有人會第四小題嗎？謝謝

Lv F=40N 60 ON 5kgw ZON do 20 53 4 例5:右圖中,物體原靜,受力 40 N後,4秒內移動了 24 m 試求: (a (1)a= 3 m/s2. (2)摩擦力= N。 (3)F 作功 480],摩擦力作功=-10 (4) 4 秒後不再施力,則物體滑行12m後停止)。(設g=10 m/s) - J。 v=0 m [中) Volt:45)= 3 x 4 = 12/4 2 F F = max 3 = ý xa Ý a=-1 v voi 2015-10-12-25 7 6 2 b=2

「高二化學2-3氣體定律」 這題的E選項怎麼解呢？我怎麼算都是309／308倍……

可用氣體體積 273 ( *( DW3. 有關查理定律:若OC時氣體體積為V,V = V. (1+ 攝氏溫度作圖,下列相關敘述哪些正確? A圖形為一通過原點的斜 直線 V. (B所得直線的斜率約為 所得直線的截距為V 直線可以利用外插法得到大約在-273 ℃時氣體體積為0 外 可以推出温度每上升1℃,體積膨脹為35 ℃時體積的 35° C = 300k Volt 3157 36 35 (14) 309 4. ~ 6. 為題組 23 1 308 |- 1) =|f 313 308 308

抱歉，再詢問一題 為何圖中紅色螢光筆這句的逗號可以加在to前面？明明沒有連接詞、沒有分詞構句，也不是同位語啊(；д；)

One of the main ideas of ecotourism is that people must take responsibility for their own actions while enjoying a natural experience. One goal is to prevent travelers" activities{like eating and washing) from hurting the environment. Ecotourism is known 1463 environmentally-friendly for encouraging environmentally-friendly activities like recycling, energy conservation, 发展 and re-using water as tools for “sustainable development.” “Sustainable development” means doing things in a responsible and careful way today, to make sure that natural 滿足需求。 resources will be able to meet the needs of future generations tomorrow. While enjoying a vacation, ecotourists have lots of opportunities to learn about the their activities affect the planet, and to think about new, more responsible ways to live on the Earth. There are sometimes chances for ecotourists to perform way voltar to 11

用節點怎麼算

CHAPTER 4 BASIC NODAL AND MESH AU sepa 92 W our structure answer. Each component of the structure is accessed rately by name as shown. and at node 2 PRACTICE 4.2 For the circuit of Fig. 4.5, compute the voltage across each current source. 212 Unfortunately, we ha is a direct result of the it is not controlled by a equation that relates i In this case, we find 12 w 432 w 3 A 352 52. TA which upon substituti Reference node FIGURE 4.5 and Eq. [10] simplifi Ans: V3A = 5.235 V; v7A = 11.47 V. Solving, we findt The previous examples have demonstrated the basic approach to nodal analysis, but it is worth considering what Thus, the power sup (-60)(-75) = 4.5k

請問一下這題怎麼算😭好難

Vout - 5 volt D2 +5 vol o Vout D OVORI = Vout == 4.310 v ok suming the diodes are | (b) tions shown in Figure P9.23 D -9.3 大 li (c) WO Do V 12 +7 V o bin 88. I won T08 . 9.24 Sketch the output waveform and the voltage transfer characteristic for the circuit of Figure P9.24. Assume an ideal diode, us (t) = & sin (it), V, = XV, Ri = 812, and R2 = $ 12. 10 4 lo lo mm ogiloc www bools boRi t V. -Vo 15 k12 15 k92 1 vs borbo R2 V 4V V c) (d) bagnozib Figure P9.24 OU 29,22, find the range of Vin 25 Repeat Problem 9.24, using the offset diode model

二極體電路問題

Given the cirsuit. All resistors are as specified. Give the voltage of r14. vsrc 11 = 4V, vsrc 12 = 4V vsrc12 vsrc11 r11 40 r13 40 r15 20 Hw d1 w r12 r14 20 40 HII

請問第十三題為什麼可以用藍筆的圖解 同樣的座標為什麼要拆開寫

abo 2 G 1. 答:11.10[道明模考】 13Exi 4 (-4, - 8) • B(-1,1). C(1, 7) teria L : 8x=y+47 74, y=3x09 feb! 一多多 上,設動點P在直線L上,求3PA + 2PB + PC之最小值為1。 PA =}x+++(32+12)? PB = T[qx+1) Jolx+4 PC? Tiox+1 3JpX44 21012H14 voltz-4, Winx to no 14試問下列各選項之加值,何者會使得方程式+3\x + 4-2 = 0 無解?(1)5 (2) - -3 (4) -6 (5) (6) 3 (30)x= 10 10 3° 答:(3) 【北區聯合模考 y=-3x+4+2 之圖形交點, y = m=- go 1 作圖如右,則無解台-3<m <- 2 m=-3 Ay 【解】:考慮 = mx -4-4-4 t ( (3+3+50510 | [ 1 - 115io!

求解第四第五題!!!電機的 謝謝😭🥺🥺🥺 拜託看到盡快><

2. The circuit in Fig. 2 represents an energy storage bank with vi(0)=-3 V, v2(0) = 5 V, v3(0)=-5 V. (a) If the switch Si is closed at tı > 0, find the values of vi, v2, and v3 after ti. (b) Following (a), if the switch S2 is closed at 12 >ti. Find the final values of vi, vz, and va after 12. 3. Let iz(0) = 0. Find the current il(t) of the circuit shown in Fig. 3 for t> 0. 22 w 22 32 1 ] 322 + 8t u(t) A 22 2H 1:6 8u(1) V 台Fv() 3i Fig. 3 Fig. 4 4. Let vo(0-) = 0. Find the voltage ve() and current i(1) denoted in Fig. 4 for t> 0. 5. The switch shown in Fig. 5 is closed for a long time before t = 0. When the switch is open at t = 0, find i(t) for t < 0 and > 0. i=0 =0 212 622 W + 392 8 V 0.510) 2=vd) - Fig. 5 FE final Sophie de EES-B22.pdf 20210611_115335.jpg 20210611_115320.jpg 18.pdf

看不懂他給的公式 位移怎麼算

mz 40 kg (3)施力期間的位移為 xz 1 1 = Volt + at = 0+,×6×0.5° = 0.75 m(向東) 2 2 (4)因兩人受力作用的時間應相同,則 VP = ap Xt=8×0.5 = 4 m/s(向西) Vz = az xt=6×0.5 = 3 m/s(向東)

請問為什麼-1不合 謝謝

3 jy平面上,一光線通過(3,4) ,經y軸反射後與實心圓盤相切,而此實心圓盤外圍方程 式為: C-D7+0y-2)" =2。設原光線打在y軸上的點坐標為4(0,o) ,則實數g= 遍 ! (3.4對y軸的對稱點為(3,4) 。 設過(一3,4)對圓的切線為一4三訪(x+3) 4 疙 監」 2 全 注 讀 (4zz二2)? ==27 +2 人zz2 十世 2 由 一14z77 二1677十2三0 人有訪 或-1(丰合) VOLtA 可知切線為 py-4=-7G+3) 了 及 中全D 人証信2在施 2 6x一42一2避 0及三直線 和信司光還 。則,在圓C 平面上有一個圓C: 二 蘆蝦 平面 關和2 2和下直線好 的距離為正整數值。