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34kurt
3 years ago
8

A researcher predicted that talking to plants enhances their growth. She gave 24 plants the same amount of water and sunlight. S

he talked to 12 of the 24 plants daily for 6 weeks and then measured their growth. In this experiment, which of the following is the dependent variable?
Physics
1 answer:
Ne4ueva [31]3 years ago
3 0

Answer:

Growth of the plants

Explanation:

In an experiment a logical set of attributes of an object is changed and the logical set of attributes which is affected by the other attribute change is noted. The set of attributes which are logical are called variables.

There are two types of variables are

The independent variable is the input. The output is observed with respect to the input.

Dependent variables are the output which is the result of the change in the independent variables

Here, the researcher wants to know how talking to plants affects their growth. The input is talking to plants and the output is the growth of the plants.

Hence, the independent variable is the growth of the plants.

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Students in an introductory physics lab are performing an experiment with a parallel-plate capacitor made of two circular alumin
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Answer:

\rm 9.186\times 10^{-7}\ C.

Explanation:

<u>Given:</u>

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  • Distance of separation between the plates, d = 1.0 cm = 0.01 m.
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When the dimensions of the plate of the capacitor is comparatively much larger than the distance of separation between the plates, then, according to the Gauss' law of electrostatics, the value of the electric field strength in the region between the plates of the capacitor is given by

\rm E=\dfrac{\sigma}{\epsilon_o}.

where,

  • \rm \sigma = surface charge density of the plate of the capacitor = \dfrac qA.
  • \rm q = magnitude of the charge on each of the plate.
  • \rm A = surface area of each of the plate =\rm \pi \times (Radius)^2=\pi \times\left ( \dfrac{D}{2}\right )^2= \pi \times \left ( \dfrac{0.21}{2}\right )^2=3.46\times 10^{-2}\ m^2.
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For the minimum value of electric field that produces spark,

\rm E = \dfrac{q}{A\epsilon_o}\\\Rightarrow q = E\ A\epsilon_o\\=3\times 10^6\times 3.46\times 10^{-2}\times 8.85\times 10^{-12}\\=9.186\times 10^{-7}\ C.

It is the maximum value of the magnitude of charge which can be added up to each of the plates of the capacitor.

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3 years ago
True or false? Thanks,
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Answer: The answer is true

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