Explanation:
Here is the complete question i guess. The jet plane travels along the vertical parabolic path defined by y = 0.4x². when it is at point A it has speed of 200 m/s, which is increasing at the rate .8 m/s^2. Determine the magnitude of acceleration of the plane when it is at point A.
→ The tangential component of acceleration is rate of increase in the speed of plane so,
→ Now we have to find out the radius of curvature at point A which is 5 Km (from the figure).
dy/dx = d(0.4x²)/dx
= 0.8x
Take the derivative again,
d²y/dx² = d(0.8x)/dx
= 0.8
at x= 5 Km
dy/dx = 0.8(5)
= 4
![p = \frac{[1+ (\frac{dy}{dx})^{2}]^{\frac{3}{2} } }{\frac{d^{2y} }{dx^{2} } }](https://tex.z-dn.net/?f=p%20%3D%20%5Cfrac%7B%5B1%2B%20%28%5Cfrac%7Bdy%7D%7Bdx%7D%29%5E%7B2%7D%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%20%7D%20%20%20%7D%7B%5Cfrac%7Bd%5E%7B2y%7D%20%7D%7Bdx%5E%7B2%7D%20%7D%20%7D)
now insert the values,
![p = \frac{[1+(4)^{2}]^{\frac{3}{2} } }{0.8} = 87.62 km](https://tex.z-dn.net/?f=p%20%3D%20%5Cfrac%7B%5B1%2B%284%29%5E%7B2%7D%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%20%7D%20%20%7D%7B0.8%7D%20%20%3D%2087.62%20km)
→ Now the normal component of acceleration is given by
= (200)²/(87.6×10³)
aₙ = 0.457 m/s²
→ Now the total acceleration is,
![a = [(a_{t})^{2} +(a_{n} )^{2} ]^{0.5}](https://tex.z-dn.net/?f=a%20%3D%20%5B%28a_%7Bt%7D%29%5E%7B2%7D%20%2B%28a_%7Bn%7D%20%29%5E%7B2%7D%20%5D%5E%7B0.5%7D)
![a = [(0.8)^{2} + (0.457)^{2}]^{0.5}](https://tex.z-dn.net/?f=a%20%3D%20%5B%280.8%29%5E%7B2%7D%20%2B%20%280.457%29%5E%7B2%7D%5D%5E%7B0.5%7D)
a = 0.921 m/s²
Answer:
Ans= 9
See attached picture for clearer solution.
Explanation:
The net electrostatic force acting on charge A = 2/ 2 + 2 /(2) 2 − 2 /(3) 2 = 2 / 2 (1 + 1/4 – 1/9 ) = 41/36 2/2 .
The net electrostatic force acting on charge B = 2/2 + 2/(2)2 − 2/2 = 1/4 2/d2 .
The net electrostatic force acting on charge C = 2/2 + 2/(2)2 + 2/2 = 2/2 (1 + 1 4 + 1) = 9/4 2/2 .
The net electrostatic force acting on charge D = 2/2+ 2 /(2)2 + 2/(3)2 = 2 /2 (1 + 1/4 + 1/9 ) = 49/36 2/ 2 .
The ratio of the largest to the smallest net force = 9/4*2/2 / 1/4 2/2 . = 9
angular vel to tangential vel
v=r omega
v = 56 x 100/60 x 2 pi
v = 56x5/3x6
v=560m/s as estimate
100 revs, 5.00m
Given that,
radius, r = 4.23 x 10∧7 m
Period, T = ?
Since, we know that,
In a geosynchronous satellite, period is equal to the period of earth that is 24 hrs.
Therefore, Time period is equal to 24 hours.
uhhhh idk cheif...
thats a big oof right there.
ged is always an option
