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padilas [110]
4 years ago
10

swimmers at a water park have a choice of two frictionless water slides as shown in the figure. although both slides drop over t

he same height, h, slide 1 is straight while slide 2 is curved, dropping quickly at first and then leveling out. how does the speed v1 of a swimmer reaching the end of slide 1 compares with v2, the speed of a swimmer reaching the end of slide 2?
Physics
1 answer:
LuckyWell [14K]4 years ago
4 0
If swimmers had a choice of the water slides shown in this figure,
they would all go home dry, since there is no figure.  I'll have to try to
answer this question based on only the words in the text, augmented
only by my training, education, life experience, and human logic.

-- Both slides are frictionless.  So no energy is lost as a swimsuit
scrapes along the track, and the swimmer's kinetic energy at the
bottom is equal to the potential energy he had at the top.

-- Both slides start from the same height.  So the same swimmer
has the same potential energy at the top of either one, and therefore
the same kinetic energy at the bottom of either one.

-- So the difference in the speeds of two different swimmers
on the slides depends only on the difference in the swimmers'
mass, and is not influenced by the shape or length of the slides
(as long as the slides remain frictionless).

If both swimmers have the same mass, then  v₁ = v₂ .
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The nuclear equation is incomplete. Superscript 239 Subscript 94 Baseline P u + Superscript 1 Subscript 0 Baseline n yields Supe
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Answer:

The particle which completes the given equation  is :_{54}^{138}\textrm{Xe}

Explanation:

The given reaction is of a fission reaction:

_{94}^{239}\textrm{Pu}+_0^1\textrm{n}\rightarrow _{40}^{100}\textrm{Zr}+_Z^A\textrm{X}+2_0^1\textrm{n}

Total mass on the reactant side is equal to the total mass on the product side:

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Sum of atomic numbers on the reactant side is equal to the sum of atomic number on the product side:

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Z = 54

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The particle which completes the given equation  is :

_{94}^{239}\textrm{Pu}+_0^1\textrm{n}\rightarrow _{40}^{100}\textrm{Zr}+_{54}^{138}\textrm{Xe}+2_0^1\textrm{n}

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