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3 years ago

10

How are all mechanical waves similar?

1 answer:

Alex17521 [72]3 years ago

6 0

Wavelength, frequency, and amplitude. Hope this helps.

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The weight of an object on Earth is 350 newtons. On Mars, the same object would weigh 134 newtons. What is the acceleration due

sineoko [7]

W = mg = 350 newton
m = W/g = 350/9.8 = 35.71 kg
on mars
W = mg = 134 newton
g = W/m = 134/35.71 = 3.75 meters/second2

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3 years ago

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Which of the following was NOT one of the top 10 reported Stressors for Teens?

olga55 [171]

I think it life changes

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3 years ago

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Which of the following describes a closed conducting loop through which an electrical current can flow?

Romashka-Z-Leto [24]

I think the correct answer is the second option. A circuit describes a closed conducting loop through which an electrical current can flow. It is a path that an electrical current could flow. A circuit could be a closed one or an open circuit. A closed circuit would be a circuit where the current could flow continuously. An open circuit would be a type of circuit where the flow current would only go once and stopped at a particular point since the current has nowhere to go. For a circuit to work, an electric supply should be available to supply the electric current.

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4 years ago

Argon gas enters steadily an adiabatic turbine at 900 kPa and 450C with a velocity of 80 m/s and leaves at 150 kPa with a veloc

Crazy boy [7]

Answer:

Temperature at the exit = $267.3 C$

Explanation:

For the steady energy flow through a control volume, the power output is given as

$W_{out}= -m_{f}(h_{2}-h_{1} + \frac{v_{2}^{2}}{2} - \frac{v_{1}^{2}}{2})$

Inlet area of the turbine = $60cm^{2}= 0.006m^{2}$

To find the mass flow rate, we can apply the ideal gas laws to estimate the specific volume, from there we can get the mass flow rate.

Assuming Argon behaves as an Ideal gas, we have the specific volume $v_{1}$

as

$v_{1}=\frac{RT_{1}}{P_{1}}=\frac{0.2081\times723}{900}=0.1672m^{3}/kg$

$m_{f}=\frac{1}{v_{1}}\times A_{1}V_{1} = \frac{1}{0.1672}\times(0.006)(80)=2.871kg/sec$

for Ideal gasses, the enthalpy change can be calculated using the formula

$h_{2}-h_{1}=C_{p}(T_{2}-T_{1})$

hence we have

$W_{out}= -m_{f}((C_{p}(T_{2}-T_{1}) + \frac{v_{2}^{2}}{2} - \frac{v_{1}^{2}}{2})$

$250= -2.871((0.5203(T_{2}-450) + \frac{150^{2}}{2\times 1000} - \frac{80^{2}}{2\times 1000})$

<em>Note: to convert the Kinetic energy term to kilojoules, it was multiplied by 1000</em>

evaluating the above equation, we have $T_{2}=267.3C$

Hence, the temperature at the exit = $267.3 C$

5 0

3 years ago

Round the number 14.587020 to 4 significant digits.

PilotLPTM [1.2K]

A. 14.59 is correctly rounded to 4 significant digits.

4 0

4 years ago

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