Answer:
Step-by-step explanation:
If you meant to solve for "z", then refer to the attachment below.
A bag of candy has a mass of 80 g. A chocolate bar has a mass of 110 g. What is the total mass of the bag of candy and the choco
1 answer:
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Answer:
233.48s
3.84 min
Step-by-step explanation:
In order to solve this problem, we can start by drawing what the situation looks like. See attached picture.
We can model this situation by making use of a trigonometric function. Trigonometric functions have the following shape:
where:
A= amplitude =-20m because the model starts at the lowest point of the trajectory.
f= the function to use, in this case we'll use cos, since it starts at the lowest point of the trajectory.
t= time
angular speed.
in this case:
where T is the period, in this case 6 min or
so:
and
= phase angle
C= vertical shift
in this case our vertical shift will be:
2m+20m=22m
in this case the phase angle is 0 because we are starting at the lowest point of the trajectory. So the equation for the ferris wheel will be:
Once we got this equation, we can figure out on what times the passenger will be higher than 13 m, so we build the following inequality:
so we can solve this inequality, we can start by turning it into an equation we can solve for t:
and solve it:
and we can take the inverse of cos to get:
which yields two possible answers: (see attached picture)
so
or
so we can solve the two equations. Let's start with the first one:
t=63.25s
and the second one:
t=296.73s
so now we can build our possible intervals we can use to test the inequality:
[0, 63.25] for a test value of 1
[63.25,296.73] for a test value of 70
[296.73, 360] for a test value of 300
let's test the first interval:
[0, 63.25] for a test value of 1
2>13 this is false
let's now test the second interval:
[63.25,296.73] for a test value of 70
15.16>13 this is true
and finally the third interval:
[296.73, 360] for a test value of 300
12>13 this is false.
We only got one true outcome which belonged to the second interval:
[63.25,296.73]
so the total time spent above a height of 13m will be:
196.73-63.25=233.48s
which is the same as:
see attached picture for the graph of the situation. The shaded region represents the region where the passenger will be higher than 13 m.
