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3 years ago

13

The bulk modulus of a fluid if it undergoes a 1% change in volume when subjected to a pressure change of 10,000 psi is (a) 0.01

psi (b) 0.001 psi (c) 0.00001 psi (d) 0.000001 psi (e) none of these

1 answer:

Veseljchak [2.6K]3 years ago

3 0

Answer:

The required bulk modulus is $10^{6}$ Psi. So, the answer is non of these.

Explanation:

Change in pressure of the fluid is directly proportional to the volumetric strain. The constant of proportionality is the bulk modulus of the fluid.

Step1

Given:

Percentage change in volume is 1%.

Change in pressure is 10000 Psi.

Calculation:

Step2

Volumetric strain is calculated as follows:

$\frac{\bigtriangleup V}{V}=\frac{1}{100}$

$\frac{\bigtriangleup V}{V}=0.01$

Step3

Bulk modulus is calculated as follows:

\frac{\bigtriangleup V}{V}=0.01

$\frac{\bigtriangleup V}{V}=0.01$

$10000=K\times0.01$

K = 1000000 Psi.

Thus, the required bulk modulus is $10^{6}$ Psi.

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Compute the volume percent of graphite, VGr, in a 3.2 wt% C cast iron, assuming that all the carbon exists as the graphite phase

Yanka [14]

Answer:

The volume percentage of graphite is 10.197 per cent.

Explanation:

The volume percent of graphite is the ratio of the volume occupied by the graphite phase to the volume occupied by the graphite and ferrite phases. The weight percent in the cast iron is 3.2 wt% (graphite) and 96.8 wt% (ferrite). The volume percentage of graphite is:

$\%V_{gr} = \frac{V_{gr}}{V_{gr}+V_{fe}} \times 100\,\%$

Where:

$V_{gr}$ - Volume occupied by the graphite phase, measured in cubic centimeters.

$V_{fe}$ - Volume occupied by the graphite phase, measured in cubic centimeters.

The expression is expanded by using the definition of density and subsequently simplified:

$\%V_{gr} = \frac{\frac{m_{gr}}{\rho_{gr}} }{\frac{m_{gr}}{\rho_{gr}}+\frac{m_{fe}}{\rho_{fe}}}\times 100\,\%$

Where:

$m_{fe}$ , $m_{gr}$ - Masses of the ferrite and graphite phases, measured in grams.

$\rho_{fe}, \rho_{gr}$ - Densities of the ferrite and graphite phases, measured in grams per cubic centimeter.

$\%V_{gr} = \frac{1}{1+\frac{\frac{m_{fe}}{\rho_{fe}} }{\frac{m_{gr}}{\rho_{gr}} } }\times 100\,\%$

$\%V_{gr} = \frac{1}{1 + \left(\frac{\rho_{gr}}{\rho_{fe}} \right)\cdot\left(\frac{m_{fe}}{m_{gr}} \right)} \times 100\,\%$

If $\rho_{gr} = 2.3\,\frac{g}{cm^{3}}$ , $\rho_{fe} = 7.9\,\frac{g}{cm^{3}}$ , $m_{gr} = 3.2\,g$ and $m_{fe} = 96.8\,g$ , the volume percentage of graphite is:

$\%V_{gr} = \frac{1}{1+\left(\frac{2.3\,\frac{g}{cm^{3}} }{7.9\,\frac{g}{cm^{3}} } \right)\cdot \left(\frac{96.8\,g}{3.2\,g} \right)} \times 100\,\%$

$\%V_{gr} = 10.197\,\%V$

The volume percentage of graphite is 10.197 per cent.

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3 years ago

Assume a steel pipe of inner radius r1= 20 mm and outer radius r2= 25 mm, which is exposed to natural convection at h = 50 W/m2.

Mekhanik [1.2K]

Answer:

98,614.82 W/m²

Explanation:

$Q = 2\pi hL(\frac{T_2-T_1}{Ln\frac{r_2}{r_1}})$

Where;

Q = the amount of heat loss from the pipe

h = the heat transfer coefficient of the pipe = 50 W/m².K

T₁ = the ambient temperature of the pipe = 30⁰C

T₂ = the outside temperature of the pipe = 100⁰C

L= the length of pipe

r₁ = inner radius of the pipe = 20mm

r₂ = outer radius of the pipe = 25mm

To determine the amount of heat loss from the pipe per unit length

From the equation above

$\frac{Q}{L} = 2\pi h(\frac{T_2-T_1}{Ln\frac{r_2}{r_1}})$

$\frac{Q}{L} = 2\pi* 50(\frac{100-30}{Ln\frac{25}{20}})$

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$\frac{Q}{L} = 314.159(313.901)$ = 98,614.82 W/m²

3 0

3 years ago

one number is 11 more than another number. find the two number if three times the larger number exceeds four times the smaller n

vaieri [72.5K]

Answer:

a = 40

b = 29

Explanation:

Give a place holder for the numbers that we don't know.

Lets call the two numbers a and b.

From the given info, we can write an expression and solve it:

"one number is 11 more than another number"

a = 11 + b

from this, we know that a > b.

''three times the larger number exceeds four times the smaller number by 4"

3a = 4b + 4

Now we have 2 equations, we can use them to solve using whatever method you want.

a = 11 + b

3a = 4b + 4

I will be using matrices RREF to solve for this.

a - b = 11

3a - 4b = 4

$\begin{bmatrix}1 & -1 & 11\\3 & -4 & 4 \end{bmatrix}$

$\begin{bmatrix}1 & 0 & 40\\0 & 1 & 29 \end{bmatrix}$

a = 40

b = 29

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A Geostationary satellite has an 8kW RF transmission pointed at the earth. How much force does that induce on the spacecraft? (N

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Answer:

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Solution:

As per the question:

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Now, the force, F is given by:

$P_{t} = Force(F)\times velocity(v) = Fv$ (1)

where

v = velocity

Now,

For a geo-stationary satellite, the centripetal force, $F_{c}$ is provided by the gravitational force, $F_{G}$ :

$F_{c} = F_{G}$

$\frac{mv^{2}}{R} = \frac{GM_{e}m{R^{2}}$

Thus from the above, velocity comes out to be:

$v = \sqrt{\frac{GM_{e}}{R}}$

$v = \sqrt{\frac{6.67\times 10^{- 11}\times 5.979\times 10^{24}}{42166\times 10^{3}}} = 3075.36 m/ s$

where

R = $R_{e} + H$

R = $\sqrt{GM_{e}(\frac{T}{2\pi})^{2}}$

where

G = Gravitational constant

T = Time period of rotation of Earth

R is calculated as 42166 km

Now, from eqn (1):

$8000 = F\times 3075.36$

F = 2.60 N

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3 years ago

Determine the dimensions for W if W = P L^3 / (M V^2) where P is a pressure, L is a length, M is a mass, and V is a velocity.

Eva8 [605]

Correct answer is option E. No dimensions

As we know formula Pressure (P) is $\frac{F}{A}$

also,

Dimensional formula of <em>Pressure is </em> $M^{1}L^{-1}T^{-2}$
Dimensional formula of <em>length is L </em>
Dimensional formula of <em>mass is M</em>
Dimensional formula of <em>velocity is </em> $L^{1} T^{-1}$

So, as given W= $\frac{P*L^{3} }{M*V^{2} }$

Dimensional formula of W = $\frac{M^{1}L^{-1}T^{-2} L^{3} }{M^{1} L^{2}T^{-2} }$

since all terms get cancelled

Work is dimensionless i.e no dimensions

Learn more about dimensions here brainly.com/question/20351712

#SPJ10

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2 years ago