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3 years ago

The main light gathering device of a refractor telescope is

Physics

1 answer:

Nataly [62]3 years ago

3 0

It's the "objective" lens ... the big one in the front.

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A charge per unit length given by l(x) = bx, where b = 12 nc/m2, is distributed along the x axis from x = +9.0 cm to x = +16 cm.

Pavel [41]

Hopefully this will help you.

4 0

3 years ago

What measures the distance between two consecutive crests of a wave?

Marina86 [1]

Answer A is incorrect
A crest is just one point. It is not the distance between 2 crests.

B is incorrect
A trough is just 1 point. It is not the distance between 2 troughs.

C is incorrect.
the amplitude measures the height of a crest from the middle of the wave to the crest (or trough).

D is the correct answer. That is the distance between 2 crests or 2 troughs

8 0

3 years ago

One end of a metal rod is in contact with a thermal reservoir at 745. K, and the other end is in contact with a thermal reservoi

Masteriza [31]

Answer:

a) $S_1=-9.65}\ J/K$

b) $S_2=71.18\ J/K$

c) 0 J/K

d)S= 61.53 J/K

Explanation:

Given that

T₁ = 745 K

T₂ = 101 K

Q= 7190 J

The entropy change of reservoir 745 K

$S_1=-\dfrac{7190}{745}\ J/K$

Negative sign because heat is leaving.

$S_1=-9.65}\ J/K$

The entropy change of reservoir 101 K

$S_2=\dfrac{7190}{101}\ J/K$

$S_2=71.18\ J/K$

The entropy change of the rod will be zero.

The entropy change of the system

S= S₁ + S₂

S = 71.18 - 9.65 J/K

S= 61.53 J/K

3 0

3 years ago

A block of mass m=9.0 kg and speed V and is behind a block of mass M= 27 kg and speed of .50 m/s. The surface is frictionless, a

sammy [17]

Answer:

2.06 m/s

Explanation:

From the law of conservation of linear momentum, the sum of momentum before and after collision are equal. Considering this case where we have frictionless surface, no momentum is lost in the process.

Momentum before collision

Momentum is given by p=mv where m and v represent mass. The initial sum of momentum will be 9v+(27*0.5)=9v+13.5

Momentum after collision

The momentum after collision will be given by (9+27)*0.9=32.4

Relating the two then 9v+13.5=32.4

9v=18.5

V=2.055555555555555555555555555555555555555 m/s

Rounded off, v is approximately 2.06 m/s

5 0

3 years ago

A spring is compressed 0.035 m inside a dart gun. (K=500 N/m). The spring has elastic energy. Calculate it.

NARA [144]

The elastic potential energy of the spring is 0.31 J

Explanation:

The elastic potential energy of a spring is given by

$E=\frac{1}{2}kx^2$

where

k is the spring constant

x is the compression/stretching of the spring

For the spring in this problem, we have:

k = 500 N/m (spring constant)

x = 0.035 m (compression)

Substituting, we find the elastic potential energy:

$E=\frac{1}{2}(500)(0.035)^2=0.31 J$

Learn more about potential energy:

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

6 0

3 years ago