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4 years ago

What us the difference between accurate data and reproducible data

1 answer:

4 0

It's just in the name! Accurate data is helpful, and correct, but reproducible data is all of that, and is able to be given to other people through different sources! At least, that's what my understanding of them are. Hope it helps!

You might be interested in

What can be inferred when an atom is electrically charged?

Elena-2011 [213]

When an atom becomes electrically charged the number of electrons or protons stops and they are not equal again. The "extra" electron or proton is not balanced by something inside the atom any longer and it starts attracting itself to othet protons or electrons in other atoms.

<h3>What is atomic structure?</h3>

An atomic structure comprises of positively charged nucleus which is surrounded by negatively charged particles called electron and neutron which is neutral charged.

Unlike charges attract each other while like charges repel each other.

Therefore, When an electron is fully charged, the number of electrons will stop to be unequal again.

Learn more about Atomic charge here.

brainly.com/question/18102056

7 0

2 years ago

Water of density 1000 kg/m3 falls without splashing at a rate of 0.373 L/s from a height of 40.5 m into a 0.64 kg bucket on a sc

Sphinxa [80]

Answer:

F_scale = 20.18 N

Explanation:

The scale reading corresponds to two factors, the first the weight of the water in the container and the second the force of the liquid that is falling at the moment of reading.

* Let's find the amount of liquid in the container for a time of t = 2.93 s

Let's use a direct proportion rule. If 0.373 l falls in one second at t = 2.93 s, how many liters are there

V_{water} = 2.93 s (0.373 l / 1s) = 1.09 l

V_{water} = 1.09 10⁻³ m³

the amount of water is

ρ = m / V

m = ρ V

m = 1000 1.09 10⁻³

m = 1.09 kg

so the weight of the liquid in the container for this time is

W = mg

W = 1.09 9.8

W = 10.68 N

* Let's look for the force of the falling jet

Let's use Bernoulli's equation, where the subscript 1 is for the container and the subscript 2 is for the water at a height h

P₁ + 1/2 ρ g v₁² + ρ g y₁ = P₂ + 1/2 ρ g v₂² + ρ g y₂

In this case, the water falls freely, so the external pressure is atmospheric.

P₂ = P_{atm}

since they indicate that the water falls, we assume that its initial velocity is zero v₂ = 0

let's use kinematics to find the speed of a drop when it reaches the container y = 0

v² = v₀² - 2 g (y-y₀)

v = $\sqrt{0 -2 g ( 0-y_o)}$

let's calculate

v = √(2 9.8 40.5)

v = 28.17 m / s

this is the speed in the container v₁ = 28.17 m / s

the height from where it falls is y₂ = 40.5 and reaches the container y₁ = 0

we substitute in Bernoulli's equation

P₁ +1/2 ρ g v₁² + 0 = P_{atm} + 0 + ρ g y₂

P₁ + ½ ρ g v₁² = P_{atm} + ρ g y₂

P₁ = P_{atm} + ρ g y₂ - ½ ρ g v₁²

P₁ = 1 10⁵ + 1000 9.8 40.5 - ½ 1000 28.17²

P₁ = 1 10⁵ + 3.97 10⁵ - 3.69 10⁵

P₁ = 1.28 10⁵ Pa

The definition of Pressure is

P = F / A

F = P A

We must suppose a time to carry out the reading suppose an average time of the modern equipment t = 0.1 s, in this time how much is now arriving

m₂ = 0.373 0.2 = 0.0746 l = 0.0746 10⁻³ m³

the volume is V = A l

if the length of l = 1 m

A = 0.0746 10⁻³ m³ = 7.45 10⁻⁵ m²

the force of this jet is

F = P A

F = 1.28 10⁵ 7.46 10⁻⁵

F = 9.5 N

with these data let's use the equilibrium equation

F_ scale -W - F = 0

F_scale = W + F

F_scale = 10.682 + 9.5

F_scale = 20.18 N

4 0

3 years ago

What is the primary outgoing radiation put off by the sun?

ira [324]

I really don’t know but I think it’s D

7 0

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

Which descriptions best fit the labels? X: kinetic energy Y: potential energy X: potential energy Y: kinetic energy X: mechanica

Pani-rosa [81]

W-APE. For example, work W done to accelerate a positive charge from rest is positive and results from a loss in PE, or a negative APE. There must be a minus sign in front of APE to make W positive. PE can be found at any point by taking one point as a reference and calculating the work needed to move a charge to the other point.

( The capital A’s in the words are supposed to be triangles ! I also hoped this helped ! Please mark me as brainliest !! )

3 0

3 years ago