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3 years ago

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(30 points) Air at 500 kPa and 400 K enters an adiabatic nozzle at a velocity of 30 m/s and leaves at 300 kPa and 350 K. Using v

ariable specific heats, determine (a) the isentropic efficiency, (b) the exit velocity, and (c) the entropy generation.

1 answer:

Sophie [7]3 years ago

8 0

Answer:

Explanation:

Check attachment for solution

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A 3-kg block, attached to a spring, executes simple harmonic motion according to x= 2cos(50t) where x is in meters and t is in s

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Explanation:

w=50=√k/m=√k/3

SO

(50²)*3 is 7500N/m

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4 years ago

The table below shows a sample wavelength of each type of electromagnetic radiation, along with its frequency and energy.

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2 years ago

A ski gondola is connected to the top of a hill by a steel cable of length 620 m and diameter 1.5 cm. As the gondola comes to th

xz_007 [3.2K]

Answer:

(a) 89 m/s

(b) 11000 N

Explanation:

Note that answers are given to 2 significant figures which is what we have in the values in the question.

(a) Speed is given by the ratio of distance to time. In the question, the time given was the time it took the pulse to travel the length of the cable twice. Thus, the distance travelled is twice the length of the cable.

$v=\dfrac{2\times 620 \text{ m}}{14\text{ s}} = \dfrac{1240\text{ m}}{14\text{ s}}=88.571428\ldots \text{ m/s}= 89\text{ m/s}$

(b) The tension, $T$ , is given by

$v =\sqrt{\dfrac{T}{\mu}}$

where $v$ is the speed, $T$ is the tension and $\mu$ is the mass per unit length.

Hence,

$T = \mu\cdot v^{2}$

To determine $\mu$ , we need to know the mass of the cable. We use the density formula:

$\rho = \dfrac{m}{V}$

where $m$ is the mass and $V$ is the volume.

$m=\rho\cdot V$

If the length is denoted by $l$ , then

$\mu = \dfrac{m}{l} = \dfrac{\rho\cdot V}{l}$

$T = \dfrac{\rho\cdot V}{l} v^{2}$

The density of steel = 8050 kg/m3

The cable is approximately a cylinder with diameter 1.5 cm and length or height of 620 m. Its volume is

$V = \pi \dfrac{d^{2}}{4} l$

$T = \dfrac{\rho\cdot\pi d^2 l}{4l}v^2 = \dfrac{\rho\cdot\pi d^2}{4}v^2$

$T = \dfrac{8050\times\pi\times0.015^2}{4} \times 88.57^2$

$T = 11159.4186\ldots \text{ N} = 11000 \text{ N}$

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3 years ago

A thermometer containing 0.10 g of mercury is cooled from 15 degrees celsius to 8.5 degrees celcius. How much energy left the me

loris [4]

To solve this exercise we will use the concept related to heat loss which is mathematically given as

$Q = mC_p \Delta T$

Where,

m = mass

$C_p$ = Specific Heat

$\Delta T =$ Change in temperature

Replacing with our values we have that

$m = 0.1g$

$C_p = 139J/Kg\cdot K \rightarrow$ Specific heat of mercury

$\Delta T = 8.5\°C-15\°C = -6.5\°C \Rightarrow -6.5K$

Replacing

$Q = (0.1*10^3)(138)(-6.5)\\Q = -0.09J$

Therefore the heat lost by mercury is 0.09J

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3 years ago