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lisov135 [29]
3 years ago
15

In a classroom demonstration, students are using a Slinky to observe and learn about wave properties. If the Slinky has a period

of 0.333 seconds, what is the frequency?
Physics
1 answer:
Andrej [43]3 years ago
3 0

Answer:

Frequency = 3.0 Hertz

Explanation:

Given the following data;

Period = 0.333 seconds

To find the frequency;

Mathematically, frequency of a wave is given by the formula;

Frequency = 1/period

Substituting into the formula, we have;

Frequency = 1/0.333

Frequency = 3.0 Hertz

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The answer to the question is shown below:

We all know that formula for solving work done is the force multiplied by distance covered:
Work done = Force x distance
Distance = 5m
Force = 500 N
Work done = 500 N * 5m
Work done = 2500 J

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During an investigation, a scientist heated 123.6 g of copper carbonate till it decomposed to form a black residue. The total ma
zloy xaker [14]

Answer:

See below explanation

Explanation:

The correspondent chemical reaction for copper carbonate decomposed by heat is:

CuCO₃ (s) → CuO (s) + CO₂ (g)

Considering all molar mass (MM) for each element ( we consider rounded numbers) :

MM CuCO₃ = 123 g/mol

MM CuO = 79 g/mol

MM CO₂ = 44 g/mol

Statement mentions that scientis heated 123.6 g of CuCO₃ (almost a MM), until a black residue is obtained, which weights 79.6 g : this solid residue is formed by CuO, and the remaining mass (approximatelly 44 g) belongs to teh second product, this is, CO₂; as it is a gas compund, it is not certainly included on the solid residue.

So, law of conservation mass is true for this case, since: 123.6 g = 79.6 g + 44 g. As explained, on the solid residue, we don not include the 44 g, which  "escaped" from our system, since it is a gas compound (CO₂)

5 0
3 years ago
You decide to roll a 0.11-kgkg ball across the floor so slowly that it will have a small momentum and a large de Broglie wavelen
Oksanka [162]

The de Broglie wavelength \lambda = 4.0\times 10^{-30}m

We know that

de Broglie wavelength = \lambda = \frac{h}{mv}\lambda = \frac{6.63\times 10^{-34}}{0.11\times 1.5 \times 10 ^{-3}}

\lambda = 4.0\times 10^{-30}m

<h3>What is de Broglie wavelength?</h3>

According to the de Broglie equation, matter can behave like waves, much like how light and radiation do, which are both waves and particles. A beam of electrons can be diffracted just like a beam of light, according to the equation. The de Broglie equation essentially clarifies the notion of matter having a wavelength.

Therefore, whether a particle is tiny or macroscopic, it will have a wavelength when examined.

The wave nature of matter can be seen or observed in the case of macroscopic objects.

To learn more about de Broglie wavelength with the given link

brainly.com/question/17295250

#SPJ4

3 0
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