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LuckyWell [14K]
3 years ago
13

Why do astronomers use spectroscopes to analyze light from distant objects?

Physics
1 answer:
RoseWind [281]3 years ago
7 0
Spectroscopy — the use of light from a distant object to work out the object is made of — could be the single-most powerful tool astronomers use, says Professor Fred Watson from the Australian Astronomical Observatory. ... "It lets you see the chemicals being absorbed or emitted by the light source.
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What does the term "speed" describe?
Blababa [14]
The cyclist who travels 20 kilometers per hour for 15 kilometers
7 0
3 years ago
PLEASE ANSWER FAST!! A scientist put red blood cells in water that contained salt. Over time, the red blood cells burst. What is
Nitella [24]
It is most likely true that there was a lower concentration of salt in the water than in the cells because when blood cells are put in a hypotonic solution such as pure water, the little to no salt concentration in the water causes the cells to swell and burst. This would occur because the water would try to dilute the solution inside of the blood cell and which would, therefore, cause it to burst. Hope this helps! 
7 0
4 years ago
Read 2 more answers
Lab: Motion
I am Lyosha [343]

Answer: In this lab we wanted to know how motion can be described. So the hypothesis is if the starting height of a sloped racetrack is increased, then the speed at which a toy car travels along the track will increase because the toy car will have a greater acceleration. My prediction is that cars travel faster on higher tracts. So the heighten the track was intentionally manipulated. So it is the independent variable the speed of the car is the dependent variable. The speed at the first quarter checkpoint is 1.09 m/s. The speed at the second quarter checkpoint is 1.95 m/s. The speed at the third quarter checkpoint is 2.373.36 m/s. The speed at the finish line is  2.803.00 m/s. The average speed increases as the height increases.

The cars on the higher track travel farther than the cars on the lower track, in the same time.

This means that the cars on the higher track have a greater average speed than those on the lower track. This is demonstrated by the

slope of the higher track line being greater than the slope of the lower track line.

Explanation: put it in notes then send it to files to compress it to submit it.

5 0
3 years ago
Read 2 more answers
Luis and Aisha conducted an experiment. They exerted different forces on four objects. Their results are shown in the table.
lesya692 [45]

Answer:

Object 3 has greatest acceleration.

Explanation:

Objects               Mass                                Force

1                            10 kg                               4 N              

2                           100 grams                       20 N

3                            10 grams                         4 N

4                             1 kg                                 20 N

Acceleration of object 1,

a_1=\dfrac{F_1}{m_1}\\\\a_1=\dfrac{4}{10}\\\\a_1=0.4\ m/s^2

Acceleration of object 2,

a_2=\dfrac{F_2}{m_2}\\\\a_2=\dfrac{20}{0.1}\\\\a_2=200\ m/s^2

Acceleration of object 3,

a_3=\dfrac{F_3}{m_3}\\\\a_3=\dfrac{4}{0.01}\\\\a_3=400\ m/s^2

Acceleration of object 4,

a_4=\dfrac{F_4}{m_4}\\\\a_4=\dfrac{20}{1}\\\\a_3=20\ m/s^2

It is clear that the acceleration of object 3 is 400\ m/s^2 and it is greatest of all. So, the correct option is (3).

4 0
3 years ago
Read 2 more answers
Constant Acceleration Kinematics: Car A is traveling at 22.0 m/s and car B at 29.0 m/s. Car A is 300 m behind car B when the dri
Ainat [17]

Answer:

The taken is  t_A  = 19.0 \ s

Explanation:

Frm the question we are told that

  The speed of car A is  v_A  =  22 \ m/s

   The speed of car B is  v_B  = 29.0 \ m/s

     The distance of car B  from A is  d = 300 \ m

     The acceleration of car A is  a_A  = 2.40 \ m/s^2

For A to overtake B

    The distance traveled by car B  =  The distance traveled by car A - 300m

Now the this distance traveled by car B before it is overtaken by A is  

          d = v_B * t_A

Where t_B is the time taken by car B

Now this can also be represented as using equation of motion as

      d = v_A t_A  + \frac{1}{2}a_A t_A^2 - 300

Now substituting values

       d = 22t_A  + \frac{1}{2} (2.40)^2 t_A^2 - 300

Equating the both d

       v_B * t_A = 22t_A  + \frac{1}{2} (2.40)^2 t_A^2 - 300

substituting values

   29 * t_A = 22t_A  + \frac{1}{2} (2.40)^2 t_A^2 - 300

   7 t_A = \frac{1}{2} (2.40)^2 t_A^2 - 300

  7 t_A =1.2 t_A^2 - 300

   1.2 t_A^2 - 7 t_A - 300  = 0

Solving this using quadratic formula we have that

     t_A  = 19.0 \ s

7 0
3 years ago
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