Answer:
B. Evaporation
Explanation:
Because I am absolutely positive
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Answer: Ax=(Vx-Vox)/(T)
Vx=Vox+Ax*T
Solving for Ax in terms of Vx, Vox, T
Vx-Vox=Ax*t
Ax=(Vx-Vox)/(T)
This is saying the acceleration in the x-direction can be found by taking the difference between the finial and initial Velocity in x-direction and dividing it by the Total Time.
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Answer:
c
Explanation:
<u>impulse</u><u> </u><u>Is</u><u> </u><u>the </u><u>product</u><u> </u><u>of </u><u>force </u><u>and </u><u>distance</u><u> </u><u>so </u><u>it's </u><u>generally</u><u> </u><u>formula</u><u> </u>
The way you will know that a figure represents a vector is that it will have magnitude and direction.
<h3>How to Identify a Vector?</h3>
A figure that is a vector will typically have magnitude and direction. This means that it will be represented by an arrow(s) with its' magnitude also showing.
Whereas, on the opposite side, the figure is termed as a scalar if it doesn’t have a direction.
Thus, we can know a vector diagram if the magnitude and direction are indicated.
Read more about Vectors at; brainly.com/question/3184914
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<span>Relative frequency histograms are important because the heights can be interpreted as probabilities. These probability histograms provide a graphical display of a </span>probability distribution<span>, which can be used to determine the likelihood of certain results to occur within a given population.
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To see the difference between frequency and relative frequency we will consider the following example. Suppose we are looking at the history grades of students in 12th grade and have the grades: A, B, C, D, F. The number of each of these grades gives us a frequency for each class:
<span>7 students with an F
9 students with a D
18 students with a C
12 students with a B
<span>4 students with an A
</span></span>
To determine the relative frequency for each class we first add the total number of data points: 7 + 9 + 18 + 12 + 4 = 50. Next we, divide each frequency by this sum 50.
<span>0.14 = 14% students with an F
0.18 = 18% students with a D</span><span>0.36 = 36% students with a C
0.24 = 24% students with a B
<span>0.08 = 8% students with an A
You can see it is very easy and convinient to analyse the data </span></span>