Q. The energy emitted from the sun is a product of ________.
A. Fusion
Given from the problem :mass m = 413 kg;coefficient of friction u = 0.0163;acceleration due to gravity g = 9.8 m/s2;inclined angle @1 = 14.3;inclined angle @2 = 4.69;distance travelled d = 175 m;applied fore F = 410 N; the component of the force from the donkey in the direction of motion isF2 = F1
[email protected]= 397.2964498768165 N
Fy = N - mg
[email protected] = 0N = mg
[email protected] = 4037.964151113007 NFx = F2 - mg
[email protected] - f = mahere f = u N=65.8188156631420141
F2 - mg
[email protected] - f = maa = F2 - mg
[email protected] - f/ m=0.31923412183075155 m/s^2
work done by donkeyW = F2 d=69526.8787284428875 J
Answer:
85.8 m/s
Explanation:
We know that the length of the circular path, L the plane travels is
L = rθ where r = radius of path and θ = angle covered
Now,its speed , v = dL/dt = drθ/dt = rdθ/dt + θdr/dt
where dθ/dt = ω = angular speed = v'/r where v' = maximum speed of plane and r = radius of circular path
Now, from θ = θ₀ + ωt where θ₀ = 0 rad, ω = angular speed and t = time,
θ = θ₀ + ωt = 0 + ωt = ωt
So, v = rdθ/dt + θdr/dt
v = rω + ωtdr/dt
v = (r + tdr/dt)ω
v = (r + tdr/dt)v'/r
v = v' + tv'/r(dr/dt)
v = v'[1 + t(dr/dt)/r]
Given that v' = 110 m/s, t = 33.0s, r = 120 m and dr/dt = rate at which line is shortened = -0.80 m/s (negative since it is decreasing)
So, v = 110 m/s[1 + 33.0 s(-0.80 m/s)/120 m]
v = 110 m/s[1 + 11.0 s(-0.80 m/s)/40 m]
v = 110 m/s[1 + 11.0 s(-0.02/s)]
v = 110 m/s[1 - 0.22]
v = 110 m/s(0.78)
v = 85.8 m/s
