gsccrepping
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Wait sorry, how do you calculate the first part to get .5 seconds?Firstly calculate the time taken to do a whole parabola from his starting point to the next time he passes the board. This gives you a value of 0.5 seconds to complete the rest of the journey. Since the velocity is symmetrical u=17.1 m/s.
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Wait sorry, how do you calculate the first part to get .5 seconds?
Just use that formula subbing in t=4, Uy=17.1 and you'll get -10.Wait sorry, how do you calculate the first part to get .5 seconds?
And yeah anybody know how to do the 1st question?How do you do his first question?
yeah thats what i got tooI tried doing your first question and got 0.012 newtons
F = mv^2/r
v = 2pi0.25/6.52
m = 0.05
r = 0.25
= 0.012 newtons.
Where did you get this question from?
I think it's wrong cause 8.5 newtons when dealing with such a slow speed and low mass doesn't sound right
or maybe i'm just doing something wrong.
sameyeah thats what i got too
Pretty sure it was cheltenham girls or some shit. But yeah that's what I got aswell, so I think we're right and the paper is wrong.I tried doing your first question and got 0.012 newtons
F = mv^2/r
v = 2pi0.25/6.52
m = 0.05
r = 0.25
= 0.012 newtons.
Where did you get this question from?
I think it's wrong cause 8.5 newtons when dealing with such a slow speed and low mass doesn't sound right
or maybe i'm just doing something wrong.
The question isn't wrong, it may just be trying to troll anyone as mentions 'closest to that required'Pretty sure it was cheltenham girls or some shit. But yeah that's what I got aswell, so I think we're right and the paper is wrong.
closest is A though (4N) but the answer was B (8N)I got 0.01160845864
The question isn't wrong, it may just be trying to troll anyone as mentions 'closest to that required'
Oh okay. I apologise then.closest is A though (4N) but the answer was B (8N)