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Anon25 [30]
3 years ago
14

A raft is made of 12 logs lashed together. Each is 45 cm in diameter and has a length of 6.5 m. How many people can the raft hol

d before they start getting their feet wet, assuming the average person has a mass of 68 kg? Do not neglect the weight of the logs. Assume the specific gravity of wood is 0.60.
Physics
1 answer:
tatyana61 [14]3 years ago
3 0

Answer:

324 people

Explanation:

radius r = diameter / 2 = 45cm /2 = 22.5 cm or 0.225 m

We can calculate the total volume of the log by computing the volume of each cylindrical log:

V = \pi r^2 L = \pi 0.225^2 * 6.5 = 4.6 m^3

So the total volume of all 12 logs is

12 * 4.6 = 55.135 m^3

The weight that the raft can support, is the buoyancy force subtracted by its own weight.

The buoyancy force is basically the weight of the water is displaced, which is water density times volume. The log weight is also log density times its volume

F = F_b - F_w = g\rho_wV - g\rho_rV = gV(\rho_w - \rho_r)

where \rho_w = 1000kg/m^3,\rho_r are the density of water and raft, respectively. But since we have the specific gravity of wood is 0.6. That measn

\rho_r / \rho_w = 0.6

\rho_r = 0.6\rho_w

Therefore \rho_w - \rho_r = \rho_w - 0.6\rho_w = 0.4\rho_w = 0.4*1000 = 400 kg/m^3

Let g = 9.8m/s2. We can now calculate the force that the raft can support

F = gV(\rho_w - \rho_r) = 9.8*55.135*400=216129.2N

Each person has a mass of 68kg, their weight would be

W = 68*g = 68*9.8 = 666.4N

So the maximum number of people that the raft can hold is

F / W = 216129.2 / 666.4 = 324 people

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False

Explanation:

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R = distance of Sun from Mercury, m = mass of Mercury

F(Merc) / F(Mars) = 4^2 / 2 = 8

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A solid conducting sphere is given a positive charge q. How is the charge q distributed in or on the sphere?
leva [86]

Answer:

Explanation:

the sphere is solid and conducting, so the charge is uniformly distributed over its volume.

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3 years ago
Read 2 more answers
Water flows through a cast steel pipe (k = 50 W m.K, ε = 0.8) with an outer diameter of 104mm and 2 mm wall thickness. Calculate
masha68 [24]

Answer:

The heat loss per unit length is   \frac{Q}{L}   = 2981 W/m

Explanation:

From the question we are told that

     The outer diameter of the pipe is d = 104mm = \frac{104}{1000} = 0.104 m

     The thickness is  D = 2mm = \frac{2}{1000} = 0.002m  

      The temperature  of water is  T = 90^oC = 90 + 273 = 363K  

      The outside air temperature is T_a = -10^oC = -10 +273 = 263K

        The water side heat transfer coefficient is z_1 = 300 W/ m^2 \cdot K

       The  heat transfer coefficient is  z_2 = 20 W/m^2 \cdot K

The heat lost per unit length is mathematically represented as

           \frac{Q}{L}   = \frac{2 \pi (T - Ta)}{ \frac{ln [\frac{d}{D} ]}{z_1}  +  \frac{ln [\frac{d}{D} ]}{z_2}}

Substituting values

         \frac{Q}{L}   = \frac{2 * 3.142 (363 - 263)}{ \frac{ln [\frac{0.104}{0.002} ]}{300}  +  \frac{ln [\frac{0.104}{0.002} ]}{20}}

           \frac{Q}{L}   = \frac{628}{0.2107}

           \frac{Q}{L}   = 2981 W/m

6 0
3 years ago
A package is dropped from an air balloon two times. In the first trial the distance between the balloon and the surface is Hand
enyata [817]

Answer:

<em>The final speed of the second package is twice as much as the final speed of the first package.</em>

Explanation:

<u>Free Fall Motion</u>

If an object is dropped in the air, it starts a vertical movement with an acceleration equal to g=9.8 m/s^2. The speed of the object after a time t is:

v=gt

And the distance traveled downwards is:

\displaystyle y=\frac{gt^2}{2}

If we know the height at which the object was dropped, we can calculate the time it takes to reach the ground by solving the last equation for t:

\displaystyle t=\sqrt{\frac{2y}{g}}

Replacing into the first equation:

\displaystyle v=g\sqrt{\frac{2y}{g}}

Rationalizing:

\displaystyle v=\sqrt{2gy}

Let's call v1 the final speed of the package dropped from a height H. Thus:

\displaystyle v_1=\sqrt{2gH}

Let v2 be the final speed of the package dropped from a height 4H. Thus:

\displaystyle v_2=\sqrt{2g(4H)}

Taking out the square root of 4:

\displaystyle v_2=2\sqrt{2gH}

Dividing v2/v1 we can compare the final speeds:

\displaystyle v_2/v_1=\frac{2\sqrt{2gH}}{\sqrt{2gH}}

Simplifying:

\displaystyle v_2/v_1=2

The final speed of the second package is twice as much as the final speed of the first package.

5 0
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