Answer:
the buoyant force on the chamber is F = 7000460 N
Explanation:
the buoyant force on the chamber is equal to the weight of the displaced volume of sea water due to the presence of the chamber.
Since the chamber is completely covered by water, it displaces a volume equal to its spherical volume
mass of water displaced = density of seawater * volume displaced
m= d * V , V = 4/3π* Rext³
the buoyant force is the weight of this volume of seawater
F = m * g = d * 4/3π* Rext³ * g
replacing values
F = 1025 kg/m³ * 4/3π * (5.5m)³ * 9.8m/s² = 7000460 N
Note:
when occupied the tension force on the cable is
T = F buoyant - F weight of chamber = 7000460 N - 87600 kg*9.8 m/s² = 6141980 N
The angle of inclination is calculated using sin
function,
sin θ = 5 m / 20 m = 0.25
θ = 14.4775°
<span>The net force exerted is then calculated:
F net = m g sin θ = 20 * 9.8 * 0.25 </span>
F net = 49N
<span>Work is product of net force and distance:
W = F net * d = 49 * 20 </span>
<span>Work = 980 J </span>
Answer:
The surface of Mercury has landforms that indicate its crust may have contracted. They are long, sinuous cliffs called lobate scarps. These scarps appear to be the surface expression of thrust faults, where the crust is broken along an inclined plane and pushed upward.
Explanation:
I hope this helps a little bit.
Answer:
![1 \times 10 { - }^{9}](https://tex.z-dn.net/?f=1%20%5Ctimes%2010%20%7B%20-%20%7D%5E%7B9%7D%20)
cubic metre or 1e-9
Explanation:
•By division. Number of cubic millimetre divided(/) by 1000000000, equal(=): Number of cubic metre.
•By multiplication. 83 mm3(s) * 1.0E-9 = 8.3E-8 m3(s)