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Elza [17]
3 years ago
6

A student examines the effect of the number of D batteries in a closed circuit on the brightness of a light bulb. In the experim

ent, four circuits were built with 1, 2, 3, and 4 batteries respectively. For each trial, the brightness of the light bulb was measured using a light meter. Which variable should the student keep constant?
the type of batteries


the number of batteries


the brightness of the light bulb


the data collected from the light meter
Physics
1 answer:
salantis [7]3 years ago
3 0

The correct answer is the Type of batteries

Explanation:

In an experiment, it is normal the researcher modifies one variable, in this case, the number of batteries (independent variable), and this variable affects the variable tested, in this case, the brightness of the bulb (dependent variable).

However, other factors or variables should be constant to guarantee the results are reliable. In this context, one factor that the student should keep constant is the type of batteries he uses because this is not one of the main variables and therefore it is not expected he changes this on purpose or this is affected by other variables. Moreover, by making any change the experiment can be negatively affected. For example, batteries might differ in their power capacity this can affect the brightness of the bulbs.

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Answer:

Electrical force, F = 90 N

Explanation:

It is given that,

Charge on sphere 1, q_1=9\ \mu C=9\times 10^{-6}\ C

Charge on sphere 2, q_1=4\ \mu C=4\times 10^{-6}\ C

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Let F is the electrical force between them. It is given by the formula of electric force which is directly proportional to the product of charges and inversely proportional to the square of distance between them such that,

F=k\dfrac{q_1q_2}{d^2}

F=9\times 10^9\times \dfrac{9\times 10^{-6}\times 4\times 10^{-6}}{(0.06)^2}

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Explanation: HOPE IT HELPED

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Explanation:

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      Volume = \frac{\text{mass}}{\text{density}}

So,       V = \frac{10^{-3}}{19.3 \times 10^{3} kg/m^{3}}

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Now, we will calculate the area as follows.

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Thus, we can conclude that the resistance of given wire is 2.71 \times 10^{6} ohm.

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