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- A marathon runner completes a 42.188-km course in 2 h, 30 min, and 12s. There is an uncertainty of 25 m in the distance traveled and an uncertainty of 1 s in the elapsed time. (a) Calculate the percent uncertainty in the distance. (b) Calculate the percent uncertainty in the elapsed time. (C) What is the average speed in meters per second? (d) What is the uncertainty in the average speed?Consider the equation s=s0+v0t+a0t2/2+j0t3/6+s0t4/24+ct5/120 , were s is a length and t is a time. What are the dimensions and SI units of (a) s0 , (b) v0 , (c) a0 , (d) j0 , (e) s0, and (f) c ?The length and width of a rectangular room are measured to be 3.9550.005m and 10500.005m . Calculate the area of the room and its uncertainty in square mete
- ipa Suppose (p | ) = e¯. What is (x | b) =? V2nh 8 (x – x') 2nh Vasi 8 (x – x') 2πh 8 (x – x') O OHello, I need help with 1.2, plus 1.3 if you can please. Thank you.ANSWER THE FOLLOWING QUESTIONS BELOW AND USE 2 TO 4 DEMICAL PLACES. SHOW YOUR COMPLETE SOLUTIONS. You may paste images of your solutions on each number DETERMINE THE mm READING OF THE FOLLOWING MICROMETER FIGURES SHOWN BELOW: 30 25 20 1. READING: 20 15 10 2. READING: BARREL 25 Det THIVDL 3. 10 READING: 00 45 40 35 4. READING: 20 15 10 10 5. READING: || | | ||
- If F1=2N, and F2= 3.4N, α1=500, α2=400. The value of FR by Cos law is:1. The following data are the measurements of the dimensions of a box: l = 8.6 ± 0.5 cm h = 20.19 ± 0.01 mm w = 0.75 ± .02 mHow should the volume of the box be reported (central value ± uncertainty) incm3? Show solution.2. If the accepted value of the volume of the box in the previous question is 1300 cm3, is the measurement acceptable? Explain.of 1 estion Given the equation f =- where = 3.43±0.78. What is the value and the absolute uncertainty in f (with the correct number of significant figures)? Select one O 0.2 Ob. 0.78 0.8 Od 1.26 O 1 Clear my choice If the length of an object is /=18.0±0.2cm What is the object's length in units of 772? There are 100cm in 1 m Select one: O 0.2±0.2m Ob. 0 1800+20m d. 0.180+0.002m Clear my choice 0.000180±0.000002m
- w = FjAc = F .Aë , W = m(v;)* - m(u,)* , v 1 zm(us)* - m(v,)* , W = Problem 1: (a) à = 3â – 2ŷ, B = 4â& – 5ŷ. à ·B =? Answer: 22 (b) à = -2â + 4ŷ, B = 3âu – lŷ. Ā·B =? Answer: –10 (c) à = 5â – 3ŷ, B = -3â – 5ŷ. ÷B =? Answer: 0 1 | F,dx , W = F. dř , F, dU W = -AU dx d = 4m F = 20 N 0 = 30° m = 4 kg Hk = 0.5) m = 2 kg F = 40 N Hk = 0.2 Ө — 30° Problem 2: In the figure above on the left we have a force F = 20 N pulling a box to the right at an angle 0 = 30° above the horizontal. The box has a mass m = 2 kg, the coefficient of kinetic firction between the box and the ground is uk = 0.2, and the block moves a distance d = 4 m to the right. (a) What is the work done by the force F? Answer: 69.3 J (b) What is the normal force from the ground? Use g = 10 m/sec² Answer: 10 N (c) What is the work done by the kinetic friction force? Answer: -8 J (d) What is Wnet, the net work done on the box? Answer: 61.3 J (e) The box has an initial speed of 4 m/sec. What is the speed of the box after it has…. Suppose [V] = L3, [ρ]=ML–3,[ρ]=ML–3, and [t] = T. (a) What is the dimension of ∫ρdV?∫ρdV? (b) What is the dimension of dV/dt? (c) What is the dimension of ρ(dV/dt)? I'm not sure what this is asking. Thank you for the assistance.One newly discovered light particle has a mass of m and property q. Suppose it moves within the vicinity of an extremely heavy (fixed in place) particle with a property Q and mass M. When the light particle is xi distance from the heavy particle, it is moving directly away from the heavy particle with a speed of vi. a) What is the lighter particle's speed when it is xf away from the heavy particle? Amım2 Consider a new expression for gravitation potential energy as: PEgrav where A is a constant, m1 and m2 are the masses of the two objects, and r r is the distance between them. Moreover, the new particle has an additional interaction with the heavy particle through the following force expression Ob 4TTE, p2 1 Fnew where Eo is a constant that is read as epsilon subscript 0, q and Q are their new properties, r is the distance between the new particle and the heavy particle. Solution: We may solve this using two approaches. One involves the Newton's Laws and the other involving Work-Energy…