Answer:
, level is rising.
Explanation:
Since liquid water is a incompresible fluid, density can be eliminated of the equation of Mass Conservation, which is simplified as follows:


By replacing all known variables:

The positive sign of the rate of change of the tank level indicates a rising behaviour.
Answer:
The uncertainty in the location that must be tolerated is 
Explanation:
From the uncertainty Principle,
Δ
Δ

The momentum P
= (mass of electron)(speed of electron)
= 
= 
If the uncertainty is reduced to a 0.0010%, then momentum
= 
Thus the uncertainty in the position would be:
Δ
Δ
Assume no air resistance, and g = 9.8 m/s².
Let
x = angle that the initial velocity makes with the horizontal.
u = 30 cos(x), horizontal velocity
v = 30 sin(x), vertical launch velocity
The horizontal distance traveled is 55 m, therefore the time of flight is
t = 55/[30 cos(x)] = 1.8333 sec(x) s
With regard to the vertical velocity, and the time of flight,obtain
[30 sin(x)]*(1.8333 sec(x)) + (1/2)*(-9.8)*(1.8333 sec(x))² = 0
55 tan(x) - 16.469 sec²x = 0
55 tan(x) - 16.469[1 + tan²x] = 0
16.469 tan²x - 55 tan(x) + 16.469 = 0
tan²x - 3.3396 tan(x) + 1 = 0
Solve with the quadratic formula.
tan(x) = 0.5[3.3396 +/- √(7.153)] = 3.007 or 0.3326
Therefore
x = 71.6° or x = 18.4°
The time of flight is
t = 1.8333 sec(x) = 5.8096 s or 1.932 s
The initial vertical velocity is
v = 30 sin(x) = 28.467 m/s or 9.468 m/s
The horizontal velocity is
u = 30 cos(x) = 9.467 m/s or 28.469 m/s
If t = 5.8096 s,
u*t = 9.467*5.8096 = 55 m (Correct)
or
u*t = 28.469*15.8096 = 165.4 m (Incorrect)
Therefore, reject x = 18.4°. The correct solution is
t = 5.8096 s
x = 71.6°
u = 9.467 m/s
v = 28.467 m/s
The height from which the ball was thrown is
h = 28.467*5.8096 - 0.5*9.8*5.8096² = -110.4 m
The ball was thrown from a height of 110.4 m
Answer: h = 110.4 m
The speed of sound in fresh water is 1482m/s.
It says ocean floor, so we should a little bit more accurate, and use the fact that the speed of sound in salt water (that has no bubbles) is 1560m/s.
speed = distance / time
Therefore Distance = speed x time = 1560 x 3.3 = 5158m
The sonar wave is sent out by the boat, reflected off the seafloor, and then is received back at the boat on the surface. So the distance 5148m is the distance from the boat to the sea bottom and then back up to the boat again.
So the depth of the water is half this distance Depth of water = 5148/2=2574m