All categories

2 years ago

What are the advantages of studying a scheduled course over a more freeform one?

Engineering

2 answers:

anyanavicka [17]2 years ago

8 0

Answer:

Taking responsibility for your own learning makes it easier to identify your strengths and weaknesses. Once these have been identified you can work on a learning plan that focuses on the areas that you need most help with, increasing the speed of your learning, and build the skills you have been trying to perfect.

Explanation:

Eva8 [605]2 years ago

4 0

Answer:

Some advantages of studying a course that is more scheduled is that it takes responsibility to remember your schedule,, so whilst having a scheduled course you are also self disciplining and learning responsibility.

Explanation:

You might be interested in

A piston–cylinder device contains a mixture of 0.5 kg of H2 and 1.2 kg of N2 at 100 kPa and 300 K. Heat is now transferred to th

Taya2010 [7]

Answer:

(a) The heat transferred is 2552.64 kJ

(b) The entropy change of the mixture is 1066.0279 J/K

Explanation:

Here we have

Molar mass of H₂ = 2.01588 g/mol

Molar mass of N₂ = 28.0134 g/mol

Number of moles of H₂ = 500/2.01588 = 248 moles

Number of moles of N₂ = 1200/28.0134 = 42.8 moles

P·V = n·R·T

V₁ = n·R·T/P = 290.8×8.3145×300/100000 = 7.25 m³

Since the volume is doubled then

V₂ = 2 × 7.25 = 14.51 m³

At constant pressure, the temperature is doubled, therefore

T₂ = 600 K

If we assume constant specific heat at the average temperature, we have

Heat supplied = m₁×cp₁×dT₁ + m₂×cp₂×dT₂

cp₁ = Specific heat of hydrogen at constant pressure = 14.50 kJ/(kg K

cp₂ = Specific heat of nitrogen at constant pressure = 1.049 kJ/(kg K

Heat supplied = 0.5×14.50×300 K+ 1.2×1.049×300 = 2552.64 kJ

b) $\Delta S = - R(n_A \times lnx_A + n_B \times ln x_B)$

Where:

$x_A$ and $x_B$ are the mole fractions of Hydrogen and nitrogen respectively.

Therefore, $x_A$ = 248 /(248 + 42.8) = 0.83

$x_B$ = 42.8/(248 + 42.8) = 0.1472

∴ $\Delta S = - 8.3145(248 \times ln0.83 + 42.8 \times ln 0.1472)$ = 1066.0279 J/K

5 0

3 years ago

The pressure gage on a 2.5-m^3 oxygen tank reads 500 kPa. Determine the amount of oxygen in the tank if the temperature is 28°C

s2008m [1.1K]

Answer:

19063.6051 g

Explanation:

Pressure = Atmospheric pressure + Gauge Pressure

Atmospheric pressure = 97 kPa

Gauge pressure = 500 kPa

Total pressure = 500 + 97 kPa = 597 kPa

Also, P (kPa) = 1/101.325 P(atm)

Pressure = 5.89193 atm

Volume = 2.5 m³ = 2500 L ( As m³ = 1000 L)

Temperature = 28 °C

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

T(K) = T( °C) + 273.15

So,

T₁ = (28.2 + 273.15) K = 301.15 K

Using ideal gas equation as:

PV=nRT

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

5.89193 atm × 2500 L = n × 0.0821 L.atm/K.mol × 301.15 K

⇒n = 595.76 moles

Molar mass of oxygen gas = 31.9988 g/mol

Mass = Moles * Molar mass = 595.76 * 31.9988 g = 19063.6051 g

7 0

3 years ago

Which statement most accurately describes Pascal's law?

marin [14]

Answer:

La C

Explanation:

7 0

3 years ago

Read 2 more answers

A 3.52 kg steel ball is tossed upward from a height of 6.93 meters above the floor with a vertical velocity of 2.99 m/s. What is

Dafna1 [17]

Answer : The final velocity of the ball is, 12.03 m/s

Explanation :

By the 3rd equation of motion,

$v^2-u^2=2as$

where,

s = distance covered by the object = 6.93 m

u = initial velocity = 2.99 m/s

v = final velocity = ?

a = acceleration = $9.8m/s^2$

Now put all the given values in the above equation, we get the final velocity of the ball.

$v^2-(2.99m/s)^2=2\times (9.8m/s^2)\times (6.93m)$

$v=12.03m/s$

Thus, the final velocity of the ball is, 12.03 m/s

7 0

3 years ago

A 800-MW steam power plant, which is cooled by a nearby river, has a thermal efficiency of 40 percent. Determine the rate of hea

Arturiano [62]

Answer:

Rate of heat transfer to river=1200MW

So the actual amount of heat rejected ti the river will be less as there will some heat loss to surrounding and in pipes

Explanation:

In order to find the actual heat transfer rate is lower or higher than its value we will first find the rate of heat transfer to power plant:

$Efficiency=\frac{work}{heat transfer to power plant}$

$Heat transfer=\frac{work}{Efficiency\\} \\\\Heat transfer=\frac{800}{0.40}\\\\Heat transfer=2000MW$

From First law of thermodynamics:

Rate of heat transfer to river=heat transfer to power plant-work done

Rate of heat transfer to river=2000-800

Rate of heat transfer to river=1200MW

So the actual amount of heat rejected ti the river will be less as there will some heat loss to surrounding and in pipes.

4 0

3 years ago