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3 years ago

Which example best illustrates the transfer of energy between two waves?

Physics

1 answer:

FromTheMoon [43]3 years ago

7 0

Answer:

A) A buoy rises in the water as a boat speeds past.

Explanation:

The passing boat transfers energy in the form of a wave. Other options illustrate other physics concepts like gravity (falling egg) or Newton's law (for every action, there is an equal and opposite reaction).

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What do nebulae have the most in common with?

alexdok [17]

Most commonly the answer is A, the sun. A nebula is a nursery for stars. The Sun itself was created in a nebula. Also,you do not need college physics to understand this question

3 0

3 years ago

A parallel combination of a 1.13-μF capacitor and a 2.85-μF one is connected in series to a 4.25-μF capacitor. This three-capaci

Nata [24]

Answer:

(a) Charge of 4.25 μF capacitor is 35.46 μC.

(b) Charge of 1.13 μF capacitor is 10.05 μC.

(c) Charge of 2.85 μF capacitor is 25.36 μC.

Explanation:

Let C₁ , C₂ and C₃ are the capacitor which are connected to the battery having voltage V. According to the problem, C₁ and C₂ are connected in parallel. There equivalent capacitance is:

C₄ = C₁ + C₂

Substitute 1.13 μF for C₁ and 2.85 μF for C₂ in the above equation.

C₄ = ( 1.13 + 2.85 ) μF = 3.98 μF

Since, C₄ and C₃ are connected in series, there equivalent capacitance is:

C₅ = $\frac{C_{3}C_{4} }{C_{3} + C_{4} }$

Substitute 4.25 μF for C₃ and 3.98 μF for C₄ in the above equation.

C₅ = $\frac{4.25\times3.98 }{4.25 + 3.98 }$

C₅ = 2.05 μF

The charge on the equivalent capacitance is determine by the relation :

Q = C₅ V

Substitute 2.05 μF for C₅ and 17.3 volts for V in the above equation.

Q = 2.05 μF x 17.3 = 35.46 μC

Since, the capacitors C₃ and C₄ are connected in series, so the charge on these capacitors are equal to the charge on the equivalent capacitor C₅.

Charge on the capacitor, C₃ = 35.46 μC

Charge on the capacitor, C₄ = 35.46 μC

Voltage on the capacitor C₄ = $\frac{Q}{C_{4} }$ = $\frac{35.46\times10^{-6} }{3.98\times10^{-6}}$ = 8.90 volts

Since, C₁ and C₂ are connected in parallel, the voltage drop on both the capacitors are same, that is equal to 8.90 volts.

Charge on the capacitor, C₁ = C₁ V = 1.13 μF x 8.90 = 10.05 μC

Charge on the capacitor, C₂ = C₂ V = 2.85 μF x 8.90 = 25.36 μC

5 0

3 years ago

A very small object with mass 8.30×10-9 kg and positive charge 6.90×10-9 C is projected directly toward a very large insulating

Rainbow [258]

Answer:

$41.4496148484\ m/s$

Explanation:

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

$\sigma$ = Surface charge density = $5.9\times 10^{-8}\ C/m^2$

$\Delta x$ = 0.57-0.26

q = Charge = $6.9\times 10^{-9}\ C$

m = Mass of object = $8.3\times 10^{-9}\ kg$

Electric field due to a sheet is given by

$E=\dfrac{\sigma}{2\epsilon_0}\\\Rightarrow E=\dfrac{5.9\times 10^{-8}}{2\times 8.85\times 10^{-12}}\\\Rightarrow E=3333.33\ V/m$

Electric field is given by

$E=\dfrac{V}{d}$

Voltage is given by

$V=E\Delta x$

Kinetic energy is given by

$K=qV$

$\dfrac{1}{2}mv^2=qE\Delta x\\\Rightarrow v=\sqrt{\dfrac{2qE\Delta x}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 6.9\times 10^{-9}\times 3333.33\times (0.57-0.26)}{8.3\times 10^{-9}}}\\\Rightarrow v=41.4496148484\ m/s$