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bonufazy [111]
2 years ago
12

Summarize two of the most common arguments made by animal rights advocates to support the discontinuation of animals in research

.
Physics
2 answers:
olasank [31]2 years ago
6 0

Animals suffer from pain and fear. Humans owe animals an obligation. Animal rights supersede human intentions and interests. Many research labs violate ethical standards for the care and use of animals. Animals cannot be owned any more than people can be owned. Animal research is never conclusive because any successful animal-based treatment will have to be tested again on humans. The use of animals in research creates a market for unethical animal breeding.

Dvinal [7]2 years ago
3 0
Animal rights<span> is the idea that some, or all, </span>non-human animals<span> are entitled to the possession of their own lives and that their most basic interests—such as the need to avoid </span>suffering—should be afforded the same consideration as similar interests of human beings. <span>They maintain that animals should no longer be viewed as property or used as food, clothing, research subjects, entertainment, or beasts of burden.</span>
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andrey2020 [161]
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2 years ago
Explain why the motto "Do whatever it takes to win!" may not be an ethical guideline to follow.
Murrr4er [49]
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2 years ago
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Two objects are placed in thermal contact and are allowed to come to equilibrium in isolation. the heat capacity of object a is
Harman [31]
Given:
Ca = 3Cb                      (1)
where
Ca =  heat capacity of object A
Cb =  heat capacity f object B

Also,
Ta = 2Tb                     (2)
where
Ta = initial temperature of object A
Tb = initial temperature of object B.

Let
Tf =  final equilibrium temperature of both objects,
Ma = mass of object A,
Mb = mass of object B.

Assuming that all heat exchange occurs exclusively between the two objects, then energy balance requires that
Ma*Ca*(Ta - Tf) = Mb*Cb*(Tf - Tb)           (3)

Substitute (1) and (2) into (3).
Ma*(3Cb)*(2Tb - Tf) = Mb*Cb*(Tf - Tb)
3(Ma/Mb)*(2Tb - Tf) = Tf - Tb

Define k = Ma/Mb, the ratio f the masses.
Then
3k(2Tb - Tf) = Tf - Tb
Tf(1+3k) = Tb(1+6k)
Tf = [(1+6k)/(1+3k)]*Tb

Answer:
T_{f} =( \frac{1+6k}{1+3k} )T_{b}= \frac{1}{2}( \frac{1+6k}{1+3k})T_{a}
where
k= \frac{M_{a}}{M_{b}} 
7 0
2 years ago
This problem explores the behavior of charge on conductors. We take as an example a long conducting rod suspended by insulating
Olegator [25]

Answer:

rod end A is strongly attracted towards the balls

rod end B is weakly repelled by the ball as it is at a greater distance

Explanation:

When the ball with a negative charge approaches the A end of the neutral bar, the charge of the same sign will repel and as they move they move to the left end, leaving the rod with a positive charge at the A end and a negative charge of equal value at end B.

Therefore rod end A is strongly attracted towards the balls and

rod end B is weakly repelled by the ball as it is at a greater distance

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