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2 years ago

6

A reversible process and an irreversible process both have the same________ between the same two states. a. Internal energy b. W

ork capacity C. Change in heat d. Change in entropy

1 answer:

Vlad1618 [11]2 years ago

5 0

Answer:

a) Internal energy

Explanation:

As we know that internal energy is a point function so it did not depends on the path ,it depends at the initial and final states of process.All point function property did not depends on the path.Internal energy is a exact function.

Work and heat is a path function so these depend on the path.They have different values for different path between two states.Work and heat are in exact function.

We know that in ir-reversible process entropy will increase so entropy will be different for reversible and ir-reversible processes.

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#198. Moment of inertia about center of a segmented bar A bar of width is formed of three uniform segments with lengths and area

zaharov [31]

Complete Complete

The complete question is shown on the first uploaded image

Answer:

The moment of inertia of the bar about the center of mass is

$I_r = 1888.80 \ kg m^2$

Explanation:

The free body diagram is shown on the second uploaded image

From the diagram we see that is

The mass of each segment is

$m_1 = \rho_1 l_1 w = 1 * 6 * 2 = 12$

$m_1 = \rho_2 l_2 w = 8 * 6 * 2 = 96$

$m_1 = \rho_2 l_2 w = 5 * 5 * 2 = 50$

The distance from the origin to the center of the segments i.e the center of masses for the individual segments

$x_2 = \frac{6}{2} + 6 = 9 m$

$x_3 = \frac{4}{2} + 12 = 14 m$

The resultant center of mass is mathematically evaluated as

$x_r = \frac{m_1 * x_1 + m_2 *x_2 + m_3 *x_3}{m_1 + m_2 + m_3}$

$= \frac{12 * 3 + 96 *9 + 50 *14}{12+ 96 + 50}$

$x_r = 10.13m$

The moment of Inertia of each segment of the bar is mathematically evaluated

$I_1 =\frac{m_1}{12}(l_1^2 + w^2)$ = $\frac{12}{12}(1^2 + 2^2)$

$I_1 = 4 \ kgm^2$

$I_2 =\frac{m_2}{12}(l_2^2 + w^2)$ = $\frac{96}{12}(6^2 + 2^2)$

$I_2 = 320 \ kgm^2$

$I_3 =\frac{m_3}{12}(l_3^2 + w^2)$ = $\frac{50}{12}(4^2 + 2^2)$

$I_2 = 83.334 \ kgm^2$

According to parallel axis theorem the moment of inertia about the center ( $x_r$ ) is mathematically evaluated as

$I_r = (I_1 + m_1 r_1^2) + (I_2 + m_2 r_2^2) +(I_3 + m_3 r_3^2)$

$I_r = (I_1 + m_1 |x_r - x_1|^2) + (I_2 + m_2 |x_r - x_2|^2) +(I_3 + m_3 |x_r - x_3|^2)$

$I_r = (4 + 12 |10.13 - 3|^2) + (320 + 96 |10.13 - 9|^2) +(83.334 + 50 |10.13 - 14|^2)$

$I_r = 1888.80 \ kg m^2$

6 0

2 years ago

10. When an adhesion bond is made by melting a filler metal and allowing it to spread into the pores of the

TEA [102]

I believe the correct answer is “A”

8 0

2 years ago

Water at 15°C is to be discharged from a reservoir at a rate of 18 L/s using two horizontal cast iron pipes connected in series

love history [14]

Answer:

The required pumping head is 1344.55 m and the pumping power is 236.96 kW

Explanation:

The energy equation is equal to:

$\frac{P_{1} }{\gamma } +\frac{V_{1}^{2} }{2g} +z_{1} =\frac{P_{2} }{\gamma } +\frac{V_{2}^{2} }{2g} +z_{2}+h_{i} -h_{pump} , if V_{1} =0,z_{2} =0\\h_{pump} =\frac{V_{2}^{2}}{2} +h_{i}-z_{1}$

For the pipe 1, the flow velocity is:

$V_{1} =\frac{Q}{\frac{\pi D^{2} }{4} }$

Q = 18 L/s = 0.018 m³/s

D = 6 cm = 0.06 m

$V_{1} =\frac{0.018}{\frac{\pi *0.06^{2} }{4} } =6.366m/s$

The Reynold´s number is:

$Re=\frac{\rho *V*D}{u} =\frac{999.1*6.366*0.06}{1.138x10^{-3} } =335339.4$

$\frac{\epsilon }{D} =\frac{0.00026}{0.06} =0.0043$

Using the graph of Moody, I will select the f value at 0.0043 and 335339.4, as 0.02941

The head of pipe 1 is:

$h_{1} =\frac{V_{1}^{2} }{2g} (k_{L}+\frac{fL}{D} )=\frac{6.366^{2} }{2*9.8} *(0.5+\frac{0.0294*20}{0.06} )=21.3m$

For the pipe 2, the flow velocity is:

$V_{2} =\frac{0.018}{\frac{\pi *0.03^{2} }{4} } =25.46m/s$

The Reynold´s number is:

$Re=\frac{\rho *V*D}{u} =\frac{999.1*25.46*0.03}{1.138x10^{-3} } =670573.4$

$\frac{\epsilon }{D} =\frac{0.00026}{0.03} =0.0087$

The head of pipe 1 is:

$h_{2} =\frac{V_{2}^{2} }{2g} (k_{L}+\frac{fL}{D} )=\frac{25.46^{2} }{2*9.8} *(0.5+\frac{0.033*36}{0.03} )=1326.18m$

The total head is:

hi = 1326.18 + 21.3 = 1347.48 m

The required pump head is:

$h_{pump} =\frac{25.46^{2} }{2*9.8} +1347.48-36=1344.55m$

The required pumping power is:

$P=Q\rho *g*h_{pump} =0.018*999.1*9.8*1344.55=236965.16W=236.96kW$

8 0

2 years ago

A MOSFET differs from a JFET mainly because

Solnce55 [7]

Answer:

The answer is option

C . the JFET has a PN junction

Explanation:

Not only is option C in the question a dissimilarity between the MOSFET and the JFET we can go on with some more dissimilarities.

1.MOSFET stands for Metal Oxide Silicon Field Effect Transistor or Metal Oxide Semiconductor Field Effect Transistor.

(JFET) stands for junction gate field-effect transistor (JFET)

2. JFET is a three-terminal semiconductor device, whereas MOFET a four-terminal semiconductor device.

3. In terms of areas of application of JFETs are used in low noise applications while MOSFETs, are used for high noise applications

5 0

2 years ago

If welding is being done in the vertical position, the torch should have a travel angle of?

siniylev [52]

Answer:

Between 35°– 45°

Explanation:

In the vertical position, Point the flame in the direction of travel. Keep the flame tip at the correct height above the base metal. An angle of 35°–45° should be maintained between the torch tip and the base metal. This angle may be varied up or down to heat or cool the weld pool if it is too narrow or too wide

4 0

1 year ago