The slope of an object is given relative to an origin.
Answer: A, C, E
Explanation:
Gamma rays, Microwaves, and Radio waves
A sample of an ideal gas is heated, and its kelvin temperature doubles. The average speed of the molecules in the sample will increases by a factor of
The root-mean square (RMS) velocity is the value of the square root of the sum of the squares of the stacking velocity values divided by the number of values. The RMS velocity is that of a wave through sub-surface layers of different interval velocities along a specific ray path.
Root mean square speed is a statistical measurement of speed.
The root mean square speed can be calculated as : V1 : ![\sqrt{3 R T / Mo}](https://tex.z-dn.net/?f=%5Csqrt%7B3%20R%20T%20%2F%20Mo%7D)
if temperature becomes double
let T1 is initial temperature
So , T2 = 2 * T1
now ,
Root mean square speed will be (V2) = ![\sqrt{(3 R (2T)) / Mo}](https://tex.z-dn.net/?f=%5Csqrt%7B%283%20R%20%282T%29%29%20%2F%20Mo%7D)
=
* ![\sqrt{3 R T / Mo}](https://tex.z-dn.net/?f=%5Csqrt%7B3%20R%20T%20%2F%20Mo%7D)
=
V1
Thus when temperature becomes double, the root mean square speed increases by a factor of
To learn more about root mean square velocity here
brainly.com/question/13751940
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Complete Question
A real (non-Carnot) heat engine, operating between heat reservoirs at temperatures of 650 K and 270 K and performs 4.3 kJ of net work and rejects 8.00 kJ of heat in a single cycle. The thermal efficiency of this heat engine is closest to A) 0.35 B) 0.31. C) 0.28. D) 0.38. E) 0.42.
Answer:
The correct option is A
Explanation:
From the question we are told that
The first operating temperature is ![T_1 = 650 \ K](https://tex.z-dn.net/?f=T_1%20%3D%20%20650%20%5C%20K)
The second operating temperature is ![T_2 = 270 \ K](https://tex.z-dn.net/?f=T_2%20%3D%20%20270%20%5C%20K)
The net workdone is
( output of the engine )
The amount of heat energy rejected is ![H = 8.00 \ kJ = 8.00 *10^{3 } \ J](https://tex.z-dn.net/?f=H%20%20%3D%20%208.00%20%5C%20kJ%20%20%3D%20%208.00%20%2A10%5E%7B3%20%7D%20%5C%20%20J)
Generally a heat engine convert heat from a high temperature to mechanical energy and then reject the remaining heat so the absorbed by the engine is
![W + H](https://tex.z-dn.net/?f=W%20%2B%20H)
Generally the thermal efficiency is mathematically represented as
![\eta = \frac{out}{In} * 100](https://tex.z-dn.net/?f=%5Ceta%20%20%3D%20%20%5Cfrac%7Bout%7D%7BIn%7D%20%20%2A%20100)
Here out is the output of the engine
and in is the input of the engine
![\eta = \frac{W}{W + H} * 100](https://tex.z-dn.net/?f=%5Ceta%20%20%3D%20%20%5Cfrac%7BW%7D%7BW%20%2B%20H%7D%20%20%2A%20100)
=> ![\eta = \frac{4.3}{4.3 + 8} * 100](https://tex.z-dn.net/?f=%5Ceta%20%20%3D%20%20%5Cfrac%7B4.3%7D%7B4.3%20%2B%208%7D%20%20%2A%20100)
=> ![\eta = 0.35](https://tex.z-dn.net/?f=%5Ceta%20%20%3D%200.35)