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Rzqust [24]
3 years ago
6

When was the Hubble telescope launched​

Physics
2 answers:
SpyIntel [72]3 years ago
5 0

Answer:

April 24, 1990 is the answer

MrRissso [65]3 years ago
4 0
April 24 1990 is the correct answer
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I WILL NAME YOU BRAINLIEST IF YOU ANSWER THIS!! This question will make you think twice, let see if you can answer this! Worth 1
11Alexandr11 [23.1K]

Answer:

He took a picture (or "shot") of the buffaloes. How cute :)

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Joanna claims that a large block of ice will cool a substance more than a small block of ice will at the same temperature. To su
Levart [38]

She puts each block of ice in the same 3000 mL beaker, each with 2000 mL of water at room temperature, and measures the temperature before and after adding ice. Therefore, small blocks of ice will have the same temperature.

Joanna puts two blocks of ice (one larger than the other) into separate cups and fills each with water. She compares the final water temperature of the two cups after each block of ice melts.

Put each block of ice in the same 3000 mL beaker, each at room temperature, put 2000 mL of water in it, and measure the temperature before and after adding ice. This way you keep the water at the same temperature in the beginning, then the temperature changes after you add the ice, giving you a better idea of ​​the final temperature reading.

Learn more about Temperature here brainly.com/question/24746268

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3 0
1 year ago
___meaning two' sugar (or saccharide) units in a chain.
ludmilkaskok [199]
The answer is c

Explanation:
3 0
3 years ago
How would this affect the electrostatic force between them?
Alja [10]

The answer is increase

Explanation:

6 0
3 years ago
Two parallel plates of area 2.34*10-3 M2 have 7.07*10-7C of charge placed on them. A6.62*10-5C charge q1 is placed between the p
ira [324]

Answer:

The force exerted on the q_1 is  F =  2.25*10^{3} \ N

Explanation:

From the question we are told that

     The area is  A =  2.34*10^{-3} \ m^2

     The magnitude of charge placed on them is  q =  7.07 * 10^{-7} C

     The charge placed between the plate is q_1 = 6.62 *10^{-5} C

   

The electric field generated around the plate  is mathematically represented as

           E =  \frac{q}{A \epsilon_o}

Substituting values

          E =  \frac{7.07*10^{-7}}{2.34*10^{-3} * 8.85 *10^{-12}}

         E = 34*10^{6} \ V/m

The force exerted the charge q_1 is  mathematically represented as

        F =  q_1 * E

Substituting values  

        F =  6.62 *10^{-5} * 34*10^{6}

        F =  2.25*10^{3} \ N

8 0
3 years ago
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