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3 years ago

Why are graphs useful when interpreting data?

Physics

2 answers:

PilotLPTM [1.2K]3 years ago

6 0

For this case what you must understand is that it represents each of the axes in a graph.

The horizontal axis represents the independent variable.

The vertical axis represents the dependent variable.

The use of graphics is useful because it represents the behavior of one variable with respect to the other.

Through the graphs we can observe what happens with the dependent variable when we add or remove values of the independent variable.

Answer:

They are useful because they represent the behavior of one variable with respect to the other.

sammy [17]3 years ago

3 0

It's a visual representation of data, so it's easier to interpret and understand.

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The first thing you should know in this case is the following definition:
PV = nRT
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P1V1 = k
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We can then equalize both equations:
P1V1 = P2V2
Substituting the values:
(1.25) * (101) = (2.25) * (P2)
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4 years ago

Consider the Uniform Circular Motion Gizmo configured as shown. Notice that, under the current settings, |a|=0.50m/s2. What chan

Eddi Din [679]

To increase the centripetal acceleration to $2.00 m/s^2$ , you can double the speed or decrease the radius by 1/4

Explanation:

An object is said to be in uniform circular motion when it is moving at a constant speed in a circular path.

The acceleration of an object in uniform circular motion is called centripetal acceleration, and it is given by

$a=\frac{v^2}{r}$

where

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r is the radius of the circular path

In the problem, the original centripetal acceleration is

$a=0.50 m/s^2$

We want to increase it by a factor of 4, i.e. to

$a'=2.00 m/s^2$

We notice that the centripetal acceleration is proportional to the square of the speed and inversely proportional to the radius, so we can do as follows:

- We can double the speed:

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This way, the new acceleration is

$a'=\frac{(2v)^2}{r}=4(\frac{v^2}{r})=4a$

so, 4 times the original acceleration

- We can decrease the radius to 1/4 of its original value:

$r'=\frac{1}{4}r$

So the new acceleration is

$a'=\frac{v^2}{(r/4)}=4(\frac{v^2}{r})=4a$

so, the acceleration has increased by a factor 4 again.

Learn more about centripetal acceleration:

brainly.com/question/2562955

brainly.com/question/6372960

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3 years ago

How do I find the cosine of theta.

NNADVOKAT [17]

The value of cos θ in the given figure is 0.98.

<h3>What is cosine of an angle?</h3>

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cos θ = sin(90 - θ)

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0.866 = 0.866

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adj² = 10² - 2²

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Thus, the value of cos θ in the given figure is 0.98.

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