A Scuba Diver And Her Gear Displace A Volume Of

A Scuba Diver And Her Gear Displace A Volume Of

A team of divers and their equipment weighs 65.0 L and a total of 68.0 kg.

Assume that the density of seawater is 1.025 x 103 kg / m3.

(a) What is the happy force in seawater?

(B) Will you drown or swim?

(of the)

According to the principle of Archimedes

Increase = weight of homeless fluid

................ = volume x density x g

From,

1 L = 10 3m 3

F = 65 x 10 3 x 1025 x 9.8 = 653 N

(B)

The average density with your team is r.

68/65 x 10 3 = 1,046 x 10 3 kg / m 3

This is more than the density of seawater. So drown

F_B = [m_ گھر homeless water} m_ {moving object}] g

m_water = 65 L × 1,025 kg / L × 9.81 m / s 2 = 653.59 N

m_r = 68 kg — 9.81 m / s 2 = 667.08 N

Answering part a) is not just a math problem.

(b) If m_r> m_water, does r sink or float (note the mark in the formula)?

(a) Buoyancy = weight of homeless fluid

=> Float = V x d (sea water)

=> Float = 65 x 10 3 x 1030 = 66.95 N

(b) Weight r = mg = 68 x 9.8 = 666.40 N.

=> Because the weight is r> buoyancy.

=> r is falling.

A Scuba Diver And Her Gear Displace A Volume Of