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3 years ago

How do astronomers best learn about distant objects in the universe

Physics

2 answers:

3241004551 [841]3 years ago

7 0

The answer will be C based on what I had researched

Brilliant_brown [7]3 years ago

5 0

They collect and analyze data from the objects' electromagnetic radiation.

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According to research, college graduates are more likely to engage in all of the following EXCEPT:

Ksivusya [100]

Strange question. I’d say B)

7 0

3 years ago

Read 2 more answers

A room has a wall with an R value of 18 F sq.ft. hr/BTU. The room is 15 feet long and 11 feet wide with walls that are 9 ft high

seropon [69]

Answer:

The heat transferred through the wall that day is 13728 BTUs

Explanation:

Here, we have the area of the wall given as

Area of wall = 2 × Length × Height + 2 × Width × Height

Length = 15 feet

Width = 11 Feet and

Height = 9 feet

Therefore, the area = 2×15×9 + 2×11×9 = 468 ft²

Temperature difference is given by

Average outside temperature - Wall temperature = 40 - 18 = 22 °F

Therefore the heat transferred through the wall that day (24 hours) at 18 sq.ft. hr/BTU is given by;

468 × 22 × 24/18 = 13728 = 13728 BTUs.

5 0

3 years ago

Which statement is true about acceleration?

Ilia_Sergeevich [38]

The answer is C is think

8 0

3 years ago

A shot-putter released the shot at an angle of 41.5 degrees and a height of 1.9 m with an initial velocity of 13.3 m/s. How far

liq [111]

Answer:

x = 17.88[m]

Explanation:

We can find the components of the initial velocity:

$(v_{x})_{o} = 13.3*cos(41.5)=9.96[m/s]\\(v_{y})_{o} = 13.3*sin(41.5)=8.81[m/s]$

We have to remember that the acceleration of gravity will be worked with negative sign, since it acts in the opposite direction to the movement in direction and the projectile upwards.

g = - 9.81[m/s^2]

Now we must find the time it takes for the projectile to hit the ground, as the problem mentions that it does not impact on the board.

$y=y_{o} +(v_{y} )_{o} *t-0.5*g*(t)^{2} \\0=1.9+(8.81*t)-(4.905*t^{2})\\-1.9=8.81*t*(1-0.5567*t)\\t=0\\t=1.796[s]$

With this time we can calculate the horizontal distance:

$x=(v_{x})_{o} *t\\x=9.96*1.796\\x=17.88[m]$

3 0

3 years ago

What kind of object are the light rays interacting with in the model below?

Ymorist [56]

Answer:

Concave lens

.............

5 0

2 years ago