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2 years ago

OExplain why the following scenario illustrates an understanding of the importance of analytical skills.

1 answer:

3 0

Analytical skills are vital in the above case as one will be able to weigh the pros and cons of choosing a person with the rights degree and they can find solutions to the problems.

<h3>How does analytical skills help solve problems?</h3>

Analytical skills is known to be those skills that helps a person to look at information or any situation and tell the elements, or strengths as well as its weaknesses and use these to solve the problem.

Note that Analytical skills are vital in the above case as one will be able to weigh the pros and cons of choosing a person with the rights degree and they can find solutions to the problems

Learn more about f analytical skills from

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I have a question.What does DIY mean?

alexdok [17]

Answer: It means "Do it yourself".

Explanation: You're welcome!

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3 years ago

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An insulated piston-cylinder device contains 0.15 of saturated refrigerant-134a vapor at 0.8 MPa pressure. The refrigerant is no

Stells [14]

Answer:

Assumption:

1. The kinetic and potential energy changes are negligible

2. The cylinder is well insulated and thus heat transfer is negligible.

3. The thermal energy stored in the cylinder itself is negligible.

4. The process is stated to be reversible

Analysis:

a. This is reversible adiabatic(i.e isentropic) process and thus $s_{1} =s_{2}$

From the refrigerant table A11-A13

$P_{1} =0.8MPa$ $\left \{ {{ {{v_{1}=v_{g} @0.8MPa =0.025645 m^{3/}/kg } } \atop { {{u_{1}=u_{g} @0.8MPa =246.82 kJ/kg } - also {{s_{1}=s_{g} @0.8MPa =0.91853 kJ/kgK } } \right.$

sat vapor

m= $\frac{V}{v_{1} } =\frac{0.15}{0.025645} =5.8491 kg\\and \\\\P_{2} =0.2MPa \left \{ {{x_{2} =\frac{s_{2} -s_{f} }{s_{fg }}=\frac{0.91853-0.15449}{0.78339} = 0.9753 \atop {u_{2} =u_{f} +x_{2} }(u_{fg}) = 38.26+0.9753(186.25)= 38.26+181.65 =219.9kJ/kg \right. \\s_{1} = s_{2}$

$T_{2} =T_{sat @ 0.2MPa} = -10.09^{o} C$

b.) We take the content of the cylinder as the sysytem.

This is a closed system since no mass leaves or enters.

Hence, the energy balance for adiabatic closed system can be expressed as:

$E_{in} - E_{out} =$ ΔE

$w_{b, out} =$ ΔU

$w_{b, out} =m([tex]u_{1} -u_{2$ )

$w_{b, out} =$ workdone during the isentropic process

=5.8491(246.82-219.9)

=5.8491(26.91)

=157.3993

=157.4kJ

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3 years ago

Evaluate to three significant figures and using appropriate prefix: (354 mg)(45 km)/(0.0356 kN)

frutty [35]

Answer:

0.447 s²

Explanation:

First, convert to SI units.

(354 mg) (45 km) / (0.0356 kN)

(0.354 g) (45000 m) / (35.6 N)

One Newton is kg m/s²:

(0.354 g) (45000 m) / (35.6 kg m/s²)

(0.000354 kg) (45000 m) / (35.6 kg m/s²)

Simplify:

0.447 s²

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What is the minimum hose diameter of an ideal vacuum cleaner that could lift a 14 kg dog off the floor?

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I don’t understand what your asking

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Two blocks A and B have a weight of 10 lb and 6 lb , respectively. They are resting on the incline for which the coefficients of

sertanlavr [38]

Answer:

Part A- $10.06^\circ$

Part B- 0.1946ft

Explanation:

Block A

Apply force equilibrium equation along x direction.

$\begin{array}{l}\\\sum {{F_x}} = 0\\\\{F_A} + {F_S} - 10\sin \theta = 0\\\end{array}$

Here, ${F_A$ } is friction force on block A and ${F_S$ }is spring force, and \theta is inclination angle.

By substitution

$0.14\,{N_A} + \left( {1.9x} \right)\,{\rm{lb}} \cdot {\rm{ft}} - 10\sin \theta = 0$

Similarly, apply force equilibrium equation along y direction.

$\begin{array}{l}\\\sum {{F_y}} = 0\\\\{N_A} - 10\cos \theta = 0\\\\{N_A} = 10\cos \theta \\\end{array}$

By substitution

$\begin{array}{l}\\0.14\,\left( {10\cos \theta } \right) + \left( {1.9x} \right) - 10\sin \theta = 0\\\\\left( {1.9x} \right) = 10\sin \theta - 1.4\cos \theta \\\end{array}$

Block B

Apply force equilibrium equation along x direction.

$\begin{array}{l}\\\sum {{F_x}} = 0\\\\{F_B} - {F_S} - 6\sin \theta = 0\\\end{array}$

By substitution

$0.24\,{N_A} - \left( {1.9x} \right) - 6\sin \theta = 0$

Apply force equilibrium equation along y direction.

$\begin{array}{l}\\\sum {{F_y}} = 0\\\\{N_B} - 6\cos \theta = 0\\\\{N_B} = 6\cos \theta \\\end{array}$

$\begin{array}{l}\\0.24\,\left( {6\cos \theta } \right) - \left( {1.9x} \right) - 6\sin \theta = 0\\\\\left( {1.9x} \right) = 1.44\cos \theta - 6\sin \theta \\\end{array}$

Calculate the inclination angle when both blocks begin to slide.

$\begin{array}{l}\\10\sin \theta - 1.4\cos \theta = 1.44\cos \theta - 6\sin \theta \\\\16\sin \theta = 2.84\cos \theta \\\\\tan \theta = \frac{{2.84}}{{16}}\\\\\theta = {\tan ^{ - 1}}\left( {\frac{{2.84}}{{16}}} \right)\\\end{array}$

Calculate the change in length of the spring.

$\begin{array}{l}\\\left( {1.9x} \right) = 1.44\cos \left( {10.06^\circ } \right) - 6\sin \left( {10.06^\circ } \right)\\\\1.9x = 0.3697\\\\x = 0.1946\,{\rm{ft}}\\\end{array}$

3 0

4 years ago