Complet the solution : Vavg Ti Te Ts Q hexp Nuexp htheo Re Nutheo Error (m/s) (*C) (*C) (*C) (W) 2.11 18.8 21.3 45.8 2.61 18.5 20.8 46.3
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- The velocity (m/s) of a body is given as a function of time (seconds) by v(t) = 200ln(1+f)-t, t20 Using Euler's method with a step size of 5 seconds, the distance traveled in meters by the body from t-2 to t=12 seconds is most nearly Select one: a. 5638.0 b. 3939.7 c. None d. 39397 e. 3133.1= MMB 241- Tutorial 1.pdf 2/3 80% + + 10. Determine a ats = 1 m v (m/s) 4 s (m) 2 11. Draw the v-t and s-t graphs if v = 0, s=0 when t=0. a (m/s²) 2 t(s) 12. Draw the v-t graph if v = 0 when t=0. Find the equation v = f(t) for each a (m/s²) 2 segment. 2 -2 13. Determine s and a when t = 3 s if s=0 when t = 0. v (m/s) 2 t(s) t(s) 2(I) [40 Points] Using centered finite difference approximations as done in class, solve the equation for O: d20 dx² + 0.010+ Q=0 subject to the boundary conditions shown in the stencil below. Do this for two values of Q: (a) Q = 0.3, and (b) Q= √(0.5 + 2x)e-sinx (cos(5x)+x-0.5√1.006-x| + e −43*|1+.001+x* | * sin (1.5 − x) + (cosx+0.001 + ex-1250+ sin (1-0.9x)|) * x - 4.68x4. For Case (a) (that is, Q = 0.3), use the stencil in Fig. 1. For Case (b), calculate with both the stencils in Fig. 1 and Fig 2. For all the three cases, show a table as well as a plot of O versus x. Discuss your results. Use MATLAB and hand in the MATLAB codes. 1 0=0 x=0 2 3 4 0=1 x=1 Fig 1 1 2 3 4 5 6 7 8 9 10 11 0=0 x=0 0=1 x=1 Fig 2
- How do you calculate the instantaneous velocity from a velocity vs time graph? Eg. From the image below, Calculate the instantaneous velocity of V(4) at t(4),etc.I am trying to compute the probability of collision with the monte carlo method in MATLAB. But I get an error in teh following code. The error is the matrix P_01 must be semi definete and P_02 must be symmetric. How should I fix this problem? % Initial Mean and Covariance mu_01 = [153446.180; 41874155.872; 0; 3066.875; -11.374; 0] / 1000; P_01 = [6494.080 -376.139 0 0.0160 -0.494 0; -376.139 22.560 0 -9.883e-4 0.0286 0; 0 0 1.205 0 0 -6.071e-5; 0.0160 -9.883e-4 0 4.437e-8 -1.212e-6 0; -0.494 0.0286 0 -1.212e-6 3.762e-5 0; 0 0 -6.071e-5 0 0 3.390e-9] * 1e-6; mu_02 = [153446.679; 41874156.372; 5.000; 3066.865; -11.374; -1.358e-6] / 1000; P_02 = [6494.224 -376.156 -4.492e-5 0.0160 -0.494 -5.902e-8; -376.156 22.561 2.550e-6 -9.885e-3 0.0286 3.419e-9; -4.492e-5 2.550e-6 1.205 -1.180e-10 3.419e-9 -6.072e-5; 0.0160 -9.883e-3 -1.180e-10 4.438e-8 -1.212e-6 -1.448e-13; -0.494 0.0286 3.419e-9 -1.212e-6 3.762e-5 4.492e-12; -5.902e-8 3.419e-9 -6.072e-5 -1.448e-13 4.492e-12 3.392e-9] * 1e-6;…Find the F thrust and F frictional force using the following info Gr - 11
- I am computing the probablility of collision with the Monte Carlo method in MATLAB. The following is my code for it. The results don't seem to be accurate. Is there something wrong with my code? Also, it takes more than 5 min to run this code. Can you tell me why? %Initial Mean and Covariance mu_01 = [153446.180; 41874155.872;0;3066.875;-11.374;0]/1000; P_01 = [6494.080 -376.139 0 0.0160 -0.494 0; -376.139 22.560 0 -9.883e-4 0.0286 0; 0 0 1.205 0 0 -6.071e-5; 0.0160 -9.883e-4 0 4.437e-8 -1.212e-6 0; -0.494 0.0286 0 -1.212e-6 3.762e-5 0; 0 0 -6.071e-5 0 0 3.390e-9]*1e-6; mu_02 = [153446.679; 41874156.372;5.000;3066.865;-11.374;-1.358e-6]/1000; P_02 = [6494.224 -376.156 -4.492e-5 0.0160 -0.494 -5.902e-8; -376.156 22.561 2.550e-6 -9.885e-3 0.0286 3.419e-9; -4.492e-5 2.550e-6 1.205 -1.180e-10 3.419e-9 -6.072e-5; 0.0160 -9.883e-3 -1.180e-10 4.438e-8 -1.212e-6 -1.448e-13; -0.494 0.0286 3.419e-9 -1.212e-6 3.762e-5 4.492e-12; -5.902e-8 3.419e-9 -6.072e-5 -1.448e-13 4.492e-12 3.392e-9]*1e-6;…Can you explain how to get the graphs and what the equations mean? I'm also confused for the fill in the blanks portion. Please and thank you!Here I'm finding final kinetic energy KE2, given initial kinetic energy, initial force, alpha N/m^3 and beta N representing F(x). I integrated F(x)dx though I'm not sure how to integrate Fo with respect to X. I tried just finding the integrand ouput of 7.5x10^4m and subtracting the initial force, since the initial force would count as x=0, or I assumed. I then followed the regular approach of adding over the initial KE1 to isolate the unknown KE2 which I keeo getting as 1.07x10^10 which seems to be off. Again, I suspect where I'm going wrong is integrating the initial force -3.5x10^6N with respect to x, but other than what I've already done, that doesn't make sense to me.