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Lisa [10]
3 years ago
7

50 points

Engineering
1 answer:
Burka [1]3 years ago
4 0

Answer:

Water vapor

Explanation:

When water is in a vapor it tends to rise to a higher point. Because of this it would be able to reach the top of a building.

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Two previously undeformed cylindrical specimens of an alloy are to be strain hardened by reducing their cross-sectional areas (w
natima [27]

Answer:

The second specimen's radius, in mm, after deformation = 9mm

Explanation:

The radii of both cylindrical specimens are missing.

I'll assume values for them.

Let r1 = initial radius of first cylinder = 20mm

Let r2 = deformed radius of the first cylinder = 15mm

Let r = Initial radius of second cylinder = 16mm.

Now, we can proceed....

For the two cylinders to have the same deformed hardness,

They must be deformed by the same percentage.

To get the deformed radius of the second cylindrical specimen, we'll need to calculate the percentage deformation of specimen 1.

This is given by

(Ao - Ad) /Ao

Where Ao = Original Area = πr²

Ao = πr1²

Ao = π20²

Ao = 400π

Ad = Deformed Area = πr²

Ad = πr2²

Ad = π15²

Ad = 225π

So, %deformation = %d =(400π - 225π)/400π

= 175π/400π

= 0.4375

%d = 43.75%

Now, we'll solve for the deformed radius of the second cylindrical specimen.

Note that, the two specimens have the same % deformation (to have the same hardness).

So, we have

rd - ro = 1 - %d

Where rd = deformed radius

ro = original radius = 16

rd/ro = 1 - %d becomes

rd / 16 = 1 - 43.75%

rd / 16 = 56.25%

rd = 16 * 56.25%

rd = 9mm

7 0
4 years ago
Tadpoles raised in water with atrazine levels of 0.1 ppb should produce a higher percentage of male frogs with gonadal abnormali
Kobotan [32]

Answer:

correct option is c. a testable prediction leading to design of an experiment

Explanation:

The results of raising tadpoles were estimated to be water with an atrazine level of 0.1 ppb compared to those grown in pure water. So, this is not the question. If this assumption can now be tested, an experiment can be made in which a certain number of tadpoles can be raised in pure water and the same number of tadpoles can be raised in water with a 0.1ppb atrazine level can. The difference between the two populations can be estimated or compared.

8 0
3 years ago
Explain by Research how a basic generator works ? using diagram<br>​
natulia [17]
Correcto no se muy bien de que se trata el tema porque está en inglés.
Sorry
8 0
3 years ago
19 million Joules of chemical energy are supplied to a boiler in a power plant. 5 x 106 Joules of energy are lost through unburn
Fiesta28 [93]

Answer:

67.89%

Explanation:

Energy Efficiency=\frac {U_{energy}}{T_{energy}}\times 100

Where U_{energy} is useful energy and T_{energy} is the total energy

The useful energy= (19-5-1.1) million= 12.9 million

The total energy= 19 million

Efficiency=\frac {12.9}{19}\times 100=67.89473684\approx 67.89

Therefore, the efficiency of the boiler is 67.89%

8 0
3 years ago
A small submarine has a triangular stabilizing fin on its stern. The fin is 1 ft tall and 2 ft long. The water temperature where
Arturiano [62]

Answer:

\mathbf{F_D \approx 1.071 \ lbf}

Explanation:

Given that:

The height of a  triangular stabilizing fin on its stern is 1 ft tall

and it length is 2 ft long.

Temperature = 60 °F

The objective is to determine the drag on the fin when the submarine is traveling at a speed of 2.5 ft/s.

From these information given; we can have a diagrammatic representation describing how the  triangular stabilizing fin looks like as we resolve them into horizontal and vertical component.

The diagram can be found in the attached file below.

If we recall ,we know that;

Kinematic viscosity v = 1.2075 \times 10^{-5} \ ft^2/s

the density of water ρ = 62.36 lb /ft³

Re_{max} = \dfrac{Ux}{v}

Re_{max} = \dfrac{2.5 \ ft/s \times 2  \  ft }{1.2075 \times 10 ^{-5} \ ft^2/s}

Re_{max} = 414078.6749

Re_{max} = 4.14 \times 10^5 which is less than < 5.0 × 10⁵

Now; For laminar flow;  the drag on  the fin when the submarine is traveling at 2.5 ft/s can be determined by using the expression:

dF_D = (\dfrac{0.664 \times \rho  \times U^2 (2-x) dy}{\sqrt{Re_x}})^2

where;

(2-x) dy = strip area

Re_x = \dfrac{2.5(2-x)}{1.2075 \times 10 ^{-5}}

Therefore;

dF_D = (\dfrac{0.664 \times 62.36  \times 2.5^2 (2-x) dy}{\sqrt{ \dfrac{2.5(2-x)}{1.2075 \times 10 ^{-5}}}})

dF_D = 1.136 \times(2-x)^{1/2} \ dy

Let note that y = 0.5x from what we have in the diagram,

so , x = y/0.5

By applying the rule of integration on both sides, we have:

\int\limits \  dF_D =  \int\limits^1_0 \  1.136 \times(2-\dfrac{y}{0.5})^{1/2} \ dy

\int\limits \  dF_D =  \int\limits^1_0 \  1.136 \times(2-2y)^{1/2} \ dy

Let U = (2-2y)

-2dy = du

dy = -du/2

F_D =  \int\limits^0_2 \  1.136 \times(U)^{1/2} \ \dfrac{du}{-2}

F_D = - \dfrac{1.136}{2} \int\limits^0_2 \ U^{1/2} \ du

F_D = -0.568 [ \dfrac{\frac{1}{2}U^{ \frac{1}{2}+1 }  }{\frac{1}{2}+1}]^0__2

F_D = -0.568 [ \dfrac{2}{3}U^{\frac{3}{2} }   ] ^0__2

F_D = -0.568 [0 -  \dfrac{2}{3}(2)^{\frac{3}{2} }   ]

F_D = -0.568 [- \dfrac{2}{3} (2.828427125)}   ]

F_D = 1.071031071 \ lbf

\mathbf{F_D \approx 1.071 \ lbf}

8 0
3 years ago
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