Question

A standard basketball (mass = 624 grams; 24.3 cm in diameter) is held fully under water. a. Calculate the buoyant force and weight. b. When released, does the ball sink into the bottom or float to the surface? c. If it floats, what percentage of it is sticking out of the water? If it sinks, what is normal force, F_N with which it sits on the bottom of the pool?

Answer 1

Answer:

Explanation:

Mass = 624 gm = .624 kg

weight = .624 x 9.8

= 6.11 N

Radius of ball = 12.15 x 10⁻² cm

volume of ball

= 4/3 x 3.14 x ( 12.15 x 10⁻²)³

= 7509.26 x 10⁻⁶ m³

Buoyant force = weight of displaced water

= 7509.26 x 10⁻⁶ x 10³ x 9.8

= 73.59 N

b ) Since buoyant force exceeds the weight of the ball , it will float .

c )

Let volume v sticks out while floating .

Volume under water

= 7509.26 x 10⁻⁶ - v

its weight

= (7509.26 x 10⁻⁶ - v ) x 10³ x 9.8

For floating

(7509.26 x 10⁻⁶ - v ) x 10³ x 9.8  =  .624 x 9.8 ( weight of ball )

(7509.26 x 10⁻⁶ - v ) x 10³ = .624

7.509 - v x 10³ = .624

v x 10³ = 7.509 - .624

v x 10³ = 6.885

v = 6.885 x 10⁻³ m³

fraction

= v / total volume

=  6.885 x 10⁻³ / 7.51 x 10⁻³

91.67 %