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Nataly [62]
2 years ago
5

Part 2: Identify the independent, dependent, and constant variables

Physics
1 answer:
diamong [38]2 years ago
4 0
ANSWER:

IV, Type of dish detergent. DV, height of foam. CV, type of container, amount of water in container, temperature of water, time the container is agitated.

Explanation:

Independent variable(IV)- what you change during the experiment.
dependent variable(DV)- what you're measuring during an experiment. The dependent variable is DEPENDENT because it's results DEPEND on the independent variable at play.
Constant variables(CV)- things that do not change in order to isolate the tested variables as much as possible.
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when the particles of the medium move back and forth along the direction of the wave motion, the wave is a
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Transverse waves are always characterized by particle motion being perpendicular to wave motion. A longitudinal wave is a wave in which particles of the medium move in a direction parallel to the direction that the wave moves.

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An AC source is connected to a series combination of a light bulb and a variable capacitor. If the capacitance is increased, the
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Answer:

Increases

Explanation:

Since power P=IV

Then it means when current increases, the power increases hence brightness increases. I represent current, P is power and v is voltage.

Current of capacitor when in series connection is given by

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3 years ago
When there is no air resistance, objects of different masses (1 points)
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Answer:

Fall at equal acceleration with similar displacements.

Explanation:

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2 years ago
ASAP HELPPPPPP POP PLEASE​
drek231 [11]

Answer:

The answer is in the attachment

Explanation:

<em>!</em><em>!</em><em>PLEASE</em><em> </em><em>PLEASE</em><em> </em><em>PLEASE</em><em> </em><em>PLEASE</em><em> </em><em>PLEASE</em><em> </em><em>MARK</em><em> </em><em>ME</em><em> </em><em>THE</em><em> </em><em>BRAINLIEST</em><em>!</em><em>!</em>

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A diver stands on a diving platform 10.0 m above the surface of a pool and leaps upward with an initial speed of 2.5 m/s. how fa
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<span>The diver is heading downwards at 12 m/s Ignoring air resistance, the formula for the distance under constant acceleration is d = VT - 0.5AT^2 where V = initial velocity T = time A = acceleration (9.8 m/s^2 on Earth) In this problem, the initial velocity is 2.5 m/s and the target distance will be -7.0 m (3.0 m - 10.0 m = -7.0 m) So let's substitute the known values and solve for T d = VT - 0.5AT^2 -7 = 2.5T - 0.5*9.8T^2 -7 = 2.5T - 4.9T^2 0 = 2.5T - 4.9T^2 + 7 We now have a quadratic equation with A=-4.9, B=2.5, C=7. Using the quadratic formula, find the roots, which are -0.96705 and 1.477251164. Now the diver's velocity will be the initial velocity minus the acceleration due to gravity over the time. So V = 2.5 m/s - 9.8 m/s^2 * 1.477251164 s V = 2.5 m/s - 14.47706141 m/s V = -11.97706141 m/s So the diver is going down at a velocity of 11.98 m/s Now the negative root of -0.967047083 is how much earlier the diver would have had to jump at the location of the diving board. And for grins, let's compute how fast he would have had to jump to end up at the same point. V = 2.5 m/s - 9.8 m/s^2 * (-0.967047083 s) V = 2.5 m/s - (-9.477061409 m/s) V = 2.5 m/s + 9.477061409 m/s V = 11.97706141 m/s And you get the exact same velocity, except it's the opposite sign. In any case, the result needs to be rounded to 2 significant figures which is -12 m/s</span>
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3 years ago
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