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3 years ago

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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Consider a point on a bicycle wheel as the wheel turns about a fixed axis, neither speeding up nor slowing down. Compare the lin

Natalija [7]

Answer:

c. Only the linear acceleration is zero.

Explanation:

The linear acceleration is defined as the rate of change of linear velocity. Since the bicycle is moving in the same direction, with the same speed, without speeding up or slowing down. Therefore, there will be no change in linear velocity and as a result, linear acceleration will be zero.

The angular acceleration is the rate of change of angular velocity. Since the angular velocity is changing its direction constantly. Therefore, it has a certain component of acceleration at all times called centripetal acceleration.

Therefore, the correct option is:

<u>c. Only the linear acceleration is zero.</u>

4 0

3 years ago

Pube Goldberg machine is a complicated contraption designed to do a very simple task, like the one shown above

Elanso [62]

Answer:

I think is 2

Explanation:

6 0

3 years ago

25 points! Please someone help me, this seems like a pretty easy question but I can’t seem to get it right, please list the give

elixir [45]

Answer:

1 my brother say that

Explanation:

i know my brother said it

4 0

3 years ago

A block of mass 0.221 kg is placed on top of a light, vertical spring of force constant 5365 N/m and pushed downward so that the

Anvisha [2.4K]

Answer:

The maximum height above the point of release is 11.653 m.

Explanation:

Given that,

Mass of block = 0.221 kg

Spring constant k = 5365 N/m

Distance x = 0.097 m

We need to calculate the height

Using stored energy in spring

$U=\dfrac{1}{2}kx^2$ ...(I)

Using gravitational potential energy

$U' =mgh$ ....(II)

Using energy of conservation

$E_{i}=E_{f}$

$U_{i}+U'_{i}=U_{f}+U'_{f}$

$\dfrac{1}{2}kx^2+0=0+mgh$

$h=\dfrac{kx^2}{2mg}$

Where, k = spring constant

m = mass of the block

x = distance

g = acceleration due to gravity

Put the value in the equation

$h=\dfrac{5365\times(0.097)^2}{2\times0.221\times9.8}$

$h=11.653\ m$

Hence, The maximum height above the point of release is 11.653 m.

3 0

3 years ago

Please help me guys never mind the calculations

vlada-n [284]

The shape is connected in parallel so;

5.1) Ans;

$\frac{1}{R} = \frac{1}{R1} + \frac{1}{R2} \\ \frac{1}{R} = \frac{1}{2} + \frac{1}{3} \\ \frac{1}{R} = \frac{3 + 2}{6} = \frac{5}{6} \\ R = \frac{6}{5} = 1.2 \: \: ohm$

5.2) Ans;

$\frac{1}{R} = \frac{1}{R1} + \frac{1}{R2} \\ \frac{1}{R} = \frac{1}{8} + \frac{1}{10} \\ \frac{1}{R} = \frac{5 + 4}{40} = \frac{9}{40} \\ R = \frac{40}{9} = 4.4 \: \: ohm$

I hope I helped you^_^

7 0

2 years ago