Answer:
A) 27209506.5 N
B) 2393640 N
The force on the underwater vehicle is about 11.37 times the weight of the jetliner for comparison.
Explanation:
Force du to depth of water is
F = pghA
P = density of salt water = 1025 kg/m3
g = acceleration due to gravity 9.81 m/s2
h = depth of water 11000 m
A = area pressure acts
Area = ¶r^2 = 3.142 x 0.280^2 = 0.246 m^2
Therefore
F = 1025 x 9.81 x 11000 x 0.246
= 27209506.5 N
Weight of a jetliner with mass 2.44 × 10^5 kg is,
2.44×10^5 x 9.81 = 2393640 N
The force on the underwater vehicle is about 11.37 times the weight of the jetliner for comparison.
Answer:
His average speed is 980 units
Answer:
hmax=81ft
Explanation:
Maximum height of the object is the highest vertical position along its trajectory.
The vertical velocity is equal to 0 (Vy = 0)
we isolate th (needed to reach the maximum height hmax)
The formula describing vertical distance is:
So, given y = hmax and t = th, we can join those two equations together:
if we launch a projectile from some initial height h all you need to do is add this initial elevation
<span>one year is 365, 1 day is 24 hours, 1 hour is 60 minutes, 60 minutes is 60 seconds, thus (365 * 24 * 60 * 60) = 31,536,000
one year is equal to 31,536,000 seconds. the plate has a speed of 4.8 cm every 31,536,000 seconds. lets find out how far it goes in 40 seconds. (4.8/31,536,000)*40 = 0.00000608828
The plate moves 0.00000608828 cm every 40 seconds</span>
T<span>he equation to be used here to determine the distance between two equipotential points is:
V = k * Q / r
where v is the voltage of the point, k is a constant, Q is charge of the point measured in coloumbs and r is the distance.
In this case, we can use ratio of proportions to determine the distance between the two points. in this respect,
Point 1:
V = k * Q / r = 290
r = k*Q/290 ; kQ = 290r
Point 2:
V = k * Q / R = 41
R = k*Q/41
from equation 10 kQ = 290r
R = 290/(41)= 7.07 m
The distance between the two points then is equal to 7.07 m.
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