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2 years ago

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A river barge, whose cross section is approximately rectangular, carries a load of grain. The barge is 27 ft wide and 93 ft long

. When unloaded its draft (depth of submergence) is 5 ft, and with the load of grain the draft is 7 ft. Determine: (a) the unloaded weight of the barge, and (b) the weight of the grain.

1 answer:

Monica [59]2 years ago

8 0

Answer:

a) for equilibrium

sum of F vertical is zero

$F_{vertical} =0$

so that

$W_{b} =F_{B}=r_{h2o} *(submerged volume)\\=(62.4lb/ft^3)(5ft*27ft*93ft)\\=783432 lb$

b)

$F_{vertical} =0\\W_{b}+ W_{mg} =r_{h2o} *(submerged volume)\\W_{g} =(62.4lb/ft^3)*(7ft*27ft*93ft)-783432\\W_{g} =313372lb$

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According to information found in an old hydraulies book, the energy loss per unit weight of fluid flowing through a nozzle conn

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Answer:

Yes equation is valid.

Explanation:

Given:

h = (0.04 to 0.09)(D/d)^4*V^2/2*g

Using SI units to assign dimensions to every quantity as follows:

Energy loss per unit weight h = J / N = kg m ^2 s^-2 / kg m s^-2 = [m]

Hose diameter D = [m]

Nozzle tip diameter d = [m]

Fluid velocity in the hose V = [ m s^-1 ]

Acceleration of gravity g = [ m s^-2 ]

Using the Given Equation and plug the SI units of respective quantities:

h = (0.04 to 0.09)(D/d)^4*V^2/2*g

[m] = (0.04 to 0.09)([m] / [m])^4*[ m s^-1 ]^2/2*[ m s^-2 ]

Simplify the equation above:

[m] = ( 1 )^4 * [ m^2 s^-2 ] / [ m s^-2 ]

[m] = [m]

Hence, SI units of RHS of given equation = LHS of given equation, we can say the equation has consistent dimensions.

3 0

2 years ago

A cylindrical specimen of a hypothetical metal alloy is stressed in compression. If its original and final diameters are 30.00 a

IrinaVladis [17]

Answer:

The original length of the specimen $l_{o} = 104.7 mm$

Explanation:

Original diameter $d_{o}$ = 30 mm

Final diameter $d_{1}$ = 30.04 mm

Change in diameter Δd = 0.04 mm

Final length $l_{1}$ = 105.20 mm

Elastic modulus E = 65.5 G pa = 65.5 × $10^{3}$ M pa

Shear modulus G = 25.4 G pa = 25.4 × $10^{3}$ M pa

We know that the relation between the shear modulus & elastic modulus is given by

$G = \frac{E}{2(1 + \mu)}$

$25.5 = \frac{65.5}{2 (1 + \mu)}$

$\mu = 0.28$

This is the value of possion's ratio.

We know that the possion's ratio is given by

$\mu = \frac{\frac{0.04}{30} }{\frac{change \ in \ length}{l_{o} } }$

${\frac{change \ in \ length}{l_{o} } = \frac{\frac{0.04}{30} }{0.28}$

${\frac{change \ in \ length}{l_{o} } =$ 0.00476

$\frac{l_{1} - l_{o} }{l_{o} } = 0.00476$

$\frac{l_{1} }{l_{o} } = 1.00476$

Final length $l_{o}$ = 105.2 m

Original length

$l_{o} = \frac{105.2}{1.00476}$

$l_{o} = 104.7 mm$

This is the original length of the specimen.

5 0

3 years ago

A cubic transmission casing whose side length is 25cm receives an input from the engine at a rate of 350 hp. If the vehicle's ve

Musya8 [376]

Answer:

The surface temperature is 921.95°C .

Explanation:

Given:

a=25 cm ,P=350 hp⇒P=260750 W

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h=230 $\frac{W}{m^2-K},\eta =0.95$

Actually heat will be transmit by the convection.

In convection Q=hA $\Delta T$

So $P=\Delta T\times Q$

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T=921.95°C

So the surface temperature is 921.95°C .

6 0

3 years ago

A DC generator turns at 2000 rpm and has an output of 200 V. The armature constant is 0.5 V-min/Wb, and the field constant of th

WITCHER [35]

Answer:

b. 10A

Explanation:

Using the formula, E= k × r×I

200= 0.5 ×2000×0.02×I

200=20×I

Dividing with 20

I = 200/20= 10A

4 0

3 years ago

Read 2 more answers

A 03-series cylindrical roller bearing with inner ring rotating is required for an application in which the life requirement is

-BARSIC- [3]

Answer:

$\mathbf{C_{10} = 137.611 \ kN}$

Explanation:

From the information given:

Life requirement = 40 kh = 40 $40 \times 10^{3} \ h$

Speed (N) = 520 rev/min

Reliability goal $(R_D)$ = 0.9

Radial load $(F_D)$ = 2600 lbf

To find C10 value by using the formula:

$C_{10}=F_D\times \pmatrix \dfrac{x_D}{x_o +(\theta-x_o) \bigg(In(\dfrac{1}{R_o}) \bigg)^{\dfrac{1}{b}}} \end {pmatrix} ^{^{^{\dfrac{1}{a}}$

where;

$x_D = \text{bearing life in million revolution} \\ \\ x_D = \dfrac{60 \times L_h \times N}{10^6} \\ \\ x_D = \dfrac{60 \times 40 \times 10^3 \times 520}{10^6}\\ \\ x_D = 1248 \text{ million revolutions}$

$\text{The cyclindrical roller bearing (a)}= \dfrac{10}{3}$

The Weibull parameters include:

$x_o = 0.02$

$(\theta - x_o) = 4.439$

$b= 1.483$

∴

Using the above formula:

$C_{10}=1.4\times 2600 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{1}{\dfrac{10}{3}}}$

$C_{10}=3640 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{3}{10}}$

$C_{10} = 3640 \times \bigg[\dfrac{1248}{0.9933481582}\bigg]^{\dfrac{3}{10}}$

$C_{10} = 30962.449 \ lbf$

Recall that:

1 kN = 225 lbf

∴

$C_{10} = \dfrac{30962.449}{225}$

$\mathbf{C_{10} = 137.611 \ kN}$

7 0

2 years ago