i think it’s B. sorry if i’m wrong
Answer:
<h2>
14.66secs</h2>
Explanation:
Given the formula for calculating the depth in metres expressed as
depth in meters = ½ (1500 m/sec × Echo travel time in seconds)
Given depth of the challenger = 10, 994 meters, we will substitute this given value into the formula given to calculate the time take for the echo to travel.
10, 994 = depth in meters = ½ * 1500 m/sec × Echo travel time in seconds
10,994 = 750 * Echo travel time in seconds
Dividing both sides by 750;
Echo travel time in seconds = 10,994 /750
Echo travel time in seconds ≈ 14.66secs (to two decimal places)
Therefore, it would take an echo sounder’s ping 14.66secs to make the trip from a ship to the Challenger Deep and back
PE = mgh
where
m = mass
g = acceleration due to gravity
h = height
The comparison of the forces in a small nucleus to the forces of a large one is the fact that they are capable of holding the protons and neutrons which made it no matter what their size may be. Therefore, as long as there is a nucleus, their forces can both hold together the two atoms tight.
Yes, it do, for a short time.
