請問3-7這題的答案是什麼？🥹

1項 9 對方 基 EX3.6F=ABC+BCD+ABCD+AB'C'化簡至最簡 AB EX 上 ai 以後<50 (0 F=BC+BD+ACD " 10/ *EX37 FLABCD)=Σ (0.2.3.5.7.8.9.10.11.13.15) EX 3.1 且 r AB CD 09 01 "\ 10 方格 00 就是可 01 示出 EX "\ 然 (So 10 ☆5变权 很困难用人工化簡(字体) K-mapstoF: (A.B.C.D.E) EX . BDE AGO A=0 BC DEA= 10 09 10 X+5= (3

這題有沒有可能只用反函數就可以解出來⋯ 為啥算不出答案，ㄚ。

Assignment (part I) (page 50) *2-8. Two forces are applied at the end of a screw eye in order to remove the post. Determine the angle 0 (0° ≤ 0 ≤ 90°) and the magnitude of force F so that the resultant force acting on the post is directed vertically upward and has a magnitude of 750 N. 30° 500 N 2025/2/25 CHIANG Yu-Chou 33

有人會這些東西嗎🥹感謝

3. 如下圖為一四連桿機構,Q2Q4為固定桿,Q2A 線段為輸入桿,設反時針方向之角速度 為100 rpm, Q.B線段長40cm,試用圖解法求A、B及C三點之速度大小(C為QB之中點), 繪圖比例1cm=10cm。 9₂ B A 60° 55⁰ 75 cm 460⁰ A, B 30 cm C 40cm La Wi=1nonyienc 4. 向量 長 25 公分,向量長 50 公分,二者之間夾60°,求Aa➔BD線段大小及方向。 25 Aa Bb E 54334 2

想請問一下L=f(～,~,~,~) F括號是什麼意思 裡面的條件用逗號隔開是什麼意思

Hence, given shape of airfoil at a given angle of attack, L = f(V∞, P, S, μloo, A∞) (5.3) and D and M are similar functions. In principle, for a given airfoil at a given angle of attack, we could find the variation of L by performing myriad wind tunnel experiments wherein V... Po and a.. are individually varied and then could twee to mala out 11

請問27題謝謝🙏

量 20 kg 的女孩 六。他們在無摩 弓孩受到200 N 度以及繩中的 200 N : 所示的滑 「施加的作 ;(b) 使它 T-10=U mi m2 120g 29 圖 5.30 27. (1) 一架 12,500 kg 的 F4 幽靈式噴射機的總 推進力為 160 kN。從航空母艦的甲板上彈 射。 在 2.1 s 內到達 80 m/s 的起飛速率。 (a) 蒸氣彈射器所施加的(固定)作用力為 何? (b) 機上 70 kg 的飛行員之視重量為何? 28. (1) 兩物塊質量為 3 kg 及 5 kg,跨懸於滑輪 兩側,如圖 5.31 所示。5 kg 的木塊起初是 高於地面4m,然後釋放。3 kg 之木塊可 達到的最大高度為何?

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

我想要知道這題的解題過程，算了好幾次都不對..

13.6474.204 練習題8.12 試求圖 8.30 電路的I。 I 答:9.546/33.8° A. 142 -j312 8 92 + 45/30° V j52 000 5 S2 1012 -j2 2 8.30 練習題 8.12 的電路

想問一下第二題的C選項是對的還是錯的 不確定答案是多少🙏

“股汽油 隨堂練習7-6 1.決定引擎性能之好壞,不包括下列哪一種? (A)燃料消耗率(B)單位重量的扭力(C)單位馬力的重量(D)單位排氣量的馬力。 有關引擎性能之敘述,下列何者錯誤? (A) 汽油引擎較柴油引擎的熱效率為低(B)柴油引擎的制動平均有效壓力較汽油引擎為 低 (C)壓縮比愈大,則容積效率愈高(D) 汽油引擎之制動熱效率約25~30%。 9 「以

該如何拉式轉換

y +4y=+ Y(0) = 4(0)=1 z L {Y'3+ 1 {y} = {{t} SYLS)-S Y ) - Y X 14/15) = 5 (374) YIG) = 1 / 2 + 5+1 1 Y (6)= 3 (5²4 + Y Stl 64 6) 413 $*$%) = { "[40:3+2 (en S184) 3 1974

想請問各位大大 我的算法是對的嗎

VER U. 試求圖 EP3.8a 中,立方結晶面的米勒指數。 解 將此平面沿y軸向右平移量單位,如圖 EP3.86 所示,接著把右後下角當成新原 點,新平面和x軸之截距是1個單位長,所以新平面和各軸之截距是(+1, -, c)。 取其倒數後可寫為(1, -, 0),最後取簡單整數比,得到平面米勒指數 (5120)。 z (1/2 - 1 = 3 - 2 子, 新原點 y X 上 Alw (b) (6)