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Sveta_85 [38]
3 years ago
7

The water in a 25-m-deep reservoir is kept inside by a 140-m-wide wall whose cross section is an equilateral triangle as shown i

n the figure Water 25 m 60° 60° Determine the total force (hydrostatic+ atmospheric) acting on the inner surface of the wall and its line of action. Take the value of g as 9.81 m/s2, the atmospheric pressure as 100,000 N/m2, and the density of water to be 1000 kg/m3 throughout. x 108 N The total force (hydrostatic + atmospheric) acting on the inner surface of the wall is The distance of the pressure center from the free surface of water along the wall surface is m.

Engineering
1 answer:
koban [17]3 years ago
7 0

Answer:  (a) 9.00 Mega Newtons or 9.00 * 10^6 N

               (b)  17.1 m

Explanation:  The length of wall under the surface can be given by

                                            b=25m/sin(60)\\=28.867

The average pressure on the surface of the wall is the pressure at the centeroid of the equilateral triangular block which can be then be calculated by multiplying it with the Plate Area which will provide us with the Resultant force.

F(resultant) = Pavg ( A) = (Patm +  \rho g h c)*A \\= [100000 N/m^2 + (1000 kg/m^3 * 9.81 m/s^2 * 25m/2)]* (140*25m/sin60)\\= 8.997*10^8 N \\= 9.0*10^8 N

Noting from the Bernoulli  equation that

Po/\rho g sin60 = 100000/1000 * 9.81* sin(60) = 11.77 m \\ \\

From the second image attached the distance of the pressure center from the free surface of the water along the surface of the wall is given by:

Yp = s+\frac{b}{2} +\frac{b^2}{s+\frac{b}{2}+Po/\rho g sin60}= 0+\frac{28.87}{2} +\frac{28.87^2}{0+\frac{28.87}{2}+100000 /1000 *9.81 sin60} = 17.1 m

Substituting the values gives us the the distance of the surface to be equal to = 17.1 m

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Magnesium sulfate has a number of uses, some of which are related to the ability of the anhydrate form to remove water from air
lara31 [8.8K]

Answer:

Magnesium sulphate is in two forms

Magnesium anhydrate and magnesium hepta hydrate

Density of MgSO4= 2.66 g/ml

Density of MgSO4. 7H2O= 1.68 g/ml

M. W of MgSO4= 120 kg/kmol

In order to form 1000 Kg 20wt% solution

A) the Kg of water required before addition of crystals for MgSO4 =(1000) (0.80)= 800 kg

Average density= (0.20) (2.66) +(0.80) (1) = 1.332 g/ml = 1332 Kg/m3

Volume of water required= (800/1332) = 600.60 litres

For hepta hydrate

Average density= (0.20) (1.68) +(0.80) (1) = 1.136 g/ml=

1136 kg/m 3

MgSO4. 7 H2O, mole fraction of water present =

7×18/(120+(7×18) )=0.512 = 51.2 mol%= 13.6 wt%

1000 kg of (20 wt%) solution required

Amount of pure MgSO4 needed= 1000(.20) = 200 Kg

But it contains 13.6% water hence total weight of

MgSO4. 7H2O= 200/(1-0.136) = 231.48 kg

Amount of water required= (1000-231.48) = 768.51 Kg

Volume of water required= 768.51/1136= 676.51 Litres

B) diameter of tank= 0.30 m

V= 3.142(d2)(H) /4

For pure MgSO4 , volume of water required=600.60 litres

H = 8.49 m

For hepta hydrate volume of water required=676.51 litres

H= 9.57 m

C) height of tank after addition of crystals

Total volume in case of pure MgSO4=

1000/(1332) = 0.7507= 750.7 litres

H= 10.62 m

Total volume of hepta hydrate =

1000/(1136) = 0.8802= 880.2 litres

H= 12.45 m

D) after dissolution of crystals

The volume after dissolution in case of pure MgSO4

= volume of water= 600.67 litres

H=8.49 m

in case of hepta hydrate

The volume after dissolution= ( volume of water + volume of water in MgSO4. 7H2O)

= (676.51+(231.48×0.136) = 707.99 Litres

H= 10.01 m

Part B question

A) Weight of house= 1×105 lb= 2.20 ×105

Force applied by house= (2.20×105) (9.81) = 2.15×106 N

Pressure of ballon= 1.05 atm= 1.063×105 Pa

Diameter= 9.5 inch= 0.2413 m

Area= (3.142) (0.2413) 2/4= 0.0457 m2

F= P×A= 4857.91 N

To float the house

Both forces should be equal

So force applied by 1 ballon= 4857.91 N

Total force required=2.15×106 N

Number of ballons required= (2.15×106) /4857.91=442.57=443 ballons

B) the ballon pressure must be greater than atmospheric pressure

If both pressures are equal then there will be no air flow

If outside pressure is higher, then air would flow from outside to inside of ballon causing deflation, hence pressure inside the ballon must be higher than outside

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3 years ago
Ratio equivalent to 12 red beans to 5 total beans
Kruka [31]

Hello!

The answer to this question as a percentage is 240%

8 0
2 years ago
The human circulatory system consists of a complex branching pipe network ranging in diameter from
Stels [109]

Answer: the average velocity decreases

Explanation:

From the provided data we have:

Vessel    avg. diameter[mm] number

Aorta                 25.0                   1

Arteries             4.0                    159

Arteioles           0.06                 1.4*10^7

Capillaries         0.012               2.9*10^9

from the information, let \hat{m} be the mass flow rate, \rho is density, n number of vessels, and A is the cross-section area for each vessel

the flow rate is constant so it is equal for all vessels,

The average velocity is related to the flow rate by:

\hat{m} = v* \rho * A * n

we clear the side where v is in:

v = \frac{\hat{m}}{\rho A n}

area is π*R^2 where R is the average radius of the vessel (diameter/2)

we get:

v = \frac{\hat{m}}{\rho \pi R^2 n}

you can directly see in the last equation that if we go from the aorta to the capillaries, the number of vessels is going to increase ( n will increase and R is going to decrease ) . From the table, R is significantly smaller in magnitude orders than n, therefore, it wont impact the results as much as n. On the other hand, n will change from 1 to 2.9 giga vessels which will dramatically reduce the average blood velocity

8 0
3 years ago
La probabilidad de que un nuevo producto tenga éxito es de 0.85. Si se eligen 10 personas al azar y se les pregunta si compraría
liq [111]

Answer:

La probabilidad pedida es 0.820196

Explanation:

Sabemos que la probabilidad de que un nuevo producto tenga éxito es de 0.85. Sabemos también que se eligen 10 personas al azar y se les pregunta si comprarían el nuevo producto. Para responder a la pregunta, primero definiremos la siguiente variable aleatoria :

X: '' Número de personas que adquirirán el nuevo producto de 10 personas a las que se les preguntó ''

Ahora bien, si suponemos que la probabilidad de que el nuevo producto tenga éxito se mantiene constante (p=0.85) y además suponemos que hay independencia entre cada una de las personas al azar a las que se les preguntó ⇒ Podemos modelar a X como una variable aleatoria Binomial. Esto se escribe :

X ~ Bi(n,p) en donde ''n'' es el número de personas entrevistadas y ''p'' es la probabilidad de éxito (una persona adquiriendo el producto) en cada caso.

Utilizando los datos ⇒ X ~ Bi(10,0.85)

La función de probabilidad de la variable aleatoria binomial es :

p_{X}(x)=P(X=x)=\left(\begin{array}{c}n&x\end{array}\right)p^{x}(1-p)^{n-x}    con x=0,1,2,...,n

Si reemplazamos los datos de la pregunta en la función de probabilidad obtenemos :

P(X=x)=\left(\begin{array}{c}10&x\end{array}\right)(0.85)^{x}(0.15)^{10-x} con x=0,1,2,...,10

Nos piden la probabilidad de que por lo menos 8 personas adquieran el nuevo producto, esto es :

P(X\geq 8)=P(X=8)+P(X=9)+P(X=10)

Calculando P(X=8), P(X=9) y P(X=10) por separado y sumando, obtenemos que P(X\geq 8)=0.820196

7 0
2 years ago
(35-39) A student travels on a school bus in the middle of winter from home to school. The school bus temperature is 68.0° F. Th
arlik [135]

Answer:

The net energy transfer from the student's body during the 20-min ride to school is 139.164 BTU.

Explanation:

From Heat Transfer we determine that heat transfer rate due to electromagnetic radiation (\dot Q), measured in BTU per hour, is represented by this formula:

\dot Q = \epsilon\cdot A\cdot \sigma \cdot (T_{s}^{4}-T_{b}^{4}) (1)

Where:

\epsilon - Emissivity, dimensionless.

A - Surface area of the student, measured in square feet.

\sigma - Stefan-Boltzmann constant, measured in BTU per hour-square feet-quartic Rankine.

T_{s} - Temperature of the student, measured in Rankine.

T_{b} - Temperature of the bus, measured in Rankine.

If we know that \epsilon = 0.90, A = 16.188\,ft^{2}, \sigma = 1.714\times 10^{-9}\,\frac{BTU}{h\cdot ft^{2}\cdot R^{4}}, T_{s} = 554.07\,R and T_{b} = 527.67\,R, then the heat transfer rate due to electromagnetic radiation is:

\dot Q = (0.90)\cdot (16.188\,ft^{2})\cdot \left(1.714\times 10^{-9}\,\frac{BTU}{h\cdot ft^{2}\cdot R^{4}} \right)\cdot [(554.07\,R)^{4}-(527.67\,R)^{4}]

\dot Q = 417.492\,\frac{BTU}{h}

Under the consideration of steady heat transfer we find that the net energy transfer from the student's body during the 20 min-ride to school is:

Q = \dot Q \cdot \Delta t (2)

Where \Delta t is the heat transfer time, measured in hours.

If we know that \dot Q = 417.492\,\frac{BTU}{h} and \Delta t = \frac{1}{3}\,h, then the net energy transfer is:

Q = \left(417.492\,\frac{BTU}{h} \right)\cdot \left(\frac{1}{3}\,h \right)

Q = 139.164\,BTU

The net energy transfer from the student's body during the 20-min ride to school is 139.164 BTU.

7 0
2 years ago
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