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Evgesh-ka [11]
3 years ago
9

What is the minimum amount of data points an experiment should gather?

Physics
1 answer:
nydimaria [60]3 years ago
7 0

The answer is "Three".

At the point when individuals take studies for factual purposes or when the Statistics Agency comes around and gets everyone's data, data from one individual is one information point for them. Toward the finish of its exploration or overview, the organization will have accumulated numerous bits of data from numerous individuals. One piece of information equals with one data point.

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An object with height h, mass M, and a uniform cross-sectional area A floats upright in a liquid with density ρ.
soldi70 [24.7K]
** Missing information: The vertical distance from surface of liquid to bottom of the object is sought in this question, with the condition that the object is at equilibrium **

Ans: The vertical distance = y = M/(ρA)

Explanation:

Support the vertical distance = y

Object's density = M/(A*h) (since A*h = volume)

By applying the condition, 

(M/(Ah))/ρ = y/h

M/(ρAh) = y/h

y = M/(ρA)  

7 0
4 years ago
Do you think aksionov is able to get justice in the end? Why?​
Mars2501 [29]
This can be seen through the fact that Aksionov has the ability to seek his own justice many times throughout the story, yet does not take it
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3 years ago
In transduction, the cochlea is part of this structure:
anyanavicka [17]

Answer:

wats up

Explanation:

8 0
3 years ago
What information in the MSDS would be most important to know at the end of an experiment?
Artemon [7]
The most important information in the MSDS that is useful at the end of an experiment is how to manage or dispose of the waste materials of the experiment. This is important especially if the materials used are toxic. They cannot just be disposed in the sink or the trash bin. They must be disposed in a waste bottle or other methods.
8 0
4 years ago
If Superman really had x-ray vision at 0.12 nm wavelength and a 4.1 mm pupil diameter, at what maximum altitude could he disting
iVinArrow [24]

Answer:

Maximum altitude to see(L) =  1.47 × 10⁶ m (Approx)

Explanation:

Given:

wavelength (λ) = 0.12 nm = 0.12 × 10⁻⁹ m

Pupil Diameter (d) = 4.1 mm = 4 × 10⁻³ m

Separation distance (D) = 5.4 cm = 0.054 m

Find:

Maximum altitude to see(L)

Computation:

Resolving power = 1.22(λ / d)

D / L = 1.22(λ / d)

0.054 / L = 1.22 [(0.12 × 10⁻⁹) / (4 × 10⁻³ m)]

0.054 / L = 1.22 [0.03 × 10⁻⁶]

L = 0.054 / 1.22 [0.03 × 10⁻⁶]

L = 0.054 / [0.0366 × 10⁻⁶]

L = 1.47 × 10⁶

Maximum altitude to see(L) =  1.47 × 10⁶ m (Approx)

8 0
3 years ago
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