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Sort the questions based on whether they are better answered by an observational study or a controlled experiment.

Zepler [3.9K]

Answer:

Following questions are better answered by observational study:

- How does the daytime temperature affect the formation of dew at night?

- How effective is a new toothpaste in reducing cavities?

Following questions are better answered by controlled experiment:

- Which is a better absorber of heat, an object painted black or the same object painted white?

- Why does a rolling ball slow down as it moves?

- Can plants prepare food even in artificial light?

Explanation:

An observational study is one in which the researcher doesn't manipulate anything.

In a controlled experiment, the researcher manipulate the outcome of the experiment and controls the factors that might affect the experiment to see different results in different conditions.

So, for questions over which research do not have any control, mostly natural processes, observational study is a better option.

However, to check why a rolling ball slow down, controlled experiments are done to find out all the factors.

Plants placed in artificial light is also an controlled experiment as the researcher will himself place the plants in artificial light and keep them away from natural light.

Similarly, to check if color of the object affects the amount of heat absorbed by that object, a controlled experiment must be performed. Same object should be painted white and observed and then painted black and observed.

7 0

3 years ago

Chemical reactions provide support for the law of conservation of matter because which of the following is true about the atoms

puteri [66]

Answer:

Explanatioit cn:

3 0

3 years ago

Two speakers placed 0.94 m apart produce pure tones in sync with each other at a frequency of 1630 Hz. A microphone can be moved

Alina [70]

Answer:

a. approximately $1.1\; \rm m$ (first minimum.)

b. approximately $2.2\; \rm m$ (first maximum.)

c. approximately $3.4\; \rm m$ (second minimum.)

d. approximately $4.7\; \rm m$ (second maximum.)

Explanation:

Let $d$ represent the separation between the two speakers (the two "slits" based on the assumptions.)

Let $\theta$ represent the angle between:

the line joining the microphone and the center of the two speakers, and
the line that goes through the center of the two speakers that is also normal to the line joining the two speakers.

The distance between the microphone and point $P_0$ would thus be $9.4\, \tan(\theta)$ meters.

Based on the assumptions and the equation from Young's double-slit experiment:

$\displaystyle \sin(\theta) = \frac{\text{path difference}}{d}$ .

Hence:

$\displaystyle \theta = \arcsin \left(\frac{\text{path difference}}{d}\right)$ .

The "path difference" in these two equations refers to the difference between the distances between the microphone and each of the two speakers. Let $\lambda$ denote the wavelength of this wave.

$\displaystyle \begin{array}{c|c} & \text{Path difference} \\ \cline{1-2}\text{First Minimum} & \lambda / 2 \\ \cline{1-2} \text{First Maximum} & \lambda \\\cline{1-2} \text{Second Minimum} & 3\,\lambda / 2 \\ \cline{1-2} \text{Second Maximum} & 2\, \lambda\end{array}$ .

Calculate the wavelength of this wave based on its frequency and its velocity:

$\displaystyle \lambda = \frac{v}{f} \approx 0.211\; \rm m$ .

Calculate $\theta$ for each of these path differences:

$\displaystyle \begin{array}{c|c|c} & \text{Path difference} & \text{approximate of $\theta$} \\ \cline{1-3}\text{First Minimum} & \lambda / 2 & 0.112 \\ \cline{1-3} \text{First Maximum} & \lambda & 0.226\\\cline{1-3} \text{Second Minimum} & 3\,\lambda / 2 & 0.343\\ \cline{1-3} \text{Second Maximum} & 2\, \lambda & 0.466\end{array}$ .