Answer: True
Explanation: Closed loop relies on feedback from PNS to make modifications in the movement, open loop allows action in the absence of feedback, 2. ... Closed loop can change the initial commands, open loop can not change the initial commands.
Explanation:
Step1
In the stress-strain curve of any material, the yield stress is the maximum stress at which material starts yielding.
Step2
Young’s modulus is the constant of proportionality of stress and strain according to hooks law. It is the slope of the slope of the stress-strain curve of the any material under proportional limit.
Step3
Ultimate tensile stress is the maximum stress that induced in the material under application of load.
Step4
Toughness is the strain energy per unit volume up to the fracture point of the stress-strain diagram of any material. This is the area under the curve of stress-strain.
Step5
Point of necking is the point where any material starts necking under application of load in necking region of the stress-strain curve.
Step6
Fracture point is the last point of the stress-strain curve where component fractures under application of load.
All the parameters are shown in below stress-strain curve:
Answer:
Q = 62 ( since we are instructed not to include the units in the answer)
Explanation:
Given that:
![n_{HCl} = 3 \ kmol\\n_{Ar} = 7 \ k mol](https://tex.z-dn.net/?f=n_%7BHCl%7D%20%3D%203%20%5C%20kmol%5C%5Cn_%7BAr%7D%20%3D%207%20%5C%20k%20mol)
![T_1 = 27^0 \ C = ( 27+273)K = 300 K](https://tex.z-dn.net/?f=T_1%20%3D%2027%5E0%20%5C%20C%20%3D%20%28%2027%2B273%29K%20%3D%20%20300%20K)
![P_1 = 200 \ kPa](https://tex.z-dn.net/?f=P_1%20%3D%20200%20%5C%20kPa)
Q = ???
Now the gas expands at constant pressure until its volume doubles
i.e if ![V_1 = x\\V_2 = 2V_1](https://tex.z-dn.net/?f=V_1%20%3D%20x%5C%5CV_2%20%3D%202V_1)
Using Charles Law; since pressure is constant
![V \alpha T](https://tex.z-dn.net/?f=V%20%5Calpha%20T)
![\frac{V_2}{V_1} =\frac{T_2}{T_1}](https://tex.z-dn.net/?f=%5Cfrac%7BV_2%7D%7BV_1%7D%20%20%3D%5Cfrac%7BT_2%7D%7BT_1%7D)
![\frac{2V_1}{V_1} =\frac{T_2}{300}](https://tex.z-dn.net/?f=%5Cfrac%7B2V_1%7D%7BV_1%7D%20%20%3D%5Cfrac%7BT_2%7D%7B300%7D)
![T_2 = 300*2\\T_2 = 600](https://tex.z-dn.net/?f=T_2%20%3D%20300%2A2%5C%5CT_2%20%3D%20600)
mass of He =number of moles of He × molecular weight of He
mass of He = 3 kg × 4
mass of He = 12 kg
mass of Ar =number of moles of Ar × molecular weight of Ar
mass of He = 7 kg × 40
mass of He = 280 kg
Now; the amount of Heat Q transferred = ![m_{He}Cp_{He} \delta T + m_{Ar}Cp_{Ar} \delta T](https://tex.z-dn.net/?f=m_%7BHe%7DCp_%7BHe%7D%20%5Cdelta%20T%20%20%2B%20m_%7BAr%7DCp_%7BAr%7D%20%5Cdelta%20T)
From gas table
![Cp_{He} = 5.9 \ kJ/Kg/K\\Cp_{Ar} = 0.5203 \ kJ/Kg/K](https://tex.z-dn.net/?f=Cp_%7BHe%7D%20%3D%205.9%20%5C%20kJ%2FKg%2FK%5C%5CCp_%7BAr%7D%20%20%3D%200.5203%20%5C%20%20kJ%2FKg%2FK)
∴ Q = ![12*5.19*10^3(600-300)+280*0.5203*10^3(600-300)](https://tex.z-dn.net/?f=12%2A5.19%2A10%5E3%28600-300%29%2B280%2A0.5203%2A10%5E3%28600-300%29)
Q = ![62.389 *10^6](https://tex.z-dn.net/?f=62.389%20%2A10%5E6)
Q = 62 MJ
Q = 62 ( since we are instructed not to include the units in the answer)
Answer:
d. low earth orbit (LEO)
Explanation:
This type of satellites form a constellation deployed as a series of “necklaces” in such a way that at any time, at least one satellite is visible by a receiver antenna, compensating the movement due to the earth rotation.
Opposite to that, a geostationary satellite is at an altitude that makes it like a fixed point over the Earth´s equator, rotating synchronously with the Earth, so it is always visible in a given area.