年級

科目

問題的種類

工程與科技 大學

各位大神 求解 拜託🙏🆘

CA 梁柱 一截面積是原來的兩倍 22. Two girders are made of the same material. Girder A is twice as long as girder B and Bo DE C B³ has a cross-sectional area that is twice as great. The ratio of the mass density of girder A to the mass density of girder B is: 質量密度 A) 4 B) 2 C) 1 D) 1/2 E) 1/4 29. During a short interval of time the speed v in m/s of an automobile is given by v=at² + bt³, where the time t is in seconds. The units of a and b are respectively: A) m-s²; m-s4 s³/m; s4/m dv=at+bt² B) C) m/s²; m/s³ D) m/s³; m/s4 E) m/s4; m/s5 34. The position y of a particle moving along the y axis depends on the time t according to the equation B) C) D) E) 質固定,体2倍 #%# 40. The coordinate of an object is given as a function of time by x = 7t-31², where x is in meters and t is in seconds. Its average velocity over the interval from t = 0 to t = 2 s is: A) 5 m/s X₂=0, X₁₂=14-12=2 B) -5 m/s 11 m/s C) D) -11 m/s (2-0)/2=1 E) -14.5 m/s 41. The coordinate of a particle in meters is given by x(t) = 16t-3.01³, where the time t is in seconds. The particle is momentarily at rest at t = A) 0.75 s A y = at -bt². The dimensions of the quantities a and b are respectively: A) L²/T, L³/T2 B) L/T², L²/T C) L/T, L/T² D) L³/T, T²/L E) none of these 1.3 s 5.3 s 7.3 s 9.3 s V(t) = 16-9+² =- [9t²-16) =-(3t+4) 13t-4) =+= 0.75 45. The velocity of an object is given as a function of time by v = 4t-312, where v is in m/s and t is in seconds. Its average velocity over the interval from t=0 to t=2 s: A) is 0 B) is -2 m/s C) is 2 m/s D) is -4 m/s E) cannot be calculated unless the initial position is given 49. A particle moves along the x axis according to the equation x = 61² where x is in meters and t is in seconds. Therefore: E B) C) D) E) A) the acceleration of the particle is 6 m/s² t cannot be negative the particle follows a parabolic path each second the velocity of the particle changes by 9.8 m/s none of the above

待回答 回答數: 0
數學與統計 大學

大一公衛系微積分,求第二題解

公衛系 微積分期末考 (28/12/2018) 1. Use the Laplace transform to solve the differential equations. (1) j(t)+2y(t) = x(t), y(0)=1, x(t)=10, t20 (20) (2) Intravenous glucose is a treatment. Disposed at a fixed rate k grams per minute inputs into the blood, while blood glucose will be converted to other substances or moved to another place, at a rate proportional to the amount of glucose in the blood, the proportionality constant is a (a> 0), the initial amount of glucose in the blood is M. A. Find the variation in the amount of glucose in the blood (15) B. Determining the equilibrium, the amount of glucose in the blood. (5) = 2. SI Epidemic Model : The size of the population, n+1, remains fixed. Let i(t) be the number of infectives at time t, and let s(t) be the number of individuals who are susceptible. Given an initial number of infectives iO), we would like to know what will happen to i(t). SI Epidemic Model is described by the differential equation. di(t) = k·i(t).s(t) ......(5.1) dt i(t)+s(t)=n+1 i(0)=i, (1) Solve this differential equation of the SI Epidemic Model (5.1). (10 h) (2) What is the peak times t of the epidemic spread? (10) 3. Consider the Two-compartment physiological models and is shown in figure 1. C1 (t) represent the drug concentration in the first compartment and C2 (t) represents the drug concentration in the second compartment. Vi and V2 represent the compartment volume. Use the first order linear differential equation general solution to solve the C1 (t) (20 ) and use the Laplace transform to solve C2 (t). 【20 分). | 世」!()

待回答 回答數: 0