How many independent variables does a good investigator use in his or her experiment?

Physics

1 answer:

MrRa [10]3 years ago

5 0

Answer:

You should generally have one independent variable in an experiment. This is because it is the variable you are changing in order to observe the effects it has on the other variables.

HOPE THIS HELPED!!!!!!!!!!!!XDDDDDDDDD

You might be interested in

A body is moving with a uniform acceleration of 2m/s² what does it mean?

Nastasia [14]

Answer:

the body has energy due to its constant motion. it means it moves in a uniform acceleration which has zero velocity

Explanation:

Uniform or constant acceleration is a type of motion in which the velocity of an object changes by an equal amount in every equal time period.

7 0

3 years ago

Explain whether a tennis ball dropped from a high distance experiences an elastic collision or inelastic collision

hodyreva [135]

I need help on that too poop head

4 0

3 years ago

According to meet one law of universal gravitation, gravity increases when?

mina [271]

Newton's law of universal gravitation states that every point mass in the universe attracts every other point mass with a force that is directly proportional to the product of their masses, and inversely proportional to the square of the distance between them. Newton's law of universal gravitation states that every point mass in the universe attracts every other point mass with a force that is directly proportional to the product of their masses, and inversely proportional to the square of the distance between them.

5 0

3 years ago

The light bulb converts electrical energy into light and ____. A) chemical B) electromagnetic C) heat D) nuclear

morpeh [17]

The light bulb converts electrical energy into light and heat.

Good luck! (:

7 0

3 years ago

Read 2 more answers

A silver wire has a cross sectional area a = 2.0 mm2. a total of 9.4 × 1018 electrons pass through the wire in 3.0 s. the conduc

marta [7]

This problem uses the relationships among current I, current density J, and drift speed vd. We are given the total of electrons that pass through the wire in t = 3s and the area A, so we use the following equation to to find vd, from J and the known electron density n, so:

$v_{d} = \frac{J}{n\left | q \right |}$

<span>The current I is any motion of charge from one region to another, so this is given by:

</span> $I = \frac{\Delta Q}{\Delta t} = \frac{9.4x1018electrons}{3s} = 3189.73(A)$

The magnitude of the current density is:

$J = \frac{I}{A} = \frac{3189.73}{2x10^{-6}} = 1594.86(A/m^{2})$

Being:

$A=2mm^{2} = 2x10^{-6}m^{2}$
<span>
Finally, for the drift velocity magnitude vd, we find:

</span> $v_{d} = \frac{1594.86}{5.8x1028\left |1.60x10^{-19}|\right } = 1.67x10^{18}(m/s)$

Notice: The current I is very high for this wire. The given values of the variables are a little bit odd

6 0

3 years ago