Physics, 24.09.2019 04:00, jdiffenbaugh115

Araft is made of a plastic block with a density of 650 kg/m 3 , and its dimensions are 2.00 m ã 3.00 m ã 5.00 m. 1. what is the volume of the raft? 2. what is its mass? 3. what is its weight? 4. what is the raft's ap parent weight in water? (hint: density of water = 1.00 ã 10 3 kg/m 3 ) 5. what is the buoyant force on the raft in water? 6. what is the mass of the displaced water? 7. what is the volume of the displaced water? 8. how much of the raft's volume is below water? how much is above? 9. answer items 5 â 8 using ethanol (density = 0.806 ã 10 3 kg/m 3 ) instead of water.

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answered: misslux

1) The volume of the raft is the product between the lenghts of its three dimensions:
V = (2.00 m)(3.00m)(5.00m)=30 m^3

2) The mass of the raft is the product between its density, d, and its volume, V:
m=dV=(650 kg/m^3)(30 m^3)=19500 kg

3) The weight of the raft is the product between its mass m and the gravitational acceleration, g=9.81 m/s^2

:
$W=mg=(19500 kg)(9.81 m/s^2)=1.91 \cdot 10^5 N$

4) The apparent weight is equal to the difference between the weight of the raft and the buoyancy (the weight of the displaced fluid):
$W_a = W- \rho_W V_{disp} g$
where $\rho _W = 1000 kg/m^3$ is the water density and $V_{disp}$ is the volume of displaced fluid.
The density of the raft ( 650 kg/m^3

) is smaller than the water density ( 1000 kg/m^3

), this means that initially the buoyancy (which has upward direction) is larger than the weight (downward direction) and so the raft is pushed upward, until it reaches a condition of equilibrium and it floats. At equilibrium, the weight and the buoyancy are equal and opposite in sign:
$W=B=\rho _W V_{disp} g$
and therefore, the apparent weight will be zero:
W_a = W-B=W-W=0

5) The buoyant force B is the weight of the displaced fluid, as said in step 4):
$B=\rho_W V_{disp} g$
When the raft is completely immersed in the water, the volume of fluid displaced $V_{disp}$ is equal to the volume of the raft, $V_{disp}=V$ . Therefore the buoyancy in this situation is
$B= \rho_W V g = (1000 kg/m^3)(30 m^3)(9.81 m/s^2)=2.94 \cdot 10^5 N$
However, as we said in point 4), the raft is pushed upward until it reaches equilibrium and it floats. At equilibrium, the buoyancy will be equal to the weight of the raft (because the raft is in equilibrium), so:
$B=W=1.91 \cdot 10^5 N$

6) At equilibrium, the mass of the displaced water is equal to the mass of the object. In fact, at equilibrium we have W=B, and this can be rewritten as
$mg = m_{disp} g$
where $m_{disp}= \rho_W V_{disp}$ is the mass of the displaced water. From the previous equation, we obtain that $m_{disp}=m=19500 kg$ .

7) Since we know that the mass of displaced water is equal to the mass of the raft, using the relationship m=dV

we can rewrite $m=m_{disp}$ as:
$d V =d_W V_{disp}$
and so
$V_{disp}= \frac{d V}{d_W}= \frac{(650 kg/m^3)(30m^3)}{1000kg/m^3}= 19.5 m^3$

8) The volume of water displaced is (point 7)

. This volume is now "filled" with part of the volume of the raft, therefore 19.5 m^3

is also the volume of the raft below the water level. We can calculate the fraction of raft's volume below water level, with respect to the total volume of the raft, 30 m^3

:
$\frac{19.5 m^3}{30 m^3}\cdot 100= 65 \%$
Viceversa, the volume of raft above the water level is 30 m^3-19.5 m^3 = 10.5 m^3

. Therefore, the fraction of volume of the raft above water level is
$\frac{10.5 m^3}{30 m^3}\cdot 100 = 35 \%$

9) Let's repeat steps 5-8 replacing $\rho _W$ , the water density, with $\rho_E=806 kg/m^3$ , the ethanol density.

9-5) The buoyant force is given by:
$B=\rho _E V_{disp} g = (806 kg/m^3)(30 m^3)(9.81 m/s^2)=2.37 \cdot 10^5 N$
when the raft is completely submerged. Then it goes upward until it reaches equilibrium and it floats: in this condition, B=W, so the buoyancy is equal to the weight of the raft.

9-6) Similarly as in point 6), the mass of the displaced ethanol is equal to the mass of the raft:
m_E = m = 19500 kg

9-7) Using the relationship $d= \frac{m}{V}$ , we can find the volume of displaced ethanol:
$V_E = \frac{m}{d_E} = \frac{19500 kg}{806 kg/m^3}=24.2 m^3$

9-8) The volume of raft below the ethanol level is equal to the volume of ethanol displaced:

. Therefore, the fraction of raft's volume below the ethanol level is
$\frac{24.2 m^3}{30 m^3}\cdot 100 = 81 \%$
Consequently, the raft's volume above the ethanol level is
30 m^3 - 24.2 m^3 = 5.8 m^3

and the fraction of volume above the ethanol level is
$\frac{5.8 m^3}{30 m^3}\cdot 100 = 19 \%$

answered: Guest

the second one and the forth one are in the purple because they are opinions and the rest are in the yellow

explanation:

50 points to answer this question drag each tile to the correct location. classify the hypotheses ab

answered: Guest

it doesn't. it was imagined up in the middle ages.

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Araft is made of a plastic block with a density of 650 kg/m 3 , and its dimensions are 2.00 m ã 3.00...