1. your ear drums feel like they pop when you go deep enough under water. kinda feels like bricks. and it hurts if you stay under to long.
2. this could be a reaction from fluid pressure yes. can could be avoided by getting wax ear plugs however even with ear plugs going deep enough under water can still make your ears "pop" due to pressure but at high surfaces it would avoid that.
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Answer:
Nuclear fission of heavy elements produces exploitable energy because the specific binding energy (binding energy per mass) of intermediate-mass nuclei with atomic numbers and atomic masses.
Explanation:
have a great day!! :)
Answer:
h = 3.10 m
Explanation:
As we know that after each bounce it will lose its 11% of energy
So remaining energy after each bounce is 89%
so let say its initial energy is E
so after first bounce the energy is
after 2nd bounce the energy is
After third bounce the energy is
here initial energy is given as
now let say final height is "h" so after third bounce the energy is given as
now from above equation we have
'H' = height at any time
'T' = time after both actions
'G' = acceleration of gravity
'S' = speed at the beginning of time
Let's call 'up' the positive direction.
Let's assume that the tossed stone is tossed from the ground, not from the tower.
For the stone dropped from the 50m tower:
H = +50 - (1/2) G T²
For the stone tossed upward from the ground:
H = +20T - (1/2) G T²
When the stones' paths cross, their <em>H</em>eights are equal.
50 - (1/2) G T² = 20T - (1/2) G T²
Wow ! Look at that ! Add (1/2) G T² to each side of that equation,
and all we have left is:
50 = 20T Isn't that incredible ? ! ?
Divide each side by 20 :
<u>2.5 = T</u>
The stones meet in the air 2.5 seconds after the drop/toss.
I want to see something:
What is their height, and what is the tossed stone doing, when they meet ?
Their height is +50 - (1/2) G T² = 19.375 meters
The speed of the tossed stone is +20 - (1/2) G T = +7.75 m/s ... still moving up.
I wanted to see whether the tossed stone had reached the peak of the toss,
and was falling when the dropped stone overtook it. The answer is no ... the
dropped stone was still moving up at 7.75 m/s when it met the dropped one.
In uniform motion, the path is a straight line, and the object r moving along it at a constant speed.