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tresset_1 [31]
4 years ago
11

Anna is conducting an experiment to determine how weather affects cell phone reception. She is trying to decide the best way to

conduct her experiment in order to collect meaningful data. Which of the following experiments would help Anna collect the best data? A. Test different cell phones in different locations on days with clear weather. B. Test different cell phones in different locations on days with rainy weather. C. Test a cell phone's reception in one location with clear weather and in another location with rainy weather. D. Test a cell phone's reception in the exact same location under various atmospheric conditions.
Physics
1 answer:
aev [14]4 years ago
4 0

Answer:

D. Test a cell phone's reception in the exact same location under various atmospheric conditions

Explanation:

Components of an experiment:

Independent variable:

A manipulated variable, in an experiment or study, whose presence or degree incurs a change in the  dependent variable.

Dependent variable:

A variable which is being studied in the experiment and is supposed to change with respect to the independent variable.

Control variable:

Control variable includes the variables which are of no concern in the experiment and must be constant so that they do not interfere with the results of the experiment. Change in control variable might affect the results of an experiment.

In Anna's experiment, the independent variable is the atmospheric conditions. The dependent variable is cell phone reception that will change according to the atmospheric conditions. All other things related to the experiment are control variables like cellphone, location of the cell phone. Both of these must stay constant to collect the best data.

Hence, the best option is D.

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a.\rm -1.49\ m/s^2.

b. \rm 50.49\ m.

Explanation:

<u>Given:</u>

  • Velocity of the particle, v(t) = 3 cos(mt) = 3 cos (0.5t) .

<h2>(a):</h2>

The acceleration of the particle at a time is defined as the rate of change of velocity of the particle at that time.

\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(3\cos(0.5\ t ))\\=3(-0.5\sin(0.5\ t.))\\=-1.5\sin(0.5\ t).

At time t = 3 seconds,

\rm a=-1.5\sin(0.5\times 3)=-1.49\ m/s^2.

<u>Note</u>:<em> The arguments of the sine is calculated in unit of radian and not in degree.</em>

<h2>(b):</h2>

The velocity of the particle at some is defined as the rate of change of the position of the particle.

\rm v = \dfrac{dr}{dt}.\\\therefore dr = vdt\Rightarrow \int dr=\int v\ dt.

For the time interval of 2 seconds,

\rm \int\limits^2_0 dr=\int\limits^2_0 v\ dt\\r(t=2)-r(t=0)=\int\limits^2_0 3\cos(0.5\ t)\ dt

The term of the left is the displacement of the particle in time interval of 2 seconds, therefore,

\Delta r=3\ \left (\dfrac{\sin(0.5\ t)}{0.05} \right )\limits^2_0\\=3\ \left (\dfrac{\sin(0.5\times 2)-sin(0.5\times 0)}{0.05} \right )\\=3\ \left (\dfrac{\sin(1.0)}{0.05} \right )\\=50.49\ m.

It is the displacement of the particle in 2 seconds.

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