For a short period of time, a m = 3400 kg truck has a frictional driving force of FD= (650- +²) N acting on the wheels, where t is in seconds. At t = 0, the truck has a speed of of v₁. V₁ t = 0 21 Values for the figure are given in the following table. Note the figure may not be to scale. Variable Value m 5- s v2 = m Using the Principle of Linear Impulse and Momentum, a. Determine the speed of the truck when t = 3 s, v2. FD = f(t) Round your final answers to 3 significant digits/figures. m S V₂ t2
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- 62. •A 5-kg object is constrained to move along a straight line. Its initial speed is 12 m/s in one direction, and its final speed is 8 m/s in the opposite Complete the graph of force versus time with direction. F (N) (s) appropriate values for both variables (Figure 7-26). Several answers are correct, just be sure that your answer is internally consistent. Figure 7-26 Problem 62Read the peak power and peak torque from the following curve. Also, find the engine speed that the peak power and peak torque occur at, and ensure this engine speed is in Rad/sec. Torque Power N.m kW 40 35 30 25 8020 70 6015 50 40 10 30 20 5 10 00 0 1000 N (C max) N(C min) N (Pmax) 2000 3000 4000 5000 6000 Engine Speed (RPM) g/kWh 400 300 200 100 0 Specific Fuel ConsumptionCalculate the ankle torque during initial contact when running from the following data set to 1 decimal place. NOTE if your answer is negative put the negative sign in the answer! Fz 1716 N Fx-560 N COP z 0 m Cop x 0.491 m Ankle z 0.069 m Ankle x 0.566 m e to search Answer: Check P Next page 7°C Light ra
- Problem Solving: )An object falls from rest in a medium offering a resistonce. The velocity of the object hefore the ohject reaches the ground is given by the differential egua- tion dv/t + V% = 32, t/sec. uhat is the veloaty of the okject one second after it falls ?The small particle of mass (800 gm) spinning with angular velocity (w₁ = 2.3 rad/s, at r₁ = 400 mm), if we use force (F) to change the Angular velocity to (w₂, and r₂ = 250 mm) the value of F is: W1 Select one: A. None B. F = 5.63 N C. F = 9.47 N D. F = 3.52 N FThe plot below of load vs. extension was obtained using a specimen (shown in the following figure) of an alloy remarkably similar to the aluminum-killed steel found in automotive fenders, hoods, etc. The crosshead speed, v, was 3.3x104 inch/second. The extension was measured using a 2" extensometer as shown (G). Eight points on the plastic part of the curve have been digitized for you. Use these points to help answer the following questions. 1. 900 800 (0.10, 630 ) 700 (0.50, 745) (0.30, 729) 600 (0.20, 699) (0.40, 741.5) 500- (0.004, 458) (0.80, 440 ) - 400 (0.0018, 405) 300 D= 3.3 " 0.03" 200 G=2.0" 100 - 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Extension, inches a. Determine the following quamtities. Do not neglect to include proper units in your answer. Young's Modulus Total elongation Yield stress Ultimate tensile strength Uniform elongation Engineering strain rate Post-uniform elongation b. Construct a table with the following headings, left-to-right: Extension, load, engineering strain,…
- The left side of this equation tells how much energy Q the cylinder gives to the water while it cools. The right side of this equation tells how much energy Q the water and aluminum cup absorb from the cylinder to warm up. Because it is the same energy, they are equal. What is known in this equation? Mcyl 411.7 g, malum 46.5 g, malum+water = 175 g Can you find: mwater =? g Twater = Talum = 20°C (water and cup of room temperature) 90°C, T; = 35°C (hot cylinder and cool "cylinder+cup+water" temperatures) Tcyl kCal Calum = 0.22, Cwater 1 (specific heat of water and aluminum, measured in units kg-°C What are we looking for is Ccul - How we find it? Plug all the numbers into the equation (1), Ccul will be one unknown which you can calculate from the equation. Important, convert all the masses from grams to kilograms! After you find Ccyl, compare it to known value for the copper 0.093(our cylinder is made out of copper). |Ceyl -0.093| % : · 100% 0.093Kindly show the COMPLETE STEP-BY-STEP SOLUTION, DON’T DO SHORTCUTS OF SOLUTION, so I can understand the proccess. I will rate you with “LIKE/UPVOTE," if it is COMPLETE STEP-BY-STEP SOLUTION. If it is INCOMPLETE SOLUTION and there are SHORTCUTS OF SOLUTION, I will rate you with “DISLIKE/DOWNVOTE.” Topics we discussed: Statics of Rigid Bodies, Force System of a Force, Moment of a Force, Moment of a Force-Scalar Formulation, Moment of a Force-Vector Formulation, and Principle of Moment. Thank you for your help.(Q-4) Balance the meter stick with two weights and a rock. 0 cm 100 cm Jrock 100g position (cm) weight (gwt) lever arm (cm) torque (gwt cm) Rock 10.60 XXXXXX XXXXXXXX Weight 1 구군, 60 150 cw Weight 2 93.5 100 (Q-5) Assume that the torques balance and use this fact to calculate the weight of the rock. Z Tecw Złcw = Wrack
- Homework Q1/ A pulley is driven by a flat belt running at a speed of 600 m/min. The coefficient of friction between the pulley and the belt is 0.3 and the angle of lap is 160°. If the maximum tension in the belt is 700 N; find the power transmitted by a belt. [Ans. P= 3.974 kW]The equation of motion of a damped, unforced pendulum is 0 + -0 + sin 0 = 0. The total energy of the pendulum is E = ml²0?/2 + mgl(1 – cos(0)). Suppose a pendulum has m = 2, g = 9.81, l = 3, and coefficient of friction y script program that uses the Modified Euler's method with h the pendulum from a starting position of (0(0), 0(0) = (.97, 0) until the energy falls below 0.01 Joules (use a while loop). Record the steps in matrices Y and T and plot the motion. Turn in your program and plot. 0.5, all in SI units. Write a well-commented 0.01 to simulate the movement of2. Try to write the motion equations of the system under the following conditions, and judge whether it is acceleration, deceleration or uniform speed(The arrows in the figure show the actual direction of torque). TX-IL IN CIL IW TW TL TO TL (4) IL TV TL TH= TL (1) Judge whether the intersection in the figure below is a stable working point? "Tx. Iz K