A space-walking astronaut has become detached from her spaceship. She's floating in space with her handy tool belt attached to h
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Answer:
Thomson's atomic model was successful in explaining the overall neutrality of the atom. However, its propositions were not consistent with the results of later experiments. In 1906, J. J. Thomson was awarded the Nobel Prize in physics for his theories and experiments on electricity conduction by gases.
Summary. J.J. Thomson's experiments with cathode ray tubes showed that all atoms contain tiny negatively charged subatomic particles or electrons. Thomson proposed the plum pudding model of the atom, which had negatively-charged electrons embedded within a positively-charged "soup."
Answer:
1.28 m
Explanation:
As shown in the diagram attached,
According to the principle of moment,
For a body at equilibrium,
Sum of clockwise moment = sum of anticlockwise moment.
Taking moment about the pivot,
W₁(1.6)+W(0.133) = W₂(x)............... Equation 1
Where W₁ = Weight of the first child, Wₓ = Weight of the seesaw, W₂ = weight of the second child, x = distance of the second child from the pivot.
But,
W = mg
Where g = 9.8 m/s², m = mass of the body
Therefore,
W₁ = 26×9.8 = 254.8 N,
Wₓ = 18×9.8 = 176.4 N
W₂ = 34.4×9.8 = 337.12 N
Substitute these values into equation 1
(254.8×1.6)+(176.4×0.133) = 337.12(x)
407.68+23.4612 = 337.12x
337.12x = 431.1412
x = 431.1412/337.12
x = 1.2789
x ≈ 1.28 m
To solve this problem we will use the work theorem, for which we have that the Force applied on the object multiplied by the distance traveled by it, is equivalent to the total work. From the measurements obtained we have that the width and the top are 14ft and 7ft respectively. In turn, the bottom of the tank is 15ft. Although the weight of the liquid is not given we will assume this value of (Whose variable will remain modifiable until the end of the equations subsequently presented to facilitate the change of this, in case be different). Now the general expression for the integral of work would be given as
Basically under this expression we are making it difficult for the weight of the liquid multiplied by the area (Top and widht) under the integral of the liquid path to be equivalent to the total work done, then replacing
Therefore the total work in the system is