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vichka [17]
3 years ago
5

What parts of an atom can change during a nuclear reaction that cannot change during a chemical reaction? Question 15 options: A

. The number of protons and neutrons B. The number of electrons only C. The number of protons and electrons D. The number of neutrons and electrons
Physics
2 answers:
krek1111 [17]3 years ago
8 0
<h3><u>Answer;</u></h3>

B. the number of protons and neutrons

<h3><u>Explanation</u>;</h3>
  • <em><u>Nuclear reactions involve a change in the nucleus of an atom, usually producing a different element. Chemical reactions, on the other hand, involve only a rearrangement of electrons and do not involve changes in the nuclei.</u></em>
  • Therefore the parts of an atom that can change during a nuclear reaction but can not change during a chemical reaction is <u><em>the number of protons and neutrons that are found in the nucleus.</em></u>
viva [34]3 years ago
6 0
Neutrons and protons cannot be removed from nucleus from chemical reactions because they are strongly held together but nuclear reactions are strong enough to separate them. Hence option A is correct.
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Is there air resistance in space?........HELP!!!!!
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Which of the following does NOT represent Newton’s second law? Question 20 options: a = m/Fnet m = Fnet/a Fnet = ma a = Fnet/m
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The train passes point A with a speed of 30 m/s and begins to decrease its speed at a constant rate of at = - 0.25 m/s^2. Determ
prisoha [69]

Explanation:

At point B, the velocity speed of the train is as follows.

          \nu^{2}_{B} = \nu^{2}_{A} + 2a_{t} (s_{B} - s_{A})

                           = (30)^{2} + 2(-0.25(412 - 0))

                           = 26.34 m/s

Now, we will calculate the first derivative of the equation of train.

          y = 200 e^{\frac{x}{1000}}

      \frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}

Now, second derivative of the train is calculated as follows.

         \frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}      

       \frac{d^{2}y}{dx^{2}} = 0.2 (10^{-3}) e^{\frac{x}{1000}}    

Radius of curvature of the train is as follows.  

   \rho = \frac{[1 + (\frac{dy}{dx})^{2}]^{\frac{3}{2}}}{\frac{d^{2}y}{dx^{2}}}

               = \frac{[1 + 0.2e^{\frac{400}{1000}}^{2}]^{\frac{3}{2}}}{0.2(10^{-3})e^{\frac{400}{1000}}}

              = 3808.96 m

Now, we will calculate the normal component of the train as follows.

            a_{n} = \frac{\nu^{2}_{B}}{\rho}

                        = \frac{(26.34)^{2}}{3808.96}

                        = 0.1822 m/s^{2}

The magnitude of acceleration of train is calculated as follows.

            a = \sqrt{(a_{t})^{2} + (a_{n})^{2}}

               = \sqrt{(-0.25)^{2} + (0.1822)^{2}}

              = 0.309 m/s^{2}

Thus, we can conclude that magnitude of the acceleration of the train when it reaches point B, where sAB = 412 m is 0.309 m/s^{2}.

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3 years ago
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