Answer:
F = 75[J]
Explanation:
We know that work is defined as the product of force by distance.
In this way we have two forces, the weight of the block down, and the force that bring about the block to rise.

where:
W = work = 50 [J]
d = distance = 2 [m]
Fweight = 50 [N]
Fupward [N]
Now replacing:
![50=-(50*2)+(F_{upward}*2)\\50+100=F_{upward}*2\\F_{upward}=150/2\\F_{upward}=75[J]](https://tex.z-dn.net/?f=50%3D-%2850%2A2%29%2B%28F_%7Bupward%7D%2A2%29%5C%5C50%2B100%3DF_%7Bupward%7D%2A2%5C%5CF_%7Bupward%7D%3D150%2F2%5C%5CF_%7Bupward%7D%3D75%5BJ%5D)
Answer:
Explanation:
Given





R for Helium 

mass of gas 


Similarly
can be found


Work done 


Since it is a polytropic Process
therefore 






From Energy balance
Neglecting kinetic and Potential Energy change

Change in Internal Energy 




i.e. Heat is being removed
On isolated ground receptacles, the metal yoke is not allowed to be integrally bonded to the equipment grounding terminal of the receptacle.
Any device with two distinct switches or receptacles is a duplex device. It can be shaped to fit a Decora opening or a typical duplex plate opening. It should be noted that they can be combination devices with a switch/outlet, switch/pilot light, etc.
Because of grounding connection removal and receptacle, it is utterly undesirable to connect the two bare equipment grounding conductors together directly.
The equipment grounding conductor associated with those circuits must be connected to the box when circuit conductors are terminated on equipment inside a metal box to prevent unneeded current discharge.
Learn more about grounding conductors here brainly.com/question/14886979
#SPJ4.
Answer:
D
Explanation:
because two vectors which align in the same line adds one to another