Which qualities does an Interactive Media worker need when creating video conference or other multimedia projects for customers?

The relatonship between Kalven and celsius______

umka2103 [35]

Answer:

$K=C+273.15$

Explanation:

Kelvin's climbing represents the <em>absolute temperature</em>. Temperature is a measure of the molecular kinetic energy of translation. If the molecules move quickly, with the same energy as in the walls of the container, which makes us feel like "heat". If the molecules do not move, the temperature is zero. 0 K.

The Celsius scale has an <em>artificial zero</em>, defined in the solidification temperature of the water. It is very useful to talk about the weather, and about some simpler technical matters. But it is artificial.

3 0

4 years ago

Air enters an adiabatic gas turbine at 1590 oF, 40 psia and leaves at 15 psia. The turbine efficiency is 80%, and the mass flow

melamori03 [73]

Answer:

a) 158.4 HP.

b) 1235.6 °F.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to set up an energy balance for the turbine's inlets and outlets:

$m_{in}h_{in}=W_{out}+m_{out}h_{out}$

Whereas the mass flow is just the same, which means we have:

$W_{out}=m_{out}(h_{out}-h_{in})$

And the enthalpy and entropy of the inlet stream is obtained from steam tables:

$h_{in}=1860.7BTU/lbm\\\\s_{in}= 2.2096BTU/lbm-R$

Now, since we assume the 80% accounts for the isentropic efficiency for this adiabatic gas turbine, we assume the entropy is constant so that:

$s_{out}= 2.2096BTU/lbm-R$

Which means we can find the temperature at which this entropy is exhibited at 15 psia, which gives values of temperature of 1200 °F (s=2.1986 BTU/lbm-K) and 1400 °F (s=2.2604 BTU/lbm-K), and thus, we interpolate for s=2.2096 to obtain a temperature of 1235.6 °F.

Moreover, the enthalpy at the turbine's outlet can be also interpolated by knowing that at 1200 °F h=1639.8 BTU/lbm and at 1400 °F h=174.5 BTU/lbm, to obtain:

$h_{out}=1659.15BUT/lbm$

Then, the isentropic work (negative due to convention) is:

$W_{out}=2500lbm/h(1659.15BUT/lbm-1860.7BUT/lbm)\\\\W_{out}=-503,875BTU$

And the real produced work is:

$W_{real}=0.8*-503875BTU\\\\W_{real}=-403100BTU$

Finally, in horsepower:

$W_{real}=-403100BTU/hr*\frac{1HP}{2544.4336BTU/hr} \\\\W_{real}=158.4HP$

Regards!

6 0

3 years ago

A civil engineer is analyzing the compressive strength of concrete. The compressive strength is approximately normal distributed

hram777 [196]

Answer:

See explanation

Explanation:

Solution:-

- A study on compressive strength of a concrete was made. The distribution of compressive strength ( experimental testing ) was normally distributed with variance ( σ^2 ).

- A random sample of n = 12 specimens were taken and the mean compressive strength ( μ ) of 3500 psi was claimed.

- We are to test the claim made by the civil engineer regarding the mean compressive strength of the concrete. The data of compressive strength of each specimen from the sample is given below:

3273, 3229, 3256, 3272, 3201, 3247, 3267, 3237,

3286, 3210, 3265, 3273

- We will conduct the hypothesis whether the mean compressive strength of the concrete conforms to the claimed value.

Null hypothesis: μ = 3500 psi

Alternate hypothesis: μ ≠ 3500 psi

- The type of test performed on the sample data will depend on the application of Central Limit Theorem.

- The theorem states that the sample can be assumed to be normally distributed if drawn from a normally distributed population. ( We are given the population is normally distributed; hence, theorem applies )

- We will approximate the mean of the population ( μ ) with the sample mean ( x ), as per the implication specified by the theorem.

- The mean of the sample ( x ) is calculated as follows:

$x = \frac{Sum ( x_i )}{n} \\\\x = \frac{Sum ( 3273+ 3229+ 3256+ 3272+ 3201+ 3247+ 3267+ 3237+ 3286+ 3210+ 3265+3273 )}{12} \\\\x = \frac{39016}{12} \\\\x = 3251.3333$

- Since, we are testing the average compressive strength of a concrete against a claimed value. Any value that deviates significantly from the claimed value is rejected. This corroborates the use of one sample two tailed test.

- The test value may be evaluated from either z or t distribution. The conditions for z-test are given below:

The population variance is known OR sample size ( n ≥ 30 )

- The population variance is known; hence, we will use z-distribution to evaluate the testing value as follows:

$Z-test = \frac{x - u}{\sqrt{\frac{sigma^2}{n} } } \\\\Z-test = \frac{3251.333 - 3500}{\sqrt{\frac{1000^2}{12} } } \\\\Z-test = -27.24$

- The rejection region for the hypothesis is defined by the significance level ( α = 0.01 ). The Z-critical value ( limiting value for the rejection region ) is determined:

Z-critical = Z_α/2 = Z_0.005

- Use the list of correlation of significance level ( α ) and critical values of Z to determine:

Z-critical = Z_0.005 = ± 2.576

- Compare the Z-test value against the rejection region defined by the Z-critical value.

Rejection region: Z > 2.576 or Z < -2.576

- The Z-test value lies in the rejection region:

Z-test < Z-critical

-27.24 < -2.576 .... Null hypothesis rejected

Conclusion: The claim made by the civil engineer has little or no statistical evidence as per the sample data available; hence, the average compressive strength is not 3500 psi.

- To construct a confidence interval for the mean compressive strength ( μ ) we need to determine the margin of error for the population.

- The margin of error (ME) is defined by the following formula:

$ME = Z^*. \frac{sigma}{\sqrt{n} }$

Where,

- The ( Z* ) is the critical value for the defined confidence level ( CI ):

- The confidence interval and significance level are related and critical value Z* is as such:

α = 1 - CI , Z* = Z_α/2

- The critical values for ( CI = 99% & 95% ) are evaluated:

α = 1 - 0.99 = 0.01 , α = 1 - 0.95 = 0.05

Z* = Z_0.005 , Z* = Z_0.025

Z* = ± 2.58 , Z* = ± 1.96

- The formulation of Confidence interval is given by the following inequality:

[ x - ME < μ < x + ME ]

[ x - Z*√σ^2 / n < μ < x + Z*√σ^2 / n ]

- The CI of 95% yields:

[ 3251.33 - 1.96*√(1000 / 12) < μ < 3251.33 + 1.96*√(1000 / 12) ]

[ 3251.333 - 17.89227 < μ < 3251.33 + 17.89227 ]

[ 3233.44 < μ < 3269.23 ]

- The CI of 99% yields:

[ 3251.33 - 2.58*√(1000 / 12) < μ < 3251.33 + 2.58*√(1000 / 12) ]

[ 3251.333 - 23.552 < μ < 3251.33 + 23.552 ]

[ 3227.78 < μ < 3274.88 ]

- We see that the width of the confidence interval increases as the confidence level ( CI ) increases. This is due to the increase in critical value ( Z* ) associated with the significance level ( α ) increases.

7 0

3 years ago

Some connecting rods have ____ to help lubricate the cylinder wall or piston pin.

Ket [755]

Answer:

some connecting rods have spit holes

4 0

4 years ago

Read 2 more answers

Describe the "Bauschinger Effect" on the stress strain behaviour of steel

Rom4ik [11]

Answer:

The Bauschinger effect is the directionally determined change in the elasticity limit of a metal or alloy after primary plastic deformation. If you first deform a metal in one direction so that it is plastically deformed and then deform it in the opposite direction, the elastic limit in the opposite direction is lower.

From the point of view of the dislocation mechanism of plastic deformation, the Bauschinger effect indicates that during repeated deformation, it is easier for dislocations blocked in front of obstacles to move in the direction opposite to their movement during preliminary plastic deformation.

7 0

4 years ago