The object will move in the direction of the applied force.
The technician A is correct. The current should be 20% or any given percentage lower than the current for which the fuse is created.
The technician B argues that orange blade fuses have a 9 Amp rating. But they usually have a 40A rating
Answer:
Explanation:
The equation for centripetal acceleration is 
. If 
is the distance from the center of the point at the tip of the blade and 
is the distance from the center of another point in the blade, and since both points are rotating at the same angular velocity 
because both of them belong to the blade, the ratio between their centripetal accelerations will be:
Let's start with the total amount of energy available for the whole scenario:
Some kind of machine gave the coaster a bunch of potential energy by
dragging it up to the top of a 45m hill,and that's the energy is has to work with.
Potential energy = (M) (G) (H) = (800) (9.8) (45) = 352,800 joules
It was then given an extra kick ... enough to give it some kinetic energy, and
start it rolling at 4 m/s.
Kinetic energy = (1/2) (M) (V)² = (1/2) (800) (4)² = 6,400 joules
So the coaster starts out with (352,000 + 6,400) =<em> </em><u><em>359,200 joules</em></u><em> </em>of energy.
There's no friction, so it'll have <u>that same energy</u> at every point of the story.
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Skip the loop for a moment, because the first question concerns the hill after
the loop. We'll come back to it.
The coaster is traveling 10 m/sat the top of the next hill. Its kinetic energy is
(1/2) (M) (V)² = (400) (10)² = 40,000 joules.
Its potential energy at the top of the hill is (359,200 - 40,000) = 319,200.
PE = (M) (G) (H)
319,200 = (800) (9.8) (H)
H = (319,200) / (800 x 9.8) = <em>40.71 meters</em>
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Now back to the loop:
You said that the loop is 22m high at the top. The PE up there is
PE = (M) (G) (H) = (800) (9.8) (22) = 172,480 joules
So the rest is now kinetic. KE = (359,200 - 172,480) = 186,720 joules.
KE = (1/2) (M) (V)² = 186,720
(400) (V)² = 186,720
V² = 186,720 / 400 = 466.8
V = √466.8 = <em>21.61 m/s</em>
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Now it looks like there should be another question ... that's why they
bothered to tell you that the end is 4m off the ground. They must
want you to find the coaster's speed when it gets to the end.
At 4m off the ground, PE = (M) (G) (H) = (800) (9.8) (4) = 31,360 joules.
The rest will be kinetic. KE = (359,200 - 31,360) = 327,840 joules
KE = (1/2) (M) (V)² = 327,840
400 V² = 327,840
V² = 327,840 / 400 = 819.6
V = √819.6 = <em>28.63 m/s</em> at the end
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If the official answers in class are a little bit different from these,
it'll be because they used some different number for Gravity.
I used '9.8' for gravity, but very often, they use '10' .
If the official answers in class are way way different from these,
then I made one or more big mistakes somewhere. Sorry.
