1. If Maggie gives her cat an unfair advantage, her experimental results will be biased.
Explanation:
Maggie using her cat in the experiment to test for intelligence gives the cat an unfair advantage, her experimental results will be biased.
- Due to her emotional attachment with the cat, the experimental results will be skewed to portraying her cat as intelligence.
- This is not a good experiment to carry out.
- Such an experiment should be carried out with an unknown cat.
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Answer:
a)1815Joules b) 185Joules
Explanation:
Hooke's law states that the extension of a material is directly proportional to the applied force provided that the elastic limit is not exceeded. Mathematically;
F = ke where;
F is the applied force
k is the elastic constant
e is the extension of the material
From the formula, k = F/e
F1/e1 = F2/e2
If a force of 60N causes an extension of 0.5m of the string from its equilibrium position, the elastic constant of the spring will be ;
k = 60/0.5
k = 120N/m
a) To get the work done in stretching the spring 5.5m from its position,
Work done by the spring = 1/2ke²
Given k = 120N/m, e = 5.5m
Work done = 1/2×120×5.5²
Work done = 60× 5.5²
Work done = 1815Joules
b) work done in compressing the spring 1.5m from its equilibrium position will be gotten using the same formula;
Work done = 1/2ke²
Work done =1/2× 120×1.5²
Works done = 60×1.5²
Work done = 135Joules
Answer:
A. Add mass to the Sun.
Explanation:
A. Add mass to the Sun.
Adding mass will make to take more time for the hydrogen to run out and hence, enough temperature will be developed to fuse helium atom into other heavier elements, and eventually get hot enough to fuse the helium in their cores into carbon.
The only hypothetical solution is that we need to add Mass to the Sun.
To solve the problem it is necessary to take into account the concepts related to energy efficiency in the engines, the work done, the heat input in the systems, the exchange and loss of heat in the soupy the radius between the work done the lost heat ( efficiency).
By definition the efficiency of the heat engine is
Where,
Temperature at the room
Temperature of the soup
The work done is defined as,
Where 
represents the input heat and at the same time is defined as
Where,
Specific Heat
The change at the work would be defined then as
On the other hand we have that the heat lost by the soup is equal to
The ratio between both would be,
Replacing with our values we have,
Therefore the fraction of heat lost by the soup that can be turned into useable work by the engine is 0.0613.
Amplitude, is the answer to the question
