Answer:
19.3m/s
Explanation:
Use third equation of motion
where v is the velocity at halfway, u is the initial velocity, g is gravity (9.81m/s^2) and h is the height at which you'd want to find the velocity
insert values to get answer
![v^2-0^2=2(9.81m/s^2)(38/2)\\v^2=9.81m/s^2 *38\\v^2=372.78\\v=\sqrt[]{372.78} \\v=19.3m/s](https://tex.z-dn.net/?f=v%5E2-0%5E2%3D2%289.81m%2Fs%5E2%29%2838%2F2%29%5C%5Cv%5E2%3D9.81m%2Fs%5E2%20%2A38%5C%5Cv%5E2%3D372.78%5C%5Cv%3D%5Csqrt%5B%5D%7B372.78%7D%20%5C%5Cv%3D19.3m%2Fs)
Answer:
The correct option is;
Force of Friction
Explanation:
As coach Hogue rode his motorcycle round in circle on the wet pavement, the motorcycle and the coach system tends to move in a straight path but due to intervention by the coach they maintain the circular path
The motion equation is
v = ωr and we have the centripetal acceleration given by
α = ω²r and therefore centripetal force is then
m×α = m × ω²r = m × v²/r
The force required to keep the coach and the motorcycle system in their circular path can be obtained by the impressed force of friction acting towards the center of the circular motion.
Answer:
38 cm from q1(right)
Explanation:
Given, q1 = 3q2 , r = 60cm = 0.6 m
Let that point be situated at a distance of 'x' m from q1.
Electric field must be same from both sides to be in equilibrium(where EF is 0).
=> k q1/x² = k q2/(0.6 - x)²
=> q1(0.6 - x)² = q2(x)²
=> 3q2(0.6 - x)² = q2(x)²
=> 3(0.6 - x)² = x²
=> √3(0.6 - x) = ± x
=> 0.6√3 = x(1 + √3)
=> 1.03/2.73 = x
≈ 0.38 m = 38 cm = x
Work done is equal to force by distance; so you take the force exerted, in newtons, and multiply that by the direction it's moved (from the starting point in a line, not along the path it's taken.)
We determine the electric potential energy of the proton by multiplying the net electric potential to the charge of the proton. The net electric potential is the difference of the final state to the that of the initial state. So, it would be 275 - 125 = 150 V.
electric potential energy = 150 (<span>1.602 × 10-19) = 2.4x10^-17 J</span>
