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A certain block is floating in water with 77% of itself submerged. It has a length of 4.3 cm, a width of 5.9 cm, and a height of

2.75 cm. What is its mass, in kilograms?
a. 1.86 kg
b. 2.77 kg
c. 0.0086 kg
d. 5.98 kg
e. 3.62 kg
f. 0.0537 kg

1 Answer

6 votes

Answer:

f. 0.0537 kg

Step-by-step explanation:

To solve this, we use the formula for force of buoyancy (Fb) = Vs × D × g

Where; Fb= Force of buoyancy

Vs= Volume of submerged

block

D= Density of water that block

submerges in

g= acceleration due to gravity

In this question, volume of block can be calculated using L×B×H

= 4.3cm × 5.9cm × 2.75cm = 69.7675cm^3

If 77% of the block is submerged in water, then 77/100 × 69.7675cm^3

= 53.72cm^3 is the Volume of submerged block

Density of water is known as 1000kg/m^3 while the gravitational force is 10m/s^2

We need to convert the Vs (volume of submerged block) to m^3 by dividing by 1,000,000

Hence, Vs = 53.72/1000000

Vs= 0.00005372m^3

Using the formula; (Fb) = Vs × D × g

Fb = 0.00005372m^3 × 1000kg/m^3 × 10m/s^2

Fb = 0.5372kg/m/s^2

Therefore, the force of buoyancy of the object on water is 0.5372kg/m/s^2

Recall that; Mass (m) = Volume (V) × Density (D)

Force of buoyancy (Fb) = Vs×D = mass (m) × acceleration due to gravity (g)

i.e. mass = Fb/g

= Mass of the block = 0.5372/10

Mass = 0.05372kg

Therefore, the mass of the block in kg is 0.05372kg

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