Pnet = Po + dgh
<span>Density of saltwater = 1030 kg/m^3. </span>
<span>Disregard the thickness. Assuming it's a circular window, then the area is pi(r^2). </span>
<span>d = 20 cm = 0.2 m </span>
<span>r = d/2 = 0.1 m </span>
<span>A = pi(r^2) </span>
<span>A = 3.14159265(.1^2) </span>
<span>A = 0.0314159265 m^2 </span>
<span>p = F/A </span>
<span>p = (1.1 x 10^6) / (0.0314159265) </span>
<span>p = 35,014,087.5 Pa </span>
<span>1 atm = 101,325 Pa </span>
<span>P = Po + dgh </span>
<span>h = (P - Po) / dg </span>
<span>h = (35,014,087.5 - 101,325) / (1030 x 9.81) </span>
<span>h = 3 455.23812 m </span>
<span>h = 3.5 km</span>
Answer:
Explanation:
a) the speed increment of the hammer as it drops past the first window, is greater than that of the speed of the hammer as it drops past the second window. This can also be translated as saying that the hammer spent more time at the second window.
b) III
The best answer would be answer III, The hammer spends more time dropping past window 1, which I had already included in my explanation in (a) above.
Thecorrect answer is B.
hope this helps brainliest would be cool if not its fine
Answer:
The percentage of its mechanical energy does the ball lose with each bounce is 23 %
Explanation:
Given data,
The tennis ball is released from the height, h = 4 m
After the third bounce it reaches height, h' = 183 cm
= 1.83 m
The total mechanical energy of the ball is equal to its maximum P.E
E = mgh
= 4 mg
At height h', the P.E becomes
E' = mgh'
= 1.83 mg
The percentage of change in energy the ball retains to its original energy,
ΔE % = 45 %
The ball retains only the 45% of its original energy after 3 bounces.
Therefore, the energy retains in each bounce is
∛ (0.45) = 0.77
The ball retains only the 77% of its original energy.
The energy lost to the floor is,
E = 100 - 77
= 23 %
Hence, the percentage of its mechanical energy does the ball lose with each bounce is 23 %
