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.

Answers: 3

Show answers

Answers

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

explanation:

v=uf/(u-f)

= 5.-10/(5+10) convex lens has negative focal length

An object is located 5.0 cm from a convex lens. the lens focuses light at a distance of 10.0 cm. wha

Answer from: Quest

jupiter is the largest planet in our solar system, having a mass and radius that are respectively, 318 and 11.2 times that of earth. suppose that an object falls from rest near the surface of each planet and that the acceleration due to gravity remains constant during the fall. each object falls the same distance before striking the ground. determine the ratio of the time of fall on jupiter to that on earth.

Another question on Physics

Physics, 22.06.2019 03:10

Athin plate moves between two parallel, horizontal, stationary flat surfaces at a constant velocity of v = 7.5 m/s. the two stationary surfaces are spaced 4 cm apart, and the medium between them is filled with oil whose viscosity is 0.9 n·s/m2. the part of the plate immersed in oil at any given time is 2 m long and 0.5 m wide. if the plate moves through the mid-plane between the surfaces, determine the force required to maintain this motion. what would your response be if the plate was 1 cm from the bottom surface (h2) and 3 cm from the top surface (h1)? if the plate moves through the mid-plane between the surfaces, the force required to maintain the motion will be n. if the plate was 1 cm from the bottom surface (h2) and 3 cm from the top (h1) surface, the force required to maintain the motion would be n.

Answers: 2

Answer

Physics, 22.06.2019 03:40

Aprojectile is launched from ground level with an initial velocity of v 0 feet per second. neglecting air resistance, its height in feet t seconds after launch is given by s equals negative 16 t squared plus v 0 t. find the time(s) that the projectile will (a) reach a height of 192 ft and (b) return to the ground when v 0equals112 feet per second. (a) find the time(s) that the projectile will reach a height of 192 ft when v 0equals112 feet per second. select the correct choice below and, if necessary, fill in the answer box to complete your choice.

Answers: 1

Answer

Physics, 22.06.2019 11:40

The ozone hole appears in antarctica during

Answers: 1

Answer

Physics, 22.06.2019 15:10

Where do magnetic lines of force enter earth?

Answers: 3

Answer

You know the right answer?

Let’s assume that the piece of furniture has an area of 1.5 m , a thickness of 5 cm, and is made of...