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Sonja [21]
2 years ago
13

A rectangular open box, 25 ft by 10 ft in plan and 12 ft deep weighs 40 tons. Sufficient amount of stones is placed in the box a

nd then it is placed in a large tank containing 12 ft of water so that it will sink just to the bottom of the tank. Determine the mass of stones placed in the box, in tons.
Engineering
1 answer:
dimulka [17.4K]2 years ago
3 0

Answer:

44.95 tonnes

Explanation:

According to principle of buoyancy the object will just sink when it's weight is more than the weight of the liquid it displaces

It is given that empty weight of box = 40 tons

Let the mass of the stones to be placed be = M tonnes

Thus the combined mass of box and stones = (40+M) tonnes..........(i)

Since the box will displace water equal to it's volume V we have volume of box = 25ft*10ft*12ft= 3000ft^{3}

Volume= 84.95m^{3}

Since 1ft^{3} =0.028m^{3}

Now the weight of water displaced = Weight =\rho \times Volumewhererho is density of water = 1000kg/m^{3}

Thus weight of liquid displaced = \frac{84.95X1000}{1000}tonnes=84.95 tonnes..................(ii)

Equating i and ii we get

40 + M = 84.95

thus Mass of stones = 44.95 tonnes

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Answer:

hello attached is the free body diagram of the missing figure

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Explanation:

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The momentum equation for the flow in the Z - direction can be expressed as

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Fr = horizontal force on the bolts

P1 = pressure of fluid at entrance

V1 = velocity of fluid at entrance

Ac = cross section area of the pipe

P2 and V2 = pressure and velocity of fluid at some distance

m = mass flow rate of fluid

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equation 1 becomes

Fr = ( P1 - P2 ) Ac + mV ( 1 - 2 )

Fr = ( P1 - P2 ) Ac - mV ---------------- equation 2

equation for mass flow rate

m = <em>p</em>AcV  

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insert this into equation 2 EQUATION 2 BECOMES

Fr = ( P1 - P2) Ac - <em>p</em>AcV^2

    = Ac [ (P1 - P2) - pV^2 ]  ---------- equation 3

Note Ac = \frac{\pi }{4} D^2

Equation 3 becomes

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6 0
2 years ago
A thin, flat plate that is 0.2 m × 0.2 m on a side is oriented parallel to an atmospheric airstream having a velocity of 40 m/s.
Leto [7]

Answer:

The rate of heat transfer from both sides of the plate to the air is 240 W

Explanation:

Given;

area of the flat plate = 0.2 m × 0.2 m = 0.04 m²

velocity of atmospheric air stream, v = 40 m/s

drag force, F =  0.075 N

The rate of heat transfer from both sides of the plate to the air:

q = 2 [h'(A)(Ts -T∞)]

where;

h' is heat transfer coefficient obtained from Chilton-Colburn analogy

h' = \frac{C_f}{2} \rho u C_p P_r^{-2/3}\\\\\frac{C_f}{2} = \frac{\tau'_s}{2*\rho u^2/2}

Properties of air at 70°C and 1 atm:

ρ = 1.018 kg/m³, cp = 1009 J/kg.K, Pr = 0.7, v = 20.22 x 10⁻⁶ m²/s

\frac{C_f}{2} = \frac{(0.075/2)/(0.2)^2}{2*(1.018)(40)^2/2} = 5.756*10^{-4}\\\\Thus,\\h' = 5.756*10^{-4} (1.018*40*1009)*(0.7)^{-2/3}\\\\h' = 30 \ W/m^2.K

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q = 2 [ 30(0.04)(120 - 20) ]

q = 240 W

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4 0
2 years ago
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Answer:

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Given data

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