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Yanka [14]
4 years ago
10

4 grams of a saturated liquid are converted to a saturated vapor by being heated in a weighted piston–cylinder device arranged t

o maintain the pressure at 200 kPa. During the phase conversion, the volume of the system increases by 1000 cm3, 5 kJ of heat are required, and the temperature of the substance stays constant at 80°C. Estimate the sfg of the substance when its temperature is 80°C.

Engineering
1 answer:
antiseptic1488 [7]4 years ago
6 0

Answer:T2= 351.6k

Explanation:

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3 years ago
A 75 ohm coaxial transmission line has a length of 2.0 cm and is terminated with a load impedance of 37.5 + j75 Ohm. If the diel
Hatshy [7]

Answer:

The load reflection coefficient, \Gamma =0.62\angle 82.875^{\circ} \Omega

Reflection coefficient at input,  \Gamma = 0.62\angle - 147.518^{\circ} \Omega

SWR = 4.26

Given:

Characteristic impedance of the co-axial cable, Z_{c} = 75 \Omega

Length of the cable, L = 2.0 cm = 0.02 m

Z_{Load} = 37.5 + j75 \Omega

Dielectric constant, K = 2.56

frequency, f = 3.0 GHz = 3.0 \times 10^{9} Hz

Explanation:

In order to calculate the reflection coefficient at load, we first calculate these:

The line input impedance Z_{i} is given by:

Z_{i} = Z_{c}\frac{Z_{Load} + jZ_{c} tan(\beta L)}{Z_{c} + jZ_{Load} tan (\beat L)}                     (1)

Now, we calculate the value of \beta:

\beta = \frac{2\pi}{\lambda'} = \farc{2\pi f\sqrt{K}}{c}

(since, \lambda' = \farc{c}{f\sqrt{K}})

\beta = \farc{2\pi f\sqrt{2.56}}{3\times 10^{8}} = 100.53

Now, Substituting the value in eqn (1):

Z_{i} = 75\frac{37.5 + j75 + j75 tan(100.53\times 0.02)}{75 + j(37.5 + j75) tan ( 100.53\times 0.02)} = 18.99 - j20.55 \Omega = 27.98\angle - 47.257^{\circ} \Omega    

Now, the load reflection coefficient is given by:

\Gamma = \frac{Z_{Load} - Z_{c}}{Z_{c} + Z_{Load}}}

Thus

\Gamma = \frac{37.5 + j75 - 75}{75 + 37.5 + j75}} = 0.077 + j0.615 = 0.62\angle 82.875^{\circ} \Omega

Similarly,

Reflection coefficient at input:

\Gamma' = \frac{Z_{i} - Z_{c}}{Z_{c} + Z_{i}}}

\Gamma' = \frac{18.99 - j20.55 - 75}{75 + 18.99 - j20.55}} = - 0.523 - j0.334 = 0.62\angle - 147.518^{\circ} \Omega

Now, the SWR is given by:

SWR, Standing Wave Ratio = \frac{1 +|\Gamma|}{1 - |\Gamma|}

SWR = \frac{1 +|0.62|}{1 - |0.62|} = 4.26

8 0
3 years ago
The blade tension is correct when you can hear a<br>O Thunk<br>O Twang<br>O Neither​
Anastaziya [24]

Answer:

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3 years ago
Lithium at 20°C is BCC and has a lattice constant of 0.35092 nm. Calculate a value for the atomic radius of a lithium atom in na
Nutka1998 [239]

Answer:

the atomic radius of a lithium atom is 0.152 nm

Explanation:

Given data in question

structure = BCC

lattice constant  (a) = 0.35092 nm

to find out

atomic radius of a lithium atom

solution

we know structure is BCC

for BCC radius formula is \sqrt{3} /4 × a

here we have known a value so we put a in radius formula

radius =  \sqrt{3} /4 × a

radius =  \sqrt{3} /4 × 0.35092

radius = 0.152 nm

so the atomic radius of a lithium atom is 0.152 nm

5 0
3 years ago
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