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elena55 [62]
3 years ago
15

Scott is making a chart to show the difference between chemical reactions and nuclear reactions. Which part of his chart is inco

rrect? Chemical Reaction Nuclear Reaction Line 1 Happens in the nucleus of atoms Happens between atoms Line 2 Forms new compounds Forms new atoms and radioactive particles Line 3 Involves small amounts of energy Involves large amounts of energy
Physics
1 answer:
Nimfa-mama [501]3 years ago
6 0

Answer:

The incorrect part of the chart is;

Line 1

Chemical {} Reaction   {}                                   Nuclear Reaction

Line 1 Happens in the nucleus of atoms   {} Happens between atoms

Explanation:

The given chart can be presented as follows;

  Chemical Reaction {}                               Nuclear Reaction

1. Happens in the nucleus of atoms    {}   Happens Between atoms

2. Forms new compounds {}                     Forms new atoms

3. Involves small amounts of energy {}    Involves large amounts of energy

From the above chart, line 1 is incorrect because a chemical reaction happens between the valence electrons of the reacting atoms while a nuclear reaction is the reaction that takes place between the nucleus of two or more atoms or the nucleus of an atom and a subatomic particle to produce one or more than new nuclide of a new atom with a change in the mass of the nucleus being represented by the large amount of energy in the reaction. Therefore, a nuclear reaction takes place in the nucleus of an atom, forms new atoms and involves large amounts of energy

The correct arrangement is presented as follows;

  Chemical Reaction {}                              Nuclear Reaction

1. Happens Between atoms    {}                Happens in the nucleus of atoms

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Answer:

\boxed{\sf Time \ in \ which \ train \ will \ come \ to \ rest = 20 \ sec}

Given:

Initial velocity (u) = 30 m/s

Final speed (v) = 0 m/s

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To Find:

Time in which train will come to rest (t).

Explanation:

\sf From \ equation \ of \ motion: \\ \sf \implies \bold{v = u + at} \\ \\ \sf Substituting \ value \ of \ v, \ u \ and \ a:  \\  \sf \implies 0 = 30 + ( - 1.5)(t) \\   \sf  \implies 0 = 30 - 1.5(t) \\  \sf \implies 30 - 1.5(t) = 0 \\  \\  \sf Subtract  \: 30  \: from  \: both  \: sides: \\  \sf \implies (30 -  \boxed{ \sf 30}) - 1.5(t) =  \boxed{ \sf  - 30} \\  \\  \sf 30 - 30 = 0 :  \\  \sf \implies  - 1.5(t) =  - 30 \\  \\  \sf Divide  \: both  \: sides \:  of \:  - 1.5(t) =  - 30 \: by \:  - 1.5 :  \\  \sf \implies  \frac{  - 1.5(t)}{ \boxed{ \sf - 1.5}}  =  \frac{ - 30}{ \boxed{ \sf -1.5 }}  \\  \\  \sf \frac{ \cancel{ \sf 1.5}}{\cancel{ \sf 1.5}}  = 1 :  \\  \sf \implies t =  \frac{ - 30}{ - 1.5}  \\  \\   \sf  \frac{ - 30}{ - 1.5}  =  \frac{\cancel{ \sf 1.5} \times 20}{\cancel{ \sf 1.5}}  = 20 :  \\  \sf  \implies t = 20 \: sec

So,

Time in which train will come to rest = 20 seconds

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Explanation:

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Length = 3.50 m

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Using conservation of energy

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\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2+U_{f}

Final potential energy is zero

So, \dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2

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v_{2}=10.57\ m/s

The speed of skier is 10.57 m/s.

Hence,  (A).The work done by friction in crossing the patch is -637.98 J.

(B).The speed of skier is 10.57 m/s.

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