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Dennis_Churaev [7]
3 years ago
14

A 144-g baseball is dropped from a tree 11.0m above the ground.A.With what speed would it hit the ground if air resistance could

be ignored?Express your answer to three significant figures and include the appropriate units.B.If it actually hits the ground with a speed of 7.50m/s , what is the magnitude of the average force of air resistance exerted on it?Express your answer to three significant figures and include the appropriate units.
Physics
1 answer:
nikitadnepr [17]3 years ago
5 0

Answer:

a. 14.68m/s

b. 1.043N

Explanation:

a.

Given

Mass = m = 144g = 0.144kg

Height = h = 11m

Using energy conservation,

½mv² = mgh where g = 9.8m/s² ----- divide both sides by m

½v² = gh ---- divide both sides by ½

v² = 2gh

v = √2gh

v = √(2 * 9.8 * 11)

v = 14.68332387438212m/s

v = 14.68m/s ----- Approximated

b.

v = 7.5m/s

By definition, “the average force of air resistance,”

When work done against air resistance is solved by multiplying height by the average force of resistance ------ by definition

i.e. W = -Fh

Minus is used because F and h points to different direction (when viewed as vectors).

This is proved by:

W = Fhcos(180) because h points toward the direction of motion (downward) while F points opposite.

W = Fh * -1

W = Fh

Using energy conservation,

mgh - Fh = ½mv² ---- Make Fh, the subject of formula

Fh = mgh - ½mv² ---- Divide through by h

F = mg - mv²/2h

F = 0.144 * 9.8 - 0.144 * 7.5²/(2 * 11)

F = 1.043N

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Two speakers placed 0.94 m apart produce pure tones in sync with each other at a frequency of 1630 Hz. A microphone can be moved
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Answer:

a. approximately 1.1\; \rm m (first minimum.)

b. approximately 2.2\; \rm m (first maximum.)

c. approximately 3.4\; \rm m (second minimum.)

d. approximately 4.7\; \rm m (second maximum.)

Explanation:

Let d represent the separation between the two speakers (the two "slits" based on the assumptions.)

Let \theta represent the angle between:

  • the line joining the microphone and the center of the two speakers, and
  • the line that goes through the center of the two speakers that is also normal to the line joining the two speakers.

The distance between the microphone and point P_0 would thus be 9.4\, \tan(\theta) meters.

Based on the assumptions and the equation from Young's double-slit experiment:

\displaystyle \sin(\theta) = \frac{\text{path difference}}{d}.

Hence:

\displaystyle \theta = \arcsin \left(\frac{\text{path difference}}{d}\right).

The "path difference" in these two equations refers to the difference between the distances between the microphone and each of the two speakers. Let \lambda denote the wavelength of this wave.

\displaystyle \begin{array}{c|c} & \text{Path difference} \\ \cline{1-2}\text{First Minimum} & \lambda / 2 \\ \cline{1-2} \text{First Maximum} & \lambda \\\cline{1-2} \text{Second Minimum} & 3\,\lambda / 2 \\ \cline{1-2} \text{Second Maximum} & 2\, \lambda\end{array}.

Calculate the wavelength of this wave based on its frequency and its velocity:

\displaystyle \lambda = \frac{v}{f} \approx 0.211\; \rm m.

Calculate \theta for each of these path differences:

\displaystyle \begin{array}{c|c|c} & \text{Path difference} & \text{approximate of $\theta$} \\ \cline{1-3}\text{First Minimum} & \lambda / 2 & 0.112 \\ \cline{1-3} \text{First Maximum} & \lambda & 0.226\\\cline{1-3} \text{Second Minimum} & 3\,\lambda / 2 & 0.343\\ \cline{1-3} \text{Second Maximum} & 2\, \lambda & 0.466\end{array}.

In each of these case, the distance between the microphone and P_0 would be 9.4\, \tan(\theta). Therefore:

  • At the first minimum, the distance from P_0 is approximately 1.1\; \rm m.
  • At the first maximum, the distance from P_0 is approximately 2.2\; \rm m.
  • At the second minimum, the distance from P_0 is approximately 3.4\; \rm m.
  • At the second maximum, the distance from P_0 is approximately 4.7\;\rm m.
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