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const2013 [10]
3 years ago
5

I need help solving the equation 2Na(s)+Cl2(g)>2NaCl(s)

Physics
1 answer:
IrinaK [193]3 years ago
3 0

it's a chemical eaction equation ... you don't solve them ... you bslance them .... it is balanced

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If the baseball and the plastic ball were moving at the same speed which ball would hit a bat harder
Lerok [7]
A baseball would hit the bat harder. This is because the baseball is a lot heavier and more dense than the plastic ball. The keyword that you're looking for is density. The baseball is dense.
7 0
3 years ago
4- by eating food only will you get energy?​
Serga [27]

Answer:

yes

Explanation:

All parts of the body (muscles, brain, heart, and liver) need energy to work. This energy comes from the food we eat. Our bodies digest the food we eat by mixing it with fluids (acids and enzymes) in the stomach.

4 0
3 years ago
I NEED HELP PLEASE, THANKS! :)
Yuliya22 [10]

Answer:

Radio waves have a wavelength between 10^{-9}m and 10^{-12}m

While,

X rays have a wavelength between 1m and 10km.

=> It is one of the condition of diffraction that the obstacle (coming in the way) must be comparable with the size of the wavelength.

=> This shows, that radio waves have a wavelength which is comparable with the size of buildings and can really easily diffract through it

=> While, X-rays are big enough to diffract through the wall.

So, if an X-ray technician stands behind a wall during the use of her machine, she will remain safe.

6 0
3 years ago
A man with a weight of 100 N is located
valentina_108 [34]

Answer:

Given,

  mass of man = 100 N = 10 kg

   height = h = 25m

   since the man does not move anything with his force, work done by him is zero

   work done on the man = gain in potential energy

   P.E=mgh

   P.E=10×9.8×25

  P.E=2.45KJ

Explanation:

so, potential energy gained by man is 2.45 KJ

5 0
3 years ago
A clam dropped by a seagull takes 3.0 seconds to hit the ground. What is the seagull's approximate height above the ground at th
ankoles [38]
<h2>The seagull's approximate height above the ground at the time the clam was dropped is 4 m</h2>

Explanation:

We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = 0 m/s

        Acceleration, a = 9.81 m/s²  

        Time, t = 3 s      

     Substituting

                      s = ut + 0.5 at²

                      s = 0 x 3 + 0.5 x 9.81 x 3²

                      s = 44.145 m

The seagull's approximate height above the ground at the time the clam was dropped is 4 m

4 0
4 years ago
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