THXXXXXXXhelpppp

tze

2011-08-15 10:53 pm

a plane of mass 4.42*10^5 kg is turning around at a radius of curvative R with a a constant speed of 900km h^-1.the wings of the plane are tilted at angle20 with the horizontal.
(a) find the upthrust acting on the wings of the plane.
(b)find the value of R

a small ball of mass m is hung by a inextensible string of lengh 0.6M.the string is now connected to a machine such that the ball is set to rotate horizontal at a speed of 200 r.p.m.the string makes an angle"o"(o係一個角度符號黎 ) with the horizontal.
(a) find the angular velocity of the ball
(b) if the mass of ball is 100g,find the size of o

更新1:

點解係COS 20!!!!!!!!!!! 唔係COS 70??? THXXXXXXXXXXXXX 佢話係 tilted at angle20 with the horizontal. 即係同HORIZONTAL係成20度角 MG係同U F成70度角才對啊???

回答 (1)

天同

2011-08-16 12:31 am

✔ 最佳答案

1. (a) Let F be the upthrust force on the wingsResolve F vertically and horizontallyIn the vertical direction: Fcos(20) = (4.42 x 10^5)gwhere g is the acceleration due to gravity, taken to be 10 m/s2Hence, solve for F gives F = 4.7x10^6 N (b) In the horizontal direction,F.sin(20) = (4.42x10^5) x (900/3.6)^2/RSolve for R gives R = 1.7 x 10^4 m = 17 km 2. (a) Angular speed = 200 x 2 x pi/60 s^-1 = 20.94 s^-1[pi = 3.14159…] (b) Let T be the tension in the stringResolve T vertically and horizontallyIn the horizontal direction,T.cos(a) = (100/1000) x (0.6cos(a)) x 20.94^2where a is the angle at which the string makes with the horizontal hence, T = 0.1 x 0.6 x 20.94^2 N = 26.31 N In the vertical direction,T.sin(a) = 0.1g, where g is the acceleration due to gravity, taken to be 10 m/s2i.e. sin(a) = 1/26.31 = 0.038a = 2.18 degrees

2011-08-16 09:21:10 補充：
The upthrusr F is perpendicular to the wings. It is thus making an angle of 20 degrees with the vertical. The vertical component of F is F.cos(20), which balances the weight of the aircraft.

👍 0 👎 0