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rosijanka [135]
4 years ago
7

What do scientist use to look at astronomical bodies?

Physics
2 answers:
ivann1987 [24]4 years ago
6 0
The answer should be C. :)
Ilya [14]4 years ago
3 0

Answer:

The Correct Answer is C

Gamma Telescope

Explanation:

  • Gamma-ray telescopes are installed at hilly altitudes like the X-ray telescope to get the best result.
  • This is also because the Gamma-ray Telescope rays sometimes get weaker when they enter to earth's atmosphere.
  • Gamma-ray telescope detects gamma rays bursts.
  • This telescope also help the astronomers to conduct mission in outer space like supernova etc.

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Samiyah pushes a cart with 20 newtons of force. The cart weighs 10 kilograms. How quickly does the cart accelerate?
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Answer:

10 newtons

Explanation: 10 of the 20 cancels out and the other 10 is force.

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3 years ago
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What is a model drawing of 2HCI?
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This is the answer just double it

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PLEASE HELP ME QUICK
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Explanation:

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3 years ago
A box of mass 3.6 kg is lifted 5.4 m above the
horsena [70]

So, the energy change that occurs is 190.512 J.

<h3>Introduction</h3>

Hello ! I am Deva from Brainly Indonesia will help you regarding energy and its transformation. In this case, it's the use of energy from the lifter to be equivalent to the change in the object's potential energy. Why potential energy? Because the box undergoes a change in height and the potential energy specializes at a certain height. Work (W) due to change in potential energy (\sf{\Delta PE}) can be realized in the equation :

\sf{W = \Delta PE}

\sf{W = m \cdot g \cdot h_2 - m \cdot g \cdot h_2}

\boxed{\sf{\bold{W = m \cdot g \cdot (h_2 - h_1)}}}

With the following condition :

  • W = work of subject (J)
  • \sf{\Delta PE} = change of potential energy (J)
  • m = mass (kg)
  • g = acceleration of the gravity (m/s²)
  • \sf{h_2} = final height (m)
  • \sf{h_1} = initial height (m)

<h3>Problem Solving</h3>

We know that :

  • m = mass = 3.6 kg
  • g = acceleration of the gravity = 9.8 m/s²
  • \sf{h_2} = final height = 5.4 m
  • \sf{h_1} = initial height = 0 m

What was asked :

  • W = work of subject = ... J

Step by step :

\sf{W = \Delta PE}

\sf{W = m \cdot g \cdot (h_2 - h_1)}

\sf{W = 3.6 \cdot 9.8 \cdot (5.4 - 0)}

\sf{W = 3.6 \cdot 9.8 \cdot 5,4}

\boxed{\sf{W = 190.512 \: J}}

So, the energy change that occurs is 190.512 J.

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3 years ago
Some steps in mitosis are shown below in the incorrect order:
AysviL [449]

Answer:

Step C, Step D, Step A, Step B i sure

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3 years ago
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