Question

a 89 kg rock with a volume of 6 l is completely submerged in water. the density of water is 1 kg/l.
a.) what volume of water is displaced by the rock?
l
b.) what mass of water is displaced by the rock?
kg
c.) what is the buoyant force on the rock?
n

Explanation:

Part a

Step1: Recall Archimedes' principle (displaced volume)

When an object is completely submerged, the volume of displaced fluid equals the object's volume. The rock's volume is 6 L.
$V_{displaced} = V_{rock} = 6\ \text{L}$

Part b

Step1: Recall density formula ($

ho = \frac{m}{V}$)
Density of water $
ho = 1\ \text{kg/L}$, displaced volume $V = 6\ \text{L}$. Rearrange for mass: $m =
ho V$.

Step2: Calculate displaced mass

$m = 1\ \text{kg/L} \times 6\ \text{L} = 6\ \text{kg}$

Part c

Step1: Recall buoyant force formula ($F_b = m_{displaced}g$), $g = 9.8\ \text{m/s}^2$

Displaced mass $m = 6\ \text{kg}$, $g = 9.8\ \text{m/s}^2$.

Step2: Calculate buoyant force

$F_b = 6\ \text{kg} \times 9.8\ \text{m/s}^2 = 58.8\ \text{N}$

Answer:

a. $\boldsymbol{6}$ L
b. $\boldsymbol{6}$ kg
c. $\boldsymbol{58.8}$ N