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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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Dmitry_Shevchenko [17]

Answer:

Doubling the voltage in this arrangement both doubles the voltage drop across the resistor and the current through it. The bulb will be much brighter.

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2 years ago
Three equal charge 1.8*10^-8 each are located at the corner of an equilateral triangle ABC side 10cm.calculate the electric pote
Arlecino [84]

Answer:

If all these three charges are positive with a magnitude of 1.8 \times 10^{-8}\; \rm C each, the electric potential at the midpoint of segment \rm AB would be approximately 8.3 \times 10^{3}\; \rm V.

Explanation:

Convert the unit of the length of each side of this triangle to meters: 10\; \rm cm = 0.10\; \rm m.

Distance between the midpoint of \rm AB and each of the three charges:

  • d({\rm A}) = 0.050\; \rm m.
  • d({\rm B}) = 0.050\; \rm m.
  • d({\rm C}) = \sqrt{3} \times (0.050\; \rm m).

Let k denote Coulomb's constant (k \approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2}.)

Electric potential due to the charge at \rm A: \displaystyle \frac{k\, q}{d({\rm A})}.

Electric potential due to the charge at \rm B: \displaystyle \frac{k\, q}{d({\rm B})}.

Electric potential due to the charge at \rm A: \displaystyle \frac{k\, q}{d({\rm C})}.

While forces are vectors, electric potentials are scalars. When more than one electric fields are superposed over one another, the resultant electric potential at some point would be the scalar sum of the electric potential at that position due to each of these fields.

Hence, the electric field at the midpoint of \rm AB due to all these three charges  would be:

\begin{aligned}& \frac{k\, q}{d({\rm A})} + \frac{k\, q}{d({\rm B})} + \frac{k\, q}{d({\rm C})} \\ &= k\, \left(\frac{q}{d({\rm A})} + \frac{q}{d({\rm B})} + \frac{q}{d({\rm C})}\right) \\ &\approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2} \\ & \quad \quad \times \left(\frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{\sqrt{3} \times (0.050\; \rm m)}\right) \\ &\approx 8.3 \times 10^{3}\; \rm V\end{aligned}.

4 0
3 years ago
During which phase of the moon do neap tides occur?
Fynjy0 [20]

Answer:

First Quarter and Third Quarter.

Explanation:

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Since gravity variates with the distance:

F = G\frac{m1\cdot m2}{r^{2}} (1)

Where m1 and m2 are the masses of the two objects that are interacting and r is the distance between them.

For example, seeing the image below, point A is closer to the Moon than point b, and at the same time the center of mass of the Earth will feel more attracted to the Moon than point B. Therefore, that creates a tidal bulge in point A and point B.

When the Sun and the Moon are alight with respect to the Earth, then the Sun tidal force contributes to the tidal force of the Moon over the Earth. That makes the high tides even higher (spring tides).

               

However, when the Sun is not in the same line than the Moon (the Moon is at 90° with respect to the Sun), then the low tides are higher and the high tides are lower. That scenario is known as neap tides.

           

Therefore, that happens when the Moon is at First Quarter and Third Quarter.

4 0
3 years ago
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vfiekz [6]

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Diameter of the rod d=2.54\ cm

length of rod is l=20\ cm

Resistivity of silicon is \rho=6.4\times 10^2\ \Omega-m

cross-section of the rod A

\Rightarrow A=\dfrac{\pi d^2}{4}\\\\\Rightarrow A=\dfrac{3.142\times 2.54^2\times 10^{-4}}{4}\\\\\Rightarrow A=5.067\times 10^{-4}\ m^2

Resistance of rod is  R

\Rightarrow R=\dfrac{\rho l}{A}

\Rightarrow R=\dfrac{640\times 0.20}{5.067\times 10^{-4}}\\\\\Rightarrow R=25.26\times 10^4\ \Omega

Current is given by

\Rightarrow I=\dfrac{V}{R}\\\\\Rightarrow I=\dfrac{1000}{25.26\times 10^4}\\\\\Rightarrow I=0.0039\ A

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