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sdas [7]
3 years ago
7

Which scientist first suggested that the sun was at the center of the solar system?

Physics
2 answers:
pogonyaev3 years ago
8 0

Answer:

Nicolaus Copernicus

Explanation:

With the development of the heliocentric model by Nicolaus Copernicus in the 16th century, the Sun was believed to be the center of the Universe, with the planets (including Earth) and stars orbiting it.

exis [7]3 years ago
6 0

#4 copernicus he was the first scientist

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Select the option that best summarizes the reason for the use of position-time graphs when studying physical science. (1 point)
Sedaia [141]

The position-time graphs show the relationship between the position of an object (shown on the y-axis) and the time (shown on the x-axis) to show velocity.

<h3>What is velocity?</h3>

Velocity is a vector quantity that tells the distance an object has traveled over a period of time.

Displacement is a vector quality showing total length of an area traveled by a particular object.

Imagine a time-position graph where the velocity of an object is constant. What will be observed on the graph concerning the slope of the line segment as well as the velocity of the object?

The slope of the line is equal to zero and the object will be stationary.

The position-time graphs show the relationship between the position of an object (shown on the y-axis) and the time (shown on the x-axis) to show velocity.

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4 0
1 year ago
Read 2 more answers
What type of electromagnetic wave does a light bulb and a radio antenna release?
andrew-mc [135]
I believe that a light bulb releases visible light and a radio antenna releases a radio waves
4 0
3 years ago
When aluminum foil is formed into a loose ball, it can float on water. But when the ball of foil is pounded flat with a hammer,
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A series of optical telescopes produced an image that has a resolution of about 0.00350 arc second.
Mila [183]

Answer:

The smallest diameter is D =122 \ m

Explanation:

From the question we are told that

       The resolution of the telescope is \theta  =  0.00350 \ arc \ second

           The wavelength is  \lambda = 1.70 \mu m = 1.70 *10^{-6} \ m

From the question we are told that

        1 arc \ sec = \frac{1}{3600^o}

So      0.00350 \ arc \ second = x

Therefore

             x =  0.00350  *  \frac{1}{3600 }

              x = ( 9.722*10^{-7} )^o

Now  1^o  =  \frac{\pi}{180}

   So  (9.722*10^{-7})^o =  \theta

  =>    \theta  =  (9.722*10^{-7}) * \frac{\pi}{180}

           \theta  =  1.69*10^{-8} rad

The smallest diameter is mathematically represented  as

          D = \frac{1.22 \lambda }{\theta  }

substituting values

           D = \frac{1.22 * 1.7 *10^{-6}} {1.69 *10^{-8}  }

           D =122 \ m

   

6 0
4 years ago
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