*Please show solution in detail so i can understand how to solveit. Thank you fo
ID: 1673968 • Letter: #
Question
*Please show solution in detail so i can understand how to solveit. Thank you for your time.Question: A 7.43 kg sphere of radius 5.02 cm is at a depth of 1.78 km inseawater that has an average density of 1025 kg/m3. Whatare the (a) gaugepressure, (b) total pressure,and (c)corresponding total force compressingthe sphere's surface? What are (d) themagnitude of the buoyant force on the sphereand (e) the magnitude of the sphere'sacceleration if it is free to move? Take atmospheric pressure to be1.01 x 105 Pa.
Question: A 7.43 kg sphere of radius 5.02 cm is at a depth of 1.78 km inseawater that has an average density of 1025 kg/m3. Whatare the (a) gaugepressure, (b) total pressure,and (c)corresponding total force compressingthe sphere's surface? What are (d) themagnitude of the buoyant force on the sphereand (e) the magnitude of the sphere'sacceleration if it is free to move? Take atmospheric pressure to be1.01 x 105 Pa.
Explanation / Answer
Depth h = 1.78 km = 1780 m
Radius r= 5.02 cm = 5.02 * 10 ^ -2 m
Volume of the sphere V = ( 4/ 3) (pi) r ^ 3
= 5.299 * 10 ^ -4 m^ 3
Mass m = 7.43 kg
Density D = 1025 kg / m^ 3
(a). Gauge pressure P = Dgh
= 17.88 * 10 ^ 6 Pa
(b). Total pressure P ‘ = Po + P
Where Po = atmospheric pressure = 101.3 * 10 ^ 3 Pa
So, P ‘ = 17.98 * 10^ 6 Pa
(c). Total force F = P ‘ * A
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