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kirza4 [7]
3 years ago
5

On a far-away (hypothetical) planet, a student is conducting an experiment to study vertical motion. The student fires a cannon

directly upward, and then records the position of the cannonball every second until it falls back to the ground. Their data is shown in the table below. Based on their data below, answer the following two questions:
1) What is the acceleration due to gravity on this planet? Clearly show how you calculated this.

2) Is there significant air resistance affecting the experiment? Write an evidence-based argument to justify your claim.

- One of the pieces of evidence that you use must be a graph based on the given data. The quantities on the graph are up to you; they can be position and time themselves, or different quantities that you calculate from the given position and time data.

- If you'd like, you can also include outside evidence from your own lab or research. This is not required; only include it if it makes your argument stronger.

Answer both questions on one document. The document can be digital, hand-written, or a mix of both (copy and paste images of your handwritten work into a doc).

Time (s) Position (m)
0 0
1 95
2 170
3 225
4 260
5 275
6 270
7 245
8 200
9 135
10 50
Physics
1 answer:
IgorC [24]3 years ago
3 0
Yeah it’s pretty much positive
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Imagine an alternate universe where the value of the Planck constant is . In that universe, which of the following objects would
HACTEHA [7]

Question: The planck constant was not given. In this calculation, planck constant of 6.62607*10^-9 Js  is used for the calculation.

Answer:

(a) A virus -------------Classical

(b) A buckyball -----Classical

(c) A mosquito ------ Quantum

(d) A turtle  ------------Quantum

Explanation:

 Calculating the wavelength using the formula;

λ= h/(mv)

where

λ= Wavelength

h = Planck Constant = 6.62607*10^-9 Js

m = mass in kg

v = velocity in m/s

Virus size = 280. nm = 2.80*10⁻⁷ m

a)

A Virus:

m = 9.4 x 10-17 g 9.4*10⁻²⁰ kg

v = 0.50 µm/s = 5 *10⁻⁷ m/s

h = 6.62607*10^-9 Js

Virus size = 280 nm = 2.80*10⁻⁷ m

Substituting into the formula; we have

λ= h/(mv)

λ= 6.62607*10^-9/ (9.4*10⁻²⁰* 5 *10⁻⁷)

  = 6.62607*10^-9/4.7*10^-26

  = 1.4*10^17 m

Classical : Wavelength is bigger than it's size

(b)

A buckyball

m = 1.2 x 10-21 g = 1.2 *10⁻²⁴ kg

V = 37 m/s

Size = 0.7 nm = 7*10⁻¹⁰ m

Substituting into the formula, we have

λ= h/(mv)

λ= 6.62607*10^-9/ ( 1.2 *10⁻²⁴* 37)

  =  6.62607*10^-9/4.44*10^-23

  = 1.49 *10^14 m

Classical : Wavelength is bigger than it's size

(c)

A mosquito

Mass = 1.0 mg = 1*10⁻⁶ kg

v = 1.1 m/s

Size =  6.3 mm = 6.3*10⁻³ m

Substituting into the formula, we have

λ= h/(mv)

λ= 6.62607*10^-9/ (  1*10⁻⁶* 1.1)

  =  6.62607*10^-9/1.1*10^-6

  = 6.02*10^-3 m

Quantum Approach: The wavelength and the size are comparable

(d)

A turtle

Mass = 710. g = 0.71 kg

Size =  22. cm = 0.22 m

V =  2.8 cm/s. = 0.028 m/s

Substituting into the formula, we have

λ= h/(mv)

λ= 6.62607*10^-9/ (  0.71* 0.028)

  = 6.62607*10^-9/0.01988

   = 3.33*10^-7 m

Quantum Approach: The wavelength and the size are comparable

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Answer:

Explanation:

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Please check attached image for solution:

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Answer:

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Explanation:

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