A train goes up a hill with a 15º incline. If the train has constant speed of 22 m/s, what are the vertical and horizontal compo
nents of the train's velocity? (Remember to use the law of sine and cosines for this problem)
1 answer:
Sinθ=opposite side/hypotenuse
cosθ=adjacent side/hypotenuse
where both the sides are relative to theta.
in this case, 22 is your hypotenuse. theta (θ) is the angle you are examining. i've included a diagram for reference because the equations may format differently depending on which angle you choose as theta.
using my diagram:
sinθ = o/h
sin 15 = o/22
sin 15(22) = o
o=5.69m/s
therefore the vertical component is about 5.69m/s.
cosθ=a/h
cos15 = a/22
cos15(22) = a
a = 21.25m/s
therefore, your horizontal component is about 21.25m/s.
cosθ=adjacent side/hypotenuse
where both the sides are relative to theta.
in this case, 22 is your hypotenuse. theta (θ) is the angle you are examining. i've included a diagram for reference because the equations may format differently depending on which angle you choose as theta.
using my diagram:
sinθ = o/h
sin 15 = o/22
sin 15(22) = o
o=5.69m/s
therefore the vertical component is about 5.69m/s.
cosθ=a/h
cos15 = a/22
cos15(22) = a
a = 21.25m/s
therefore, your horizontal component is about 21.25m/s.
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<h2>Answer with explanation:</h2>
Given : A standardized exam's scores are normally distributed.
Mean test score :
Standard deviation :
Let x be the random variable that represents the scores of students .
z-score :
We know that generally , z-scores lower than -1.96 or higher than 1.96 are considered unusual .
For x= 1900
Since it lies between -1.96 and 1.96 , thus it is not unusual.
For x= 1240
Since it lies between -1.96 and 1.96 , thus it is not unusual.
For x= 2190
Since it is greater than 1.96 , thus it is unusual.
For x= 1240
Since it lies between -1.96 and 1.96 , thus it is not unusual.