$

So,

$v_y=0.8\times 5\times (1-2)$

$v_y=-4\ m/s$

Resultant velocity, $v=\sqrt{v_x^2+v_y^2}$

$v=\sqrt{5^2+(-4)^2}$

v = 6.4 m/s

So, the resultant velocity at point (2,3) is 6.4 m/s. Hence, this is the required solution.

8 0

3 years ago

Guys I need the answer ASAP look at the picture and please give me the correct answer ASAP! PLZ

VLD [36.1K]

Answer:

Explanation:

7 0

3 years ago