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2 years ago

If x < 5 and x >c, give a value of c such that there

Engineering

1 answer:

Arlecino [84]2 years ago

7 0

we have

x<5

x>c

we know that

The solution is the intersection of both solution sets of the given inequalities.

The solutions of the compound inequality must be solutions of both inequalities.

The value of c could be 5 or any number greater than 5, such that there are no solutions to the compound inequality

Because

A number cannot be both less than 5 and greater than 5 at the same time

therefore

the answer is

for c_> there are no solutions to the compound inequality

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Which of the following placards allows a driver to park in a handicapped parking space.

malfutka [58]

All of the above ......

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3 years ago

Read 2 more answers

What is the activation energy (Q) for a vacancy formation if 10 moles of a metal have 2.3 X 10^13 vacancies at 425°C?

Yakvenalex [24]

Answer:

$Activation\ Energy=2.5\times 10^{-19}\ J$

Explanation:

Using the expression shown below as:

$N_v=N\times e^{-\frac {Q_v}{k\times T}$

Where,

$N_v$ is the number of vacancies

N is the number of defective sites

k is Boltzmann's constant = $1.38\times 10^{-23}\ J/K$

${Q_v}$ is the activation energy

T is the temperature

Given that:

$N_v=2.3\times 10^{13}$

N = 10 moles

1 mole = $6.023\times 10^{23}$

So,

N = $10\times 6.023\times 10^{23}=6.023\times 10^{24}$

Temperature = 425°C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (425 + 273.15) K = 698.15 K

T = 698.15 K

Applying the values as:

$2.3\times 10^{13}=6.023\times 10^{24}\times e^{-\frac {Q_v}{1.38\times 10^{-23}\times 698.15}$

$ln[\frac {2.3}{6.023}\times 10^{-11}]=-\frac {Q_v}{1.38\times 10^{-23}\times 698.15}$

$Q_v=2.5\times 10^{-19}\ J$

4 0

2 years ago

Determine the pressure difference in N/m2,between two points 800m apart in horizontal pipe-line,150 mm diameter, discharging wat

zlopas [31]

Answer: $10.631\times 10^3\ N/m^2$

Explanation:

Given

Discharge is $Q=12.5\ L$

Diameter of pipe $d=150\ mm$

Distance between two ends of pipe $L=800\ m$

friction factor $f=0.008$

Average velocity is given by

$\Rightarrow v_{avg}=\dfrac{12.5\times 10^{-3}}{\frac{\pi }{4}(0.15)^2}\\\\\Rightarrow v_{avg}=\dfrac{15.9134\times 10^{-3}}{2.25\times 10^{-2}}\\\\\Rightarrow v_{avg}=7.07\times 10^{-1}\\\Rightarrow v_{avg}=0.707\ m/s$

Pressure difference is given by

$\Rightarrow \Delta P=f\ \dfrac{L}{d}\dfrac{\rho v_{avg}^2}{2}\\\\\Rightarrow \Delta P=0.008\times \dfrac{800}{0.15}\times \dfrac{997\times (0.707)^2}{2}\\\\\Rightarrow \Delta P=10,631.45\ N/m^2\\\Rightarrow \Delta P=10.631\ kPa$