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elena55 [62]
3 years ago
11

Five different students in Mr. Garcia's class have done an experiment once, and each one got similar results. What should the st

udents do to increase the validity of their results?
Physics
2 answers:
trapecia [35]3 years ago
7 0

Repeat the experiment

Explanation:

To increase the validity of the results obtained from the single experiment, the students should be encouraged to repeat the experiments more number of times as much as possible.

In an experiment, scientist always try to limit errors by making precise and accurate observation.  A single observation does not really represent a precise and accurate finding. When an experiment is repeated as often as possible, the reliability of the conclusion drawn from the hypothesis testing will improve and the results can be accepted to be valid.

A single observation/experiment is not valid enough.

Learn more:

Experiments brainly.com/question/5096428

#learnwithBrainly

grandymaker [24]3 years ago
5 0

<em>Answer:</em>

<em>To increase the validity of the results, all the students of Mr. Gracia's class must perform a few more experimental trials.</em>

<em>Answer:  A</em>

<em>Explanation:</em>

<em>While dealing with the scientific analysis, one must always commit number of experiments to gain the most accurate and valid result. By doing number of trials, we get different approximate values that can be used to estimate the average result.</em>

<em>This average will be the nearest value of the exact answer if we add more and more number of experiments or trials. Since, the results are similar for each student in the class, they all should attempt more trials or experiments to get the most accurate result.</em>

<em />

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Student swings a small rubber stopper attached to a string over her head in a horizontal, circular path. The string is 1.50 mete
harkovskaia [24]

Answer:

v = 18.84 m/s

Explanation:

Given that,

The length of the string, r = 1.5 m (it will act as radius)

The rubber stopper makes 120 complete circles every minute.

Since, 1 minute = 60 seconds

It means, its frequency is 2 circles every second.

Let we need to find the average speed of the rubber stopper. It can be calculated as follows :

v=\dfrac{d}{T}

d is distance, d=2\pi r and 1/T = f (frequency)

v=2\pi rf\\\\=2\pi \times 1.5\times 2\\\\=18.84\ m/s

So, the average speed of the rubber stopper is 18.84 m/s.  

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Which of the following statements best describes ergonomics?
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3 years ago
Your cousin Jannik skis down a blue square ski slope, with an initial speed of 3.6 m/s. He travels 15 m down the mountain side b
fenix001 [56]

Answer: The loss of energy due to friction is equal to 1,253 J.

Explanation:

The problem tells us that the skier has an initial speed of 3.6 m/s, which means that his initial kinetic energy is as follows:

K₁ = 1/2 m v₁² = 1/2 . 58.0 Kg. (3.6)² (m/s)² =  376 J

After coming to a  flat landing, his final speed is 7.8 m/s, so the final kinetic energy is as follows:

K₂ = 1/2 m v₂² = 1/2. 58.0 Kg. (7.8)² (m/s)² = 1,764 J

Now, when skying down the slope the increase in kinetic energy only can come from another type of energy, in this case, gravitational potential energy.

If we take the ground flat level as a Zero reference, the initial gravitational potential energy, can be written as follows, by definition:

U₁ = m.g. h (1)

Now, we don't know the value of the height h, but we know that the incline has a 18º angle above the horizontal, and that the distance travelled along the incline is 15 m.

By definition, the sinus of an angle, is equal to the proportion between the height and the hypotenuse , so we can write the following equation:

sin 18º = h / 15 m ⇒ h = 15 m. sin 18º = 4.6 m

Replacing in (1), we get:

U₁ = 58.0 Kg. 9.8 m/s². 4.6 m = 2,641 J

So, we can get the total initial mechanical energy, as follows:

E₁ = K₁ + U₁ = 376 J + 2,641 J = 3,017 J

After arriving to the flat zone, all potential energy has become in kinetic energy, even though not completely, due to the effect of friction.

This remaining kinetic energy can be written as follows:

E₂ = K₂ = 1,764 J

The difference E₂-E₁, is the loss of energy due to friction forces acting during the travel along the 15 m path, and is as follows:

ΔE= E₂ - E₁ = 1,764 J - 3,017 J = -1,253 J

8 0
3 years ago
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