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Please help me with this question (1 Viewer)
- Thread starter craig93
- Start date
Hmm...very interesting question. But first I have to ask
a) is this the whole question?
b) do you have the answer (or even solution) to this?
c) where is this question from?
d) how many mark question would this be?
Because it seems to be referencing to 3u maths and rates of change, but knowing Engineering you shouldn't need to use this, and it should be logically very simple.
a) is this the whole question?
b) do you have the answer (or even solution) to this?
c) where is this question from?
d) how many mark question would this be?
Because it seems to be referencing to 3u maths and rates of change, but knowing Engineering you shouldn't need to use this, and it should be logically very simple.
your right it is pritty easy actually firstly with a max power of 300 and a efficiency of 82% you need to find the "real power" 300 x 82 = 246kw
100
now finding the max velocity of the plane 275 x 246 = 541.2
125
as the plane is travelling 275kmh to the left it is left with 541.2-275 = 266.2kmh it can use this velovity in the up direction. this drawing a force triangle it has 275 to the left and 266.2 up. then both the angle and resultant can be found( angle = 44 resultant = 383kmh
100
now finding the max velocity of the plane 275 x 246 = 541.2
125
as the plane is travelling 275kmh to the left it is left with 541.2-275 = 266.2kmh it can use this velovity in the up direction. this drawing a force triangle it has 275 to the left and 266.2 up. then both the angle and resultant can be found( angle = 44 resultant = 383kmh
I didn't understand your notation there, why is there a 100 and a 125 on separate lines? And what symbol is supposed to go between 275 and 246 to get 541.2?boss302 said:your right it is pritty easy actually firstly with a max power of 300 and a efficiency of 82% you need to find the "real power" 300 x 82 = 246kw
100
now finding the max velocity of the plane 275 x 246 = 541.2
125
as the plane is travelling 275kmh to the left it is left with 541.2-275 = 266.2kmh it can use this velovity in the up direction. this drawing a force triangle it has 275 to the left and 266.2 up. then both the angle and resultant can be found( angle = 44 resultant = 383kmh
Also, what is the question asking? It says find the rate of climb, is that referring to velocity??? That doesn't sound right
the 100 and 125 are on sperate lines due to a justification error they are ment to be below the underline (deviding the number on top) and secondly the rate of clime refers to the angle of attack so the max rate of climb is 44 degrees. i just found the velocity as it would usually ask for both