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777dan777 [17]
4 years ago
10

One useful tool that may help a scientist interpret data by revealing unexpected patterns is a

Physics
2 answers:
Paul [167]4 years ago
8 0

For this case we have a very useful tool is to make use of a graph.

To make use of a graphic you need two things fundamentally:

1) Define the independent variable: normally placed on the horizontal axis

2) Define the dependent variable: normally placed on the vertical axis

Then, when plotting the obtained data, the behavior of both variables can be observed, revealing the unexpected pattern.

Answer:

One useful tool that may help a scientist interpret data by revealing unexpected patterns is a:

b. graph.

Yanka [14]4 years ago
7 0
<span>b. graph

When data are graphed then the unexpected pattern can be revealed.</span>
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Which statement about work and power correctly describes an automobile race?
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Well this question looks like it makes some assumptions.  So assuming that both cars have the same mass and experience the same wind resistance regardless of speed and same internal frictions, then we could say "The car that finishes last has the lowest power".  The reason is that for a given race the cars must overcome losses associated with motion.  Since they all travel the same distance, the amount of work will be the same for both.  This is because work is force times distance.  If the force applied is the same in both cases (identical cars with constant wind resistance) and the distance is the same for both (a fair race track) then W=F·d will be the same.
Power, however, is the work done divided by the time over which it is done.  So for a slower car, time t will be larger.  The power ratio W/t will be smaller for the longer time (slower car).
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The Achilles tendon connects the muscles in your calf to the back of your foot. When you are sprinting, your Achilles tendon alt
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Explanation:

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Length of the tendon, l = 15 cm = 0.15 m

Area of cross section, A=110\ mm^2=0.00011\ m^2

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Let the runner's Achilles tendon stretch if the force on it is 8.0 times his weight, F = 8 mg

Young's modulus for tendon is, Y=0.15\times 10^{10}\ N/m^2

The formula of the Young modulus is given by :

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0.15\times 10^{10}=\dfrac{8\times 70\times 9.8/0.00011}{\dfrac{\Delta L}{0.15}}

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Part B,

The fraction of the tendon's length does this correspond is given by :

\dfrac{\Delta L}{L}=\dfrac{0.0049}{0.15}

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