Answer:
while (userNum >=0) {...}
Explanation:
In a <u>while loop</u>, <u>the loop is executed until the condition is false</u>.
Since the loop will execute while the user enters a number greater than or equal to 0 (and that number is declared as <em>userNum</em>), we need to check if <em>userNum</em> is greater than or equal to 0.
Answer:
[inaudible]
Explanation:
TranscribeMe is an online transcription company. It employs people all over the world to render transcription services to clients. They have a style guideline which must be strictly adhered to by the transcribers. The guidelines specifically states the "do's " and "dont's" during transcription. Now, when transcribing a word or phrase that cannot be heard or understood due to poor audio or difficult answer, the transcriber uses the tag [inaudible].
Answer:
Audience knowledge level
Explanation:
The Key issue that Mario failed to consider when preparing for her presentation is known as Audience knowledge level.
Because Audience knowledge level comprises educating your audience with extensive information about the topic been presented. she actually tried in doing this but she was educating/presenting to the wrong audience ( People who are more knowledgeable than her on the topic ). she should have considered presenting a topic slightly different in order to motivate the audience.
The first thing we are going to do is find the equation of motion:
ωf = ωi + αt
θ = ωi*t + 1/2αt^2
Where:
ωf = final angular velocity
ωi = initial angular velocity
α = Angular acceleration
θ = Revolutions.
t = time.
We have then:
ωf = (7200) * ((2 * pi) / 60) = 753.60 rad / s
ωi = 0
α = 190 rad / s2
Clearing t:
753.60 = 0 + 190*t
t = 753.60 / 190
t = 3.97 s
Then, replacing the time:
θ1 = 0 + (1/2) * (190) * (3.97) ^ 2
θ1 = 1494.51 rad
For (10-3.97) s:
θ2 = ωf * t
θ2 = (753.60 rad / s) * (10-3.97) s
θ2 = 4544,208 rad
Number of final revolutions:
θ1 + θ2 = (1494.51 rad + 4544.208 rad) * (180 / π)
θ1 + θ2 = 961.57 rev
Answer:
the disk has made 961.57 rev 10.0 s after it starts up
If you take the photo with one person on the ground that way when they take the photo it will look like the person jumping is jumping higher then they really are.
I hope this helps
