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nata0808 [166]
3 years ago
12

Describe how you expect the waveform and the sound you hear changes when you hit the tuning fork harder.

Physics
1 answer:
myrzilka [38]3 years ago
6 0

Answer:

In a tuning fork, two basic qualities of sound are considered, they are

1) The pitch of the waveform: This pitch depends on the frequency of the wave generated by hitting the tuning fork.

2) The loudness of the waveform: This loudness depends on the intensity of the wave generated by hitting the tuning fork.

Hitting the tuning fork harder will make it vibrate faster, increasing the number of vibrations per second. The number of vibration per second is proportional to the frequency, so hitting the tuning fork harder increase the frequency. From the explanation on the frequency above, we can say that by increasing the frequency the pitch of the tuning fork also increases.

Also, hitting the tuning fork harder also increases the intensity of the wave generated, since the fork now vibrates faster. This increases the loudness of the tuning fork.

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A battery establishes a voltage V on a parallel-plate capacitor. After the battery is disconnected, the distance between the pla
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d.

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therefore voltage between the plates(increases ).

Hence, option d is correct.

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What would we do if we didn't have solar energy?
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We would have to use wind or another source

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A 1kg box is pushed on a flat surface that is 250m long. The box is initially at rest and then pushed with a constant Net force
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Answer:

C) 50 m/s

Explanation:

With the given information we can calculate the acceleration using the force and mass of the box.

Newton's 2nd Law: F = ma

  • 5 N = 1 kg * a
  • a = 5 m/s²

List out known variables:

  • v₀ = 0 m/s
  • a = 5 m/s²
  • v = ?
  • Δx = 250 m

Looking at the constant acceleration kinematic equations, we see that this one contains all four variables:

  • v² = v₀² + 2aΔx

Substitute known values into the equation and solve for v.

  • v² = (0)² + 2(5)(250)
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The final velocity of the box is C) 50 m/s.

7 0
3 years ago
b) The distance of the red supergiant Betelgeuse is approximately 427 light years. If it were to explode as a supernova, it woul
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Answer:

b) Betelgeuse would be \approx 1.43 \cdot 10^{6} times brighter than Sirius

c) Since Betelgeuse brightness from Earth compared to the Sun is \approx 1.37 \cdot 10^{-5} } the statement saying that it would be like a second Sun is incorrect

Explanation:

The start brightness is related to it luminosity thought the following equation:

B = \displaystyle{\frac{L}{4\pi d^2}} (1)

where B is the brightness, L is the star luminosity and d, the distance from the star to the point where the brightness is calculated (measured). Thus:

b) B_{Betelgeuse} = \displaystyle{\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2}} and B_{Sirius} = \displaystyle{\frac{26L_{Sun}}{4\pi (26\ ly)^2}} where L_{Sun} is the Sun luminosity (3.9 x 10^{26} W) but we don't need to know this value for solving the problem. ly is light years.

Finding the ratio between the two brightness we get:

\displaystyle{\frac{B_{Betelgeuse}}{B_{Sirius}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (26\ ly)^2}{26L_{Sun}} \approx 1.43 \cdot 10^{6} }

c) we can do the same as in b) but we need to know the distance from the Sun to the Earth, which is 1.581 \cdot 10^{-5}\ ly. Then

\displaystyle{\frac{B_{Betelgeuse}}{B_{Sun}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (1.581\cdot 10^{-5}\ ly)^2}{1\ L_{Sun}} \approx 1.37 \cdot 10^{-5} }

Notice that since the star luminosities are given with respect to the Sun luminosity we don't need to use any value a simple states the Sun luminosity as the unit, i.e 1. From this result, it is clear that when Betelgeuse explodes it won't be like having a second Sun, it brightness will be 5 orders of magnitude smaller that our Sun brightness.

4 0
3 years ago
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