Physics, 10.12.2019 01:31 vanessajimenez361

Let’s assume that the piece of furniture has an area of 1.5 m , a thickness of 5 cm, and is made of pine (ρ = 350 kg/m3). assuming rose has a mass of 46 kg and jack has a mass of 58 kg, could they have both managed to stay afloat on the raft together? assume the density of seawater is approximately 1025 kg/m3 and that the raft is completely submerged when rose is on top of it.

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Answer from: Nicolas9979

They could not have managed to stay afloat on the raft together

Explanation:

A body is able to float in a liquid if its weight is less than the maximum buoyant force , when this condition is not fulfilled the body is submerged.

So, the buoyant force for the piece of furniture in seawater is given by:

$B=\rho_{w}V_{f}g$ (1)

Where:

$\rho_{w}=1025 kg/m^{3}$ is the density of seawater

$V_{f}$ is the displaced volume of the furniture in seawater, which can be found knowing the dimensions of the piece: Area: $A=1.5 m^{2}$ , depth: d=5 cm=0.05 m . Then $V_{f}=A.d=(1.5 m^{2})(0.05 m)=0.075 m^{3}$

$g=9.8 m/s^{2}$ is the acceleration due gravity

$B=(1025 kg/m^{3})(0.075 m^{3})(9.8 m/s^{2})$ (2)

B=753.375 N (3) This is the buoyant force with the table alone, now let's calculate the weight of Jack and Rose and evaluate if their weights added to te piece of furniture fulfill the condition given above

Furniture's weight:

$W_{f}=m_{f}g$ (4)

Where $m_{f}$ is the mass of the piece of furniture, which can be calculated knowing its density $\rho_{f}=350 kg/m^{3}$ and volume $V_{f}=0.075 m^{3}$

$m_{f}=\rho_{f}V_{f}$ (5)

$m_{f}=(350 kg/m^{3})(0.075 m^{3})$ (6)

$m_{f}=26.25 kg$ (7)

Hence $W_{f}=(26.25 kg)(9.8 m/s^{2})=257.25 N$ (8) As we can see $W_{f}$ which means the weight of the furniture is less than the buoancy force and is able to float

Rose's weight:

$W_{R}=m_{R}g$ (9)

Where $m_{R}=46 kg$ is Rose's mass

$W_{R}=(46 kg)(9.8 m/s^{2})$ (10)

$W_{R}=450.8 N$ (11)

Jack's weight:

$W_{J}=m_{J}g$ (12)

Where $m_{J}=58 kg$ is Jack's mass

$W_{J}=(58 kg)(9.8 m/s^{2})$ (13)

$W_{J}=568.4 N$ (14)

Rose + Furniture weight:

$W_{R+f}=W_{R}+W_{f}$ (15)

$W_{R+f}=450.8 N+257.25 N$ (16)

$W_{R+f}=708.05 N$ (17) As we can see $W_{R+f}$ which means the weight of both Rose and the furniture is less than the buoancy force, so Rose would not drown

Jack + Furniture weight:

$W_{J+f}=W_{J}+W_{f}$ (18)

$W_{J+f}=568.4 N+257.25 N$ (19)

$W_{J+f}=825.65 N$ (20) As we can see $W_{J+f}B$ which means the weight of both Jack and the furniture is greater than the buoancy force and do not fulfill the condition.

Hence Jack would drown

Jack + Rose + Furniture weight:

$W_{J+R+f}=W_{J}+W_{R}+W_{f}$ (21)

$W_{J+R+f}=568.4 N+450.8 N+257.25 N$ (22)

$W_{J+R+f}=1276.45 N$ (23) As we can see $W_{J+R+f}B$ which means the combined weight of Jack, Rose and the furniture is greater than the buoancy force and do not fulfill the condition.

Hence, Jack and Rose could not have managed to stay afloat on the raft together.

In fact, according to the calculations Jack did not have options to survive anyway :(

Answer from: Quest

Egirls? nah. galaxy or a milky way. i think

Answer from: Quest

found in the nucleus

has mass of one amu

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