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aivan3 [116]
3 years ago
7

Energy cannot be created nor destroyed. A. Ella said "a lot of energy in the hot water has disappeared." Explain why her stateme

nt is wrong?
Physics
1 answer:
harkovskaia [24]3 years ago
6 0

Answer:

See Explanation

Explanation:

The principle of conservation of energy states that; energy can neither be created nor destroyed but is converted from one form to another.

In view of this principle, Ella can not be correct when she says that a lot of energy has disappeared. The use of the term "disappeared" connotes the idea that the energy no longer exists which does not happen.

Hence, energy can not "disappear" from hot water rather the energy in the water may be transferred to the surroundings.

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A voltaic cell with an aqueous electrolyte is based on the reaction between Cd2+(aq) and Mg(s), producing Cd(s) and Mg2+(aq). Wr
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Answer : The balanced two-half reactions will be,

Oxidation half reaction (anode) : Mg(s)\rightarrow Mg^{2+}(aq)+2e^-

Reduction half reaction (cathode) : Cd^{2+}(aq)+2e^-\rightarrow Cd(s)

Thus the overall reaction will be,

Mg(s)+Cd^{2+}(aq)\rightarrow Mg^{2+}(aq)+Cd(s)

Explanation :

Voltaic cell : It is defined as a device which is used for the conversion of the chemical energy produces in a redox reaction into the electrical energy. It is also known as the galvanic cell or electrochemical cell.

The given redox reaction occurs between the magnesium and cadmium.

In the voltaic cell, the oxidation occurs at an anode which is a negative electrode and the reduction occurs at the cathode which is a positive electrode.

The balanced two-half reactions will be,

Oxidation half reaction (anode) : Mg(s)\rightarrow Mg^{2+}(aq)+2e^-

Reduction half reaction (cathode) : Cd^{2+}(aq)+2e^-\rightarrow Cd(s)

Thus the overall reaction will be,

Mg(s)+Cd^{2+}(aq)\rightarrow Mg^{2+}(aq)+Cd(s)

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3 years ago
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(7)Figure 4 shows three charges: Q₁, Q₂ and Q3 . Determine the net force (Fnet) acting on Q3. (Hint: Draw a free body diagram of
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Remember Coulomb's law: the magnitude of the electric force F between two stationary charges q₁ and q₂ over a distance r is

F = \dfrac{kq_1q_2}{r^2}

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8.1. The diagram is simple, since only two forces are involved. The particle at Q₂ feels a force to the left due to the particle at Q₁ and a downward force due to the particle at Q₃.

8.2. First convert everything to base SI units:

0,02 µC = 0,02 × 10⁻⁶ C = 2 × 10⁻⁸ C

0,03 µC = 3 × 10⁻⁸ C

0,04 µC = 4 × 10⁻⁸ C

300 mm = 300 × 10⁻³ m = 0,3 m

600 mm = 0,6 m

Force due to Q₁ :

F_{Q_2/Q_1} = \dfrac{k (6 \times 10^{-16} \,\mathrm C)}{(0,3 \, \mathrm m)^2} \approx \boxed{6,0 \times 10^{-5} \,\mathrm N} = 0,06 \,\mathrm{mN}

Force due to Q₃ :

F_{Q_2/Q_3} = \dfrac{k (12 \times 10^{-16} \,\mathrm C)}{(0,6 \, \mathrm m)^2} \approx \boxed{3,0 \times 10^{-5} \,\mathrm N} = 0,03 \,\mathrm{mN}

8.3. The net force on the particle at Q₂ is the vector

\vec F = F_{Q_2/Q_1} \, \vec\imath + F_{Q_2/Q_3} \,\vec\jmath = \left(-0,06\,\vec\imath - 0,03\,\vec\jmath\right) \,\mathrm{mN}

Its magnitude is

\|\vec F\| = \sqrt{\left(-0,06\,\mathrm{mN}\right)^2 + \left(-0,03\,\mathrm{mN}\right)^2} \approx 0,07 \,\mathrm{mN} = \boxed{7,0 \times 10^{-5} \,\mathrm N}

and makes an angle θ with the positive horizontal axis (pointing to the right) such that

\tan(\theta) = \dfrac{-0,03}{-0,06} \implies \theta = \tan^{-1}\left(\dfrac12\right) - 180^\circ \approx \boxed{-153^\circ}

where we subtract 180° because \vec F terminates in the third quadrant, but the inverse tangent function can only return angles between -90° and 90°. We use the fact that tan(x) has a period of 180° to get the angle that ends in the right quadrant.

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