I dont know if I did the problem right check please.

  • Thread starter Thread starter yaho8888
  • Start date Start date
AI Thread Summary
The problem involves calculating the height from which a jet must start a quarter turn to avoid exceeding a maximum acceleration of 4G while diving at 1200 km/hr. The equation used is A = V^2/r, leading to the conclusion that h, the radius of the circular motion, can be derived from the acceleration. The initial calculation yielded a height of approximately 2.834 km, but there were concerns about unit conversions, particularly with acceleration and speed. Corrections were made regarding the conversion of 4G and the need to ensure all units were consistent. The final consensus confirms that the calculations are correct, provided the units are properly managed.
yaho8888
Messages
62
Reaction score
0
[SOLVED] I don't know if I did the problem right check please.

Homework Statement


A jet is diving vertically downward at 1200km/hr
the pilot can withstand a maximum acceleration of 4G, before losing consciousness, at what height must the plane start a quarter turn to pull out of the dive? Assume the speed constant.


Homework Equations



A=V^2/r

The Attempt at a Solution



for this circular motion we know that h = radius of the circle
so
4G=(1200km/hr)^2/h
4G=4*9.8m/s=39.2m/s=508032km/hr^2(Not sure if convert it right)
508032=1200^2/h
h=2.834Km
Is this right, the answer has to be in Km so I convert the 4G.
 
Physics news on Phys.org
If you want the ans in km/hr, convert everything to those units.

4g = v^2/h =>
4*(9.8/1000)/(1/3600)^2 = 1200^2/h. The value of h will be in kms.

EDIT: 1/3600 corrected (1/3600)^2, after being pointed out by yaho8888.
 
Last edited:
Looks good; you incorrectly typed m/s instead of m/s^2 in your 4G calculation, but the math came out OK.
 
Shooting star said:
If you want the ans in km/hr, convert everything to those units.

4g = v^2/h =>
4*(9.8/1000)/(1/3600) = 1200^2/h. The value of h will be in kms.

wouldn't that be 4*(9.8/1000)/(1/3600)^2
 
yaho8888 said:
wouldn't that be 4*(9.8/1000)/(1/3600)^2

Yes. Thanks for pointing out the typo. I'll edit the post.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top