Answer:
230.4 N
Explanation:
From the question given above, the following data were obtained:
Charge (q) of each protons = 1.6×10¯¹⁹ C
Distance apart (r) = 1×10¯¹⁵ m
Force (F) =?
NOTE: Electric constant (K) = 9×10⁹ Nm²/C²
The force exerted can be obtained as follow:
F = Kq₁q₂ / r²
F = 9×10⁹ × (1.6×10¯¹⁹)² / (1×10¯¹⁵)²
F = 9×10⁹ × 2.56×10¯³⁸ / 1×10¯³⁰
F = 2.304×10¯²⁸ / 1×10¯³⁰
F = 230.4 N
Therefore, the force exerted is 230.4 N
Explanation:
Given that,
Initial speed of the airfield, u = 0
Final speed, v = 27.8 m/s
Acceleration of the airfield, ![a=2\ m/s^2](https://tex.z-dn.net/?f=a%3D2%5C%20m%2Fs%5E2)
Length of the runway, d = 150 m
Let v' is the speed of the airplane to reach the required speed for takeoff. Finding v' using third equation of motion as :
![v'^2-u^2=2ad\\\\v'=\sqrt{2ad} \\\\v'=\sqrt{2\times 2\times 150} \\\\v'=24.49\ m/s](https://tex.z-dn.net/?f=v%27%5E2-u%5E2%3D2ad%5C%5C%5C%5Cv%27%3D%5Csqrt%7B2ad%7D%20%5C%5C%5C%5Cv%27%3D%5Csqrt%7B2%5Ctimes%202%5Ctimes%20150%7D%20%5C%5C%5C%5Cv%27%3D24.49%5C%20m%2Fs)
This speed is not enough as the airfield must reach a speed before takeoff of at least 27.8 m/s. Now, the required length of the runways is :
![v^2=2ax\\\\x=\dfrac{v^2}{2a}\\\\x=\dfrac{(27.8)^2}{2\times 2}\\\\x=193.21\ m](https://tex.z-dn.net/?f=v%5E2%3D2ax%5C%5C%5C%5Cx%3D%5Cdfrac%7Bv%5E2%7D%7B2a%7D%5C%5C%5C%5Cx%3D%5Cdfrac%7B%2827.8%29%5E2%7D%7B2%5Ctimes%202%7D%5C%5C%5C%5Cx%3D193.21%5C%20m)
So, the minimum length of the runways is 193.21 meters.
Answer:
diagram: see image, x-component: 84.3 N, acceleration: 4.38 m/s^2
Explanation:
(see image for further explanation)
Answer:
where is the graph I can't see it how can I solve the problem if I don't see the graph can you show the graph please