<span>Depth = 5.0 Ă— 10^2 m
Density of sea water = 1.025 x 10^3
Pd = Po + pgh
Atmospheric pressure is standard Patm = 1.01325 x 10^5 Pa
Since the normal pressure is retained in the hull, no need to bother about Po
Pd = pgh = 1.025 x 10^3 x 9.8 x 5.0 x 10^2 = 50.225 x 10^5
So now Pd / Patm = 50.225 x 10^5 / 1.01325 x 10^5 = 49.56
So it is 49.56 times larger.</span>
Answer:
x = 0.176 m
Explanation:
For this exercise we will take the condition of rotational equilibrium, where the reference system is located on the far left and the wire on the far right. We assume that counterclockwise turns are positive.
Let's use trigonometry to decompose the tension
sin 60 =
/ T
T_{y} = T sin 60
cos 60 = Tₓ / T
Tₓ = T cos 60
we apply the equation
∑ τ = 0
-W L / 2 - w x + T_{y} L = 0
the length of the bar is L = 6m
-Mg 6/2 - m g x + T sin 60 6 = 0
x = (6 T sin 60 - 3 M g) / mg
let's calculate
let's use the maximum tension that resists the cable T = 900 N
x = (6 900 sin 60 - 3 200 9.8) / (700 9.8)
x = (4676 - 5880) / 6860
x = - 0.176 m
Therefore the block can be up to 0.176m to keep the system in balance.
Work = (force) x (distance)
80 J = (force) x (4 m)
Force = (80 J) / (4 m) = 20 N
That's IF the force was in the same direction as the 4m of motion.
If the force was kind of slanted, then it had to be stronger, and
it had a component of 20N in the direction of the motion.
D.
The reading between 7N and 8N would have to be 7.5N. Answers A and B are much to small and answer C is way to big.