Efficiency is the minimum use of energy to accomplish the task. The wasted energy will be 375 J when 750 J of energy is given.
<h3>What is wasted energy?</h3>
Wasted energy is energy that is not useful when the transformation in the system occurs.
Total energy = 750 J
The efficiency of the system = 50 %
Output work (OW) is calculated as:
Efficiency = output work ÷ input work × 100%
750 × 50 = 100 OW
OW = 375 J
Wasted energy = Total energy - output work
= 750 - 375
= 375 J
Therefore, the machine is 50 % inefficient and has wasted energy of 375 J.
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Answer:
The hardenability increases with increasing austenite grain size, because the grain boundary area is decreasing. This means that the sites for the nucleation of ferrite and pearlite are being reduced in number, with the result that these transformations are slowed down, and the hardenability is therefore increased.
Answer:
Below see details
Explanation:
A) It is attached. Please see the picture
B) First to calculate the overall mean,
μ=65∗25/75+80∗25/75+95∗25/75
μ=65∗25/75+80∗25/75+95∗25/75 = 80
Next to calculate E(MSTR) = σ2+(1/r−1) ∑ni(μi−μ)^2 = 5634
And E(MSE) = σ^2= 9
C) Yes, it is substantially large than E(MSE) in this case.
D) If we sampled 25 employees from each group, we are likely to get a F statistics to indicate differences of job satisfactions among three types of length of service of employees.
Answer:
The velocity of flow is 10.0 m/s.
Explanation:
We shall use Manning's equation to calculate the velocity of flow
Velocity of flow by manning's equation is given by
![V=\frac{1}{n}R^{2/3}S^{1/2}](https://tex.z-dn.net/?f=V%3D%5Cfrac%7B1%7D%7Bn%7DR%5E%7B2%2F3%7DS%5E%7B1%2F2%7D)
where
n = manning's roughness coefficient
R = hydraulic radius
S = bed slope of the channel
We know that for an asphalt channel value of manning's roughness coefficient = 0.016
Applying values in the above equation we obtain velocity of flow as
![V=\frac{1}{0.016}\times 3.404^{2/3}\times 0.005^{1/2}\\\\\therefore V=10.000m/s](https://tex.z-dn.net/?f=V%3D%5Cfrac%7B1%7D%7B0.016%7D%5Ctimes%203.404%5E%7B2%2F3%7D%5Ctimes%200.005%5E%7B1%2F2%7D%5C%5C%5C%5C%5Ctherefore%20V%3D10.000m%2Fs)
Answer:
b) Commonly used to transmit network signals over great distances.
Explanation:
The transmission of information or data by using microwave radio waves is known as microwave transmission. Microwave transmitter is commonly used to transmit network signals over great distances. It is an electronic device that transmits and receives radio frequency signals ranging from 1GHz to 100GHz.
The microwave transmitter has a wide range of applications and these includes, radio stations, television stations, mobile phones, radio astronomy, radar,