Dhibaatooyinka lagu xalliyay Fiisigiska Aasaasiga ah

1. A capacitor-ka is barbar socda leeyahay bilowga ah awoodda C, ogolaanshaha booska xorta ah is ε_o, aagga saxanka is A, masaafada u dhaxaysa saxamada is d. Haddiie Bedka saxanka ayaa kordhay 4 jeer, masaafada u dhaxaysa saxanka wuxuu noqonayaa 2d iyo ogolaanshaha booska xorta ah waa 5ε_o, waa maxay awoodda ugu dambeysa ee capacitor-ka saxanka barbar socda.

La yaqaan:

Awoodda capacitor-ka = C

Togolaanshaha booska xorta ah = ε_o

Paag dambe =A

The masaafada inta u dhaxaysa saxamada = d

SE buska: Awoodda kaabsoosha (C)

Xalka:

2. Kaaliyaha leh awoodda ugu weyn iyadoo lagu saleynayo garaafka hoose is...

Xalka:

Qaacidada kaabsoosha-saxanka barbar socda:

Awoodda capacitors:

Capacitor-ka leh awoodda ugu weyny waa capacitor C_3.

3.

Ka fiirso arrimaha soo socda!
(1) Joogtada Dielectric
(2) farqiga suurtagalka ah inta u dhaxaysa taarikada
(3) Saxanka thick
(4) Bedka dusha sare ee saxanka
(5) Masaafada u dhaxaysa saxamada
(6) Tirada kharash koronto
Arrimaha saameeya awoodaha taarikada barbar socda capacitors-ku waa…

Xalka:

(1), (4) iyo (5)

4. Isbarbardhigga awoodda ee capacitor-ka 1 iyo 2 waa…

La yaqaan:

Capacitor 1:

Aagga dusha sare = 2A

Masaafada u dhaxaysa saxanadaha =d₁

Capacitor 2:

Aagga dusha sare =A

Masaafada u dhaxaysa saxanadaha = 2 maalmood₁

SE buska: Isbarbardhigga awoodda ee capacitors 1 iyo 2

Xalka:

5.

Go'aami isbarbardhigga awoodda Kaabsadeyaal is barbar socda I iyo II.

Xalka:

Isbarbardhigga awoodda Kaadhadhka is barbar socda ee I iyo II:

6. Laba kaabayaal oo is barbar socda oo lagu muujiyay sawirka hoose.

Haddii A₁ = ½ A₂ iyo d₂ = 3 maalmood₁ ka dibna go'aamiso saamiga awoodaha ee barbar-capacitor-ka saxanka inta u dhaxaysa sawirka 2 iyo sawirka 1.

La yaqaan:

Kaaliyaha saxanka is barbar socda I:

A₁ = 1

d₁ = 1

Kondensarka is barbar socda II:

A₂ = 2

d₂ = 3

SE buska: saamiga awoodaha capacitor-ka saxanka barbar socda ee u dhexeeya sawirka 2 iyo sawirka 1

Xalka:

Kaaliyaha saxanka is barbar socda I:

Kondensarka is barbar socda II:

Tsaamiga awoodaha ee barbar-dhigga-capacitor-ka saxanka II iyo I:

Akhri wax dheeraad ah

1. Go'aaminta awoodda korantada meel ku taal 1 cm marka loo eego dallacaadda 5.0 μC. Joogtada Coulomb (k) = 9 x 10⁹ Nm²C^-2, 1 μC = 10^-6 C.

La yaqaan:

Masaafada laga joogo kharashka (r) = 1 cm = 1/100 m = 0.01 m = 10^-2 m

eedayn (q) = 5.0 μC = 5.0 x 10^-6 C

Joogtada ah ee Coulomb (k) = 9 x 10⁹ Nm²C^-2

La doonayo: Awoodda korontada (V)

Xalka:

Awoodda korontada:

Awoodda korontada waa 4.5 x 10⁶ Waat

2. eedayn Q₁ = 5.0 μC iyo lacag Q₂ = 6.0 μC. Joogtada ah ee Coulomb (k) = 9 x 10⁹ Nm²C^-2, 1 μC = 10^-6 C. Barta A waxay ku taal inta u dhaxaysa lacagaha. Go'aami awoodda korantada ee barta A.

La yaqaan:

eedayn Q₁ = -5.0 μC = -5.0 x 10^-6 C

The masaafada dhibicda A laga bilaabo Q₁ = 10 cm = 0.1 m = 10^-1 m

eedayn Q₂ = 6.0 μC = 6.0 x 10^-6 C

Masaafada barta laga bilaabo Q₂ = 10 cm = 0.1 m = 10^-1 m

Joogtada ah ee Coulomb (k) = 9 x 10⁹ Nm²C^-2

La doonayo: Awoodda korantada ee barta A

Xalka:

Awoodda korontada 1 :

Awoodda korontada 2 :

Awoodda korantada ee barta A :

V = V₂ - V₁

V = (54 – 45) x 10⁴

V = 9 x 10⁴

3. eedayn q₁ = 5.0 μC iyo lacag q₂ = 6.0 μC. Joogtada ah ee Coulomb (k) = 9 x 10⁹ Nm²C^-2, 1 μC = 10^-6 C. Go'aami awoodda korantada ee barta A.

La yaqaan:

eedayn Q₁ = -5.0 μC = -5.0 x 10^-6 C

Masaafada barta A laga bilaabo Q₁ = 40 cm = 0.4 m = 4 x 10^-1 m

eedayn Q₂ = 6.0 μC = 6.0 x 10^-6 C

Masaafada barta A laga bilaabo Q₂ = 50 cm = 0.5 m = 5 x 10^-1 m

Joogtada ah ee Coulomb (k) = 9 x 10⁹ Nm²C^-2

La doonayo: Awoodda korantada ee meesha A

Xalka:

Awoodda korontada 1 :

Awoodda korontada 2 :

Awoodda korantada ee barta A :

V = V₁ + V₂

V = (-11.25 + 10.8) x 10⁴

V = -0.45 x 10⁴

V = -4.5 x 10³

Awoodda korantada ee barta A waa -4.5x10³ Waat

Akhri wax dheeraad ah

1. Elektaroonku wuxuu ka soo kacaa nasashada iyada oo loo marayo farqi suurtagal ah oo ah 12 V. Waa maxay isbeddelka ku jira tamarta awoodda korontada elektarooniga?

La yaqaan:

The waa lacag elektaroon (e) = -1.60 x 10^-19 Coulomb

Awoodda korontada = danab (V) = 12 Volt

SE buska: Isbeddelka tamarta korontada ee elektaroonada (ΔPE)

Xalka:

ΔPE = q V = (-1.60 x 10^-19 C)(12 V) = -19.2 x 10^-19 Joule

Calaamadda laga jaray waxay muujinaysaa in tamarta suurtagalka ah ay hoos u dhacayso.

2. Laba saxan oo is barbar socda ayaa la dallacayaa. Kala-soocidda u dhaxaysa saxannada waa 2 cm iyo cabbirka korontada Inta u dhaxaysa saxannada waa 500 Volt/mitir. Waa maxay isbeddelka tamarta suurtagalka ah ee proton-ka marka laga soo kaco saxanka togan ee lagu dallacay una gudbo saxanka taban ee lagu dallacay.

La yaqaan:

Cabbirka garoonka korantada ee u dhexeeya taarikada (E) = 500 Volt/mitir

The masaafada inta u dhaxaysa saxanadaha (yada) = 2 cm = 0,02 m

Kharashka ku jira proton = +1.60 x 10^-19 Coulomb

La doonayo: Isbeddelka tamarta korontada ee suurtogalka ah (ΔPE)

Xalka:

Awoodda korontada:

V = E s

V = (500 Volt/m)(0.02 m)

V = 10 Volt

Isbeddelka ku yimaada tamarta korontada ku shaqaysa:

ΔPE = q V

ΔPE = (1,60 x 10^-19 C) (10 V)

ΔPE = 16 x 10^-19 Joule

ΔPE = 1.6 x 10^-18 Joule

3. Laba dhibcood oo lagu dallaco waxaa loo kala saaraa masaafo dhan 10 cm. Dalacaadda barta A =+9 μC oo ku dallac barta B = -4 μC. k = 9 x 10⁹ Nm²C^-2, 1 μC = 10^-6 C. Waa maxay isbeddelka tamarta korontada ee korantada ee ku jirta heerka B haddii loo dedejiyo heerka A?

La yaqaan:

Dalac A (q)₁) = +9 μC = +9 x 10^-6 C

eedayn B (q₁) = -4 μC = -4 x 10^-6 C

k = 9 x 10⁹ Nm²C^-2

Masaafada u dhaxaysa khidmadaha A iyo B (r) = 10 cm = 0.1 m = 10^-1 m

La doonayo: Isbeddelka tamarta korontada ee suurtogalka ah (ΔEP)

Xalka:

Akhri wax dheeraad ah

1. Dharka korontada E = 8000 N/C oo dhex marta dhul siman oo laba jibbaaran A = 10 m². Go'aaminta qulqulka korantada.

La yaqaan:

Baaxadda goobta korantada (E) = 8000 N/C

Bedka (A) = 10 m²

θ = 0^o (xagasha u dhaxaysa jihada goobta korantada iyo xariiq si toosan u jiidan aagga)

SE buska: Dareeraha korantada ((Φ)

Xalka:

Qaacidada qulqulka korantada:

Φ = EA cos q

Φ = qulqulka korantada (Nm²/C), E = garoon koronto (N/C), A = bedka (m²), q = xagal u dhexeeya khadka garoonka korontada iyo khadka caadiga ah.

Dareeraha korantada:

Φ = EA cos q = (8000)(10)(cos 0) = (8000)(10)(1)) = = 80,000 8 x 10⁴ Nm²/C

2. Goob koronto oo isku mid ah E = 5000 N/C ayaa dhex marta dhul siman oo laba jibbaaran A = 2 m². Go'aami qulqulka korantada.

La yaqaan:

Goobta korontada (E) = 5000 N/C

Bedka (A) = 2 m²

θ = 60^o (xagasha u dhaxaysa jihada goobta korantada iyo xariiq si toosan u jiidan aagga)

La doonayo: Dareeraha korantada ((Φ)

Xalka:

Dareeraha korantada:

Φ = EA cos q = (5000)(2)(cos 60) = (5000)(2)(0.5)= 5000 = 5 x 10³ Nm²/C

3. Kubbad adag oo leh gacan 0.5 mitir ah waxay leedahay koronto 10 μC ah oo ku jirta bartamaha. Go'aami qulqulka korantada ee dhex maraya kubbadda adag.

La yaqaan:

Gacanka kubbadda (r) = 0.5 m

Kharashka korontada (Q) = 10 μC = 10 x 10^-6 C

La doonayo: Dareeraha korantada ((Φ)

Xalka:

Goobta korontada:

E = kq/r²

E = (9 x 10⁹ Nm²/C²)(10 x 10^-6 C) / 0.5²

E = (90 x 10³) / 0,25

E = 360 x 10³

E = 3.60 x 10⁵ N / C

Bedka dusha sare:

A=4 π r² = 4 (3.14)(0.5)² = (12.56)(0.25) = 3.14 m²

Dareeraha korantada:

Xariiqyada garoonka korontada ee ku toosan aagga, sidaas darteed xagasha u dhaxaysa jihada goobta korantada iyo xariiq si toosan u jiidan aagga, waa 0^o.

Φ = EA cos q

Φ = (3.60 x 10⁵)(3.14)(cos 0)

Φ = (11.304 x) 10⁵) (1)

Φ = 11.304x 10⁵

Φ = 1.13x 10⁶ Nm²/C

Akhri wax dheeraad ah

1. Xisaabi baaxadda iyo jihada goobta korantada barta A oo ku taal 5 cm laga bilaabo dallacaadda dhibicda Q = + 10 μC. k = 9 x 10⁹ Nm²C^-2, 1 μC = 10^-6 C)

La yaqaan:

Kharashka korontada (Q) = + 10 μC = + 10 x 10^-6 C

The masaafada inta u dhaxaysa barta A iyo barta dallacaadda Q (r)_A) = 5 cm = 0.05 m = 5 x 10^-2 m

k = 9 x 10⁹ Nm²C^-2

SE buska: Baaxadda iyo jihada korontada barta A

Xalka:

Jihada garoonka korontada ee barta A:

Dareeraha korontadu waa mid togan sidaa darteed jihada goobta korontadu waxay ka fog tahay dareeraha korontada iyo dhibcaha A.

2. Xisaabi baaxadda iyo jihada goobta korontada ee barta P oo ku taal 10 cm laga bilaabo dallacaadda dhibicda Q = -20 μC. k = 9 x 10⁹ Nm²C^-2, 1 μC = 10^-6 C.

La yaqaan:

Kharashka korontada (q) = -20 μC = -20 x 10^-6 C

Masaafada u dhaxaysa barta P iyo dallacaadda korantada (r)_P) = 10 cm = 0.1 m = 1 x 10^-1 m

k = 9 x 10⁹ Nm²C^-2

SE buska: Cabbirka iyo jihada goobta korantada ee barta P

Xalka:

Jihada garoonka korontada ee barta A:

Dareeraha korontadu waa taban sidaas darteed jihada garoonka korontadu u jeedo dareeraha korontadu waa taban.

3. Laba dalac oo dhibco ah ayaa kala fogaada masaafada 40 cm. Waa maxay baaxadda iyo jihada goobta korontada ee barta P ee u dhaxaysa labada dalac oo ah 20 cm u jirta barta A?

La yaqaan:

Dalac A (q)_A) = -2 μC = -2 x 10^-6 C

Dalac B (q)_B) = +4 μC = +4 x 10^-6 C

Masaafada u dhaxaysa dallacaadda A iyo barta P (r)_AP) = 20 cm = 0.2 m = 2 x 10^-1 m

Masaafada u dhaxaysa dallacaadda B iyo barta P (r)_BP) = 20 cm = 0.2 m = 2 x 10^-1 m

La doonayo: Cabbirka iyo jihada goobta korantada ee barta P.

Xalka:

Kharash A waa taban sidaa darteed jihada dhibcaha goobta korantada waxay u socotaa Q_A (bidixda).

Kharash B waa mid togan si jihada dhibcaha goobta korontada ay uga fogaato Q_B (bidixda).

Wadarta garoon koronto ee barta A:

E = E_A +E_B

E = (4.5 x 10⁵) + (9 x 10⁵)

E = 13.5 x 10⁵ N / C

Jihada garoonka korontada ayaa u jeedda dhanka Q_A (bidixda).

4. Baaxadda goobta korontada waa eber marka loo eego...

Kharash A waa togan yahay, kharash B-na waa togan yahay si baaxadda goobta korontada loogu dhigo eber. barta P, labada dacwadoodba.

La yaqaan:

eedayn A (q)_A) = +20 μC = +20 x 10^-6 C

eedayn B (q)_B) = +40 μC = +40 x 10^-6 C

k = 9 x 10⁹ Nm²C^-2

Masaafada u dhaxaysa khidmadaha A iyo kharashka B = 20 cm

Kharashka u dhexeeya kharashka A iyo dhibicda P (r_AP) = a

Masaafada u dhaxaysa dallacaadda B iyo barta P (r_BP) = 20 – a

La doonayo: Baaxadda goobta korontada ku shaqeysa waa eber oo ku taal….

Xalka:

Baaxadda goobta korantada ee ay soo saarto dallacaadda A barta P

Dalacaadda A waa mid togan si jihada meelaha korontada laga helo ay uga fogaato dalacaadda A (midigta).

Baaxadda goobta korantada ee ay soo saarto dallacaadda B ee barta P:

Dalac B waa togan si jihada meelaha korontada laga helo ay uga fogaato dalac B (bidixda).

Wadarta garoon koronto ee barta P = 0:

Waxaan isticmaalnaa qaacidada labajibbaaran si aan u go'aamino a.

Baaxadda goobta korontada waa eber oo ku taal 8 cm laga bilaabo dallacaadda A ama 12 cm laga bilaabo dallacaadda B.

5Iyada oo ku saleysan sawirka hoose, wWaa kan qodobka P si garoonka korontada ee barta P uu eber u noqdo? (k = 9 x 10⁹ Nm²C^-2, 1 μC = 10^-6 C)

Solution

To xisaabi xoogga goobta korantada ee barta P, oo la filayo barta P inay jirto kharash tijaabo oo togan. S₁ waa mid togan iyo Q₂ waa taban, sidaa darteed dhibicda P waa inay ku taal dhinaca midig ee Q₂ ama bidixda Q₁Haddii dhibicda P ay bidixda ka tahay Q₁; garoon koronto oo ay soo saarto Q₁ barta P waa bidix (ka fog Q)₁) iyo garoon koronto oo ay soo saarto Q₂ barta P ee dhanka midig (dhinaca Q₁Jihada garoonka korontadu waa mid ka soo horjeeda si labada goobood ee korontadu ay isu baabi'iyaan si xoogga goobta korontadu ee barta P uu noqdo eber.

La yaqaan:

Q₁ = +9 μC = +9 x 10^-6 C

Q₂ = -4 μC = -4 x 10^-6 C

k = 9 x 10⁹ Nm²C^-2

Masaafada u dhaxaysa dallacaadda 1 iyo dallacaadda 2 = 3 cm

Masaafada u dhaxaysa Q₁ iyo dhibic P (r)_1P) = a

Masaafada u dhaxaysa Q₂ iyo dhibic P (r)_2P) = 3 + a

La doonayo: goobta barta P si goobta korantada ee barta P waa eber

Xalka:

PBarta P waxay ku taal bidixda Q₁.

Goobta korontada ee uu soo saaray Q₁ barta P:

Kharashka baaritaanka waa mid togan iyo Q₁ waa mid togan si jihada garoon koronto ay u noqoto dhanka bidix.

Goobta korontada ee uu soo saaray Q₂ barta P:

Kharashka baaritaanka waa mid togan iyo Q₂ waa taban si jihada garoon koronto ay u noqoto midig.

Garoon koronto oo saafi ah oo ku yaal barta A:

Isticmaal qaacidada labajibbaaran si aad u go'aamiso:

a = -1.25, b = -13.5, c = -20.25

Masaafada u dhaxaysa Q₂ iyo dhibic P (r)_2P) = 3 + a = 3 – 1.8 = 1.2 cm.

Barta P waxay ku taal 1.2 cm midig ee Q₁.

Akhri wax dheeraad ah

1. Laba dhibcood oo dacwad ah, Q_A = +8 μC iyo Q_B = -5 μC, waxaa kala soocaya masaafo r = 10 cm. Waa maxay baaxadda xoogga korontadaJoogtada ah k = 8.988 x 10⁹ Nm²C^-2 = 9x10⁹ Nm²C^-2.

La yaqaan:

eedayn A (q)_A) = +8 μC = +8 x 10^-6 C

Dalac B (q)_B) = -5 μC = -5 x 10^-6 C

k = 9 x 10⁹ Nm²C^-2

The masaafada inta u dhaxaysa khidmadda A iyo B (r)_AB) = 10 cm = 0.1 m

rabay : Baaxadda xoogga korontada

Xalka:

Qaacidada Sharciga Coulomb :

Baaxadda xoogga korontada:

2. Laba qaybood oo la dallacay sida ku cad sawirka hoose. S_P = +10 μC iyo Q_q = +20 μC waxaa kala fogeeya masaafada r = 10 cm. Waa maxay baaxadda xoogga korontada ku shaqeeya.

La yaqaan:

Dalac P (Q)_P) = +10 μC = +10 x 10^-6 C

Kharashka Q (Q)_Q) = +20 μC = +20 x 10^-6 C

k = 9 x 10⁹ Nm²C^-2

Masaafada u dhaxaysa khidmadaha P iyo Q (r_PQ) = 12 cm = 0.12 m = 12 x 10^-2 m

La doonayo: Tbaaxadda xoogga korontada

Xalka:

3. Saddex walxood oo la dallacay ayaa loo habeeyay xariiq sida ku cad jaantuska hoose. Dalacsiinta A = -5 μC, dalacsiinta B = +10 μC iyo dalacsiinta C = -12 μC. Xisaabi xoogga elektaroonigga saafiga ah ee ku jira walxaha B sababtoo ah labada dalacsiinta kale.

La yaqaan:

Dalac A (q)_A) = -5 μC = -5 x 10^-6 C

Dalac B (q)_B) = +10 μC = +10 x 10^-6 C

Dalacaadda C (q)_C) = -12 μC = -12 x 10^-6 C

k = 9 x 10⁹ Nm²C^-2

Masaafada u dhaxaysa walxaha A iyo B (r)_AB) = 6 cm = 0.06 m = 6 x 10^-2 m

Masaafada u dhaxaysa walxaha B iyo C (r)_BC) = 4 cm = 0.04 m = 4 x 10^-2 m

La doonayo: Baaxadda iyo jihada xoogga elektrostaatigga saafiga ah ee ku jira walxaha B

Xalka:

Xoogga saafiga ah ee ku jira walxaha B waa wadarta vector-ka ee xoogga F_BA ku dul sabbaynaya walaxda B iyadoo la adeegsanayo walaxda A iyo xoogga F_BC ku shubtay walxaha B iyadoo la adeegsanayo walxaha C.

Ciidanka F_BA lagu sameeyay walxaha B iyadoo la adeegsanayo walxaha A:

 

Jihada xoogga korontada ku shaqeeya waxay tilmaamaysaa walxaha A (tilmaan bidix).

Ciidanka F_BC lagu sameeyay walxaha B iyadoo la adeegsanayo walxaha A:

 

Jihada xoogga korontada ku shaqeeya waxay tilmaamaysaa walxaha C (tilmaan dhanka midig).

Xoogga elektaroonigga saafiga ah ee ku jira walxaha B :

F_B =F_AB - F_BC = 675 N – 125 N = 550 Newtons.

Jihada xoogga elektrostaatigga saafiga ah ee ku yaal walxaha B waxay tilmaamaysaa walxaha C (waxay tilmaamaysaa dhinaca midig).

4. +Q₁ = 10 μC, +Q₂ = 50 μC iyo Q₃ waxaa loo kala saaraa sida ku cad sawirka hoose. Waa maxay kharashka korontada ku shaqeeya ee ku jira walaxda 3 haddii xoogga korontada ku shaqeeya ee ku jira walaxda 2 uu eber yahay.

La yaqaan:

Kharash 1 (q₁) = +10 μC = +10 x 10^-6 C

Kharash 2 (q₂) = +50 μC = +50 x 10^-6 C

Masaafada u dhaxaysa kharashyada 1 iyo 2 (r₁₂) = 2 cm = 0.02 m = 2 x 10^-2 m

Masaafada u dhaxaysa charge 2 iyo charge 3 (r)₂₃) = 6 cm = 0.06 m = 6 x 10^-2 m

Xoogga elektaroonigga saafiga ah ee ku jira walxaha 2 (F)₂= 0

rabay : khidmad 3 (q₃)

Xalka:

Xoogga saafiga ah ee ku jira walxaha 2 waa vector wadarta xoogga F₂₁ lagu sameeyay walaxda 2 iyadoo la adeegsanayo walaxda 1 iyo xoogga F₂₃ lagu sameeyay walaxda 2 iyadoo la adeegsanayo walaxda 3.

Ciidanka F₂₁ lagu sameeyay walxaha 2 iyadoo la adeegsanayo walxaha 1:

Jihada xoogga korontada ku shaqeeya waxay tilmaamaysaa walxaha 3 (tilmaan dhanka midig).

Ciidanka F₂₃ lagu sameeyay walxaha 2 iyadoo la adeegsanayo walxaha 3: 

Jihada xoogga korontada ku shaqeeya waxay tilmaamaysaa walxaha 1 (tilmaan bidix).

Xoogga elektaroonigga saafiga ah ee ku jira walxaha 2 = 0:

Akhri wax dheeraad ah

1. Shay leh daqiiqada ofertia of 2 kg m² wuxuu ku wareegayaa 1 rad/s. Waa maxay dhaqdhaqaaq xagal ah shayga?

La yaqaan:

Waqtiga inertia (I) = 2 kg m²

Xawaaraha xagasha (ω) = 1 rad/s

La doonayo: Dhaqdhaqaaq xagal (L)

Xalka:

Qaacidada dhaqdhaqaaqa xagasha:

L = Aniga ω

L= dhaqdhaqaaq xagal ah (kg m²/ s), I = daqiiqada ofertia (kg m²), ω = xawaaraha xagal (raad/s)

Xawaaraha xagasha:

L = Aniga ω = (2)(1) = 2 kg m²/s

2. 2-kg boolal dhululubo ah oo leh gacan 0.1 m wuxuu ku wareegaa xawaare xagal joogto ah oo ah 2 rad/s. Waa maxay xawaaraha xagasha ee jiidaha?

La yaqaan:

Mass ee boolal (M) = 2kg

Gacanka jiidista (r) = 0.1 m

Xawaaraha xagasha (ω) = 2 rxayeysiis/yada

La doonayo: Dhaqdhaqaaq xagal

Xalka:

Qaacidada daqiiqadda firfircoonida ee dhululubada adag:

I = 1/2 mr²

I = daqiiqada ofertia (kg m²), m = mass (kg), r = radius (M)

Daqiiqadda firfircoonida:

I = 1/2 (2)(0.1)² = (1)(0.01) = 0.01 kg m²

Xawaaraha xagasha:

L = Aniga ω = (0.01)(2) = 0.02 kg m²/s

3. Goob wareegsan oo isku mid ah oo 2-kg ah oo leh gacan 0.2 m wuxuu ku wareegayaa 4 rad/s. Waa maxay xawaaraha xagasha ee kubbadda.

La yaqaan:

Mass kubadda (M) = 2 kg

Gacanka kubbadda (r) = 0.2 m

Xawaaraha xagasha (ω) = 4 rad/s

La doonayo: Dhaqdhaqaaq xagal

Xalka:

Qaacidada daqiiqadda firfircoonida ee wareegga isku midka ah:

I = (2/5) Mudane.²

Aniga = daqiiqadt ee firfircoonida (kg m²), m = cufnaanta (kg), r = radius (M)

Daqiiqadda firfircoonida ee wareegga isku midka ah:

I = (2/5)(2)(0.2)² = (4/5)(0.04) = 0.032 kg m²

Dhaqdhaqaaqa xagasha ee kubbadda:

L = I ω = (0.032)(4) = 0.128 kg m²/s

4. A 1 kg Walaxdu waxay ku wareegtaa xawaare xagal joogto ah oo ah 2 rad/s. Waa maxay xawaaraha xagasha haddii gacanka goobadu yahay 10 cm.

La yaqaan:

Mass shayga (M) = 1kg

Gacanka goobada (r) = 10 cm = 10/100 = 0.1 m

Xawaaraha xagasha (ω) = 2 rad/s

La doonayo: Dhaqdhaqaaq xagal

Xalka:

Qaacidada daqiiqadda firfircoonida ee walxaha:

I = Mudane.² = (1)(0.1)² = (1)(0.01) = 0.01 kg m²

Dhaqdhaqaaqa xagasha:

L = I ω = (0.01)(2) = 0.02 kg m²/s

Akhri wax dheeraad ah

1. Shay ayaa leh daqiiqada ofertia oo ah 1 kg m² wuxuu ku wareegaa xawaare xagal joogto ah oo ah 2 rad/s. Waa maxay tamarta dhaqdhaqaaqa wareega shayga?

La yaqaan:

Daqiiqadda firfircoonida (I) = 1 kg m²

The xawaaraha xagasha (ω) = 2 rad/s

SE buska: Tamarta dhaqdhaqaaqa wareega (KE)

Xalka:

Qaacidada tamarta wareegga wareegga:

KE = 1/2 I ω²

KE = tamarta dhaqdhaqaaqa wareega (kg m²/s²), aniga = daqiiqadda firfircoonida (kg m²), ω = xawaaraha xagasha (raad/s)

Tamarta dhaqdhaqaaqa wareega:

KE = 1/2 I ω² = 1/2 (1)(2)² = 1/2 (1)(4) = 2 Joule

2. 20-kg khaanadaha dhululubada iyadoo gacan 0.2 m ah ay ku wareegeyso xawaare xagal joogto ah oo ah 4 rad/s. Waa maxay tamarta dhaqdhaqaaqa wareega ee jiidista?

La yaqaan:

Mass ee boolal dhululubo ah (M) = 20kg

Gacanka dhululubada (r) = 0.2 m

Xawaaraha xagasha (ω) = 4 rad/s

La doonayo: Waa maxay tamarta dhaqdhaqaaqa wareega

Xalka;

Qaacidada daqiiqadda inertia ee silinda:

I = 1/2 mr²

I = daqiiqadda firfircoonida (kg m²), m = cufnaanta (kg), r = radius (mitir)

Daqiiqada firfircoonida ee boolal silinda ah:

I = 1/2 (20)(0.2)² = (10)(0.04) = 0.4 kg m²

Tamarta dhaqdhaqaaqa wareega ee boolalka:

KE = 1/2 I ω² = 1/2 (0.4)(4)² = (0.2)(16) = 3.2 Joule

3. A-10 kg kubad leh gacanka 0.1 m wuxuu ku wareegaa heer joogto ah oo ah 10 rad/s. Waa maxay tamarta dhaqdhaqaaqa ee kubadda.

La yaqaan:

Mass kubadda (M) = 10 kg

Gacanka kubbadda (r) = 0.1 m

Xawaaraha codka (ω) = 10 rxayeysiis/yada

La doonayo: Tamarta dhaqdhaqaaqa wareega

Xalka:

Qaacidada daqiiqadda inertia:

I = (2/5) Mudane.²

I = daqiiqada ofertia (kg m²), m = cufnaanta (kg), r = radius (M)

Daqiiqadda firfircoonida kubadda:

I = (2/5)(10)(0.1)² = (4)(0.01) = 0.04 kg m²

Tamarta dhaqdhaqaaqa wareega ee kubbadda:

KE = 1/2 I ω² = 1/2 (0.04)(10)² = (0.02)(100) = 2 Joule

4. A 0.5-walxaha kg waxay ku wareegaan xawaare xagal joogto ah oo ah 2 rad/s. Waa maxay tamarta wareegga ee walaxda haddii gacanka goobada uu yahay 10 cm.

La yaqaan:

Mass walxaha (M) = 0.5kg

Gacanka kubadda (r) = 10 cm = 10/100 = 0.1 m

Xawaaraha xagasha (ω) = 2 rad/s

La doonayo: Tamarta dhaqdhaqaaqa wareega

Xalka:

Daqiiqadda firfircoonida walxaha:

I = Mudane.² = (0.5)(0.1)² = (0.5)(0.01) = 0.005 kg m²

Tamarta dhaqdhaqaaqa wareega:

KE = 1/2 I ω² = 1/2 (0.005)(2)² = 1/2 (0.005)(4) = (0.005)(2) = 0.01 Joule

Akhri wax dheeraad ah

1. Xoogga F waxaa lagu dhejiyaa xadhig ku duuban boolal dhululubo ah. Xuddunta is 2 N m iyo daqiiqada ofertia is 1 kg m², waa maxay dardargelinta xagal dhululubada.

La yaqaan:

Xuddunta (τ) = 2 N m

Daqiiqadda firfircoonida (I) = 1 kg m²

SE buska: Dardargelinta xagasha ee dhululubada

Xalka:

Στ = Aniga α

Στ = wareegga saafiga ah, I = daqiiqadda firfircoonida, α = dardargelinta xagasha

Dardargelinta xagasha ee dhululubada:

α = Στ / I = 2/1 = 2 rad/s²

2. Xoogga F waxaa lagu dabaqaa xarig ku duuban boolal dhululubo ah. Cabbirka xoogga waa 10 N, gacanka dhululubada waa 0.2 m iyo daqiiqadda inertia waa 1 kg m2, WKoofigu waa xawaaraha xagasha ee dhululubada?

La yaqaan:

Xoogga (F) = 10 N

Radius-ka dhululubada (R) = 0.2 m

Daqiiqadda firfircoonida (I) = 1 kg m²

SE buska: Xawaaraha xagasha ee dhululubada.

Xalka:

τ = FR

τ = xoog, F = xoog, R = gacanka dhululubada

Xuddunta:

τ = FR = (1)0 N)(0.2 m) = 2 N m

Στ = Aniga α

Στ = wareegga saafiga ah, I = daqiiqadda firfircoonida, α = dardargelinta xagasha

Dardargelinta xagasha ee dhululubada:

α = Στ / I = 2 / 1 = 2 rad/s²

3. Xoogga F waxaa lagu dhejiyaa xarig ku duuban boolal dhululubo ah. Cabbirka xoogga waa 10 N, gacanka dhululubada waa 0.2 m, cufka dhululubadana waa 20 kg m^2,. WKoofiyaddu waa xawaaraha xagasha ee silinda.

La yaqaan:

Xoogga (F) = 10 N

Radius-ka dhululubada (R) = 0.2 m

Cufka dhululubada (M) = 20 kg

La doonayo: Dardargelinta xagasha ee dhululubada

Xalka:

τ = FR = (1)0 N)(0.2 m) = 2 N m

Daqiiqadda firfircoonida:

I = 1⁄2 MR² = 1⁄2 (20)(0.2)² = 1⁄2 (20)(0.04) = 0.4 kg m²

Dardargelinta xagasha ee dhululubada:

α = Στ / I = 2 / 0.4 = 5 rad/s²

4. Baloog 1-kg ah oo ka laalaada xadhig ku duuban baloog dhululubo ah. Daqiiqadda firfircoonida ee baloogga waa 1 kg. m² iyo gacanka galka galka waa 0.2 m. Waa maxay xawaaraha xagasha ee galka. Dardargelinta cufjiidka awgeed waa 10 m/s².

La yaqaan:

Daqiiqadda firfircoonida ee khaanadaha (I) = 1 kg m²

Mass baloogga (m) = 1 kg

Dardargelinta cufisjiidadka awgeed (g) = 10 m/s²

Miisaanka (w) = mg = (1 kg)(10 m/s²) = 10 kg m/s² = 10N

Gacanka roogga (R) = 0.2 m

La doonayo: Dardargelinta xagasha

Xalka:

Xuddunta:

τ = FR = w R = (1)0 N)(0.2 m) = 2 N m

Daqiiqadda firfircoonida:

I = 1 kg m²

Dardargelinta xagasha:

α = Στ / I = 2 / 1 = 2 rad/s²

5. Baloog 1-kg ah oo ka laalaada xadhig ku duuban baloog dhululubo ah. Cufka baloogga waa 20 kg. iyo gacanka galka galka waa 0,2 m. Waa maxay xawaaraha xagasha ee galka iyo free-dhaco dardargelinta baloogga. Dardargelinta cufisjiidadka awgeed waa 10 m/s².

La yaqaan:

Cufka roogga (M) = 20 kg

Gacanka roogga (R) = 0,2 m

Cufka baloogga (m) = 1 kg

Dardargelinta cufisjiidadka awgeed (g) = 10 m/s²

Miisaanka (w) = mg = (1 kg)(10 m/s²) = 10 kg m/s² = 10N

La doonayo: dardargelinta xagasha ee jiidista iyo dardargelinta dhicitaanka xorta ah ee baloogga.

Xalka:

Xoogga:

τ = FR = w R = (1)0 N)(0.2 m) = 2 N m

Daqiiqada firfircoonida ee boolal silinda ah:

I = 1⁄2 MR² = 1⁄2 (20)(0.2)² = (10)(0.04) = 0.4 kg m²

Dardargelinta xagasha ee boolalku:

α = Στ / I = 2 / 0.4 = 5 rad/s²

Dardargelinta dhicitaanka xorta ah ee baloogga:

a = R α = (0.2)(5) = 1 m/s²

Akhri wax dheeraad ah

Daqiiqada firfircoonida walaxda

1. Kubbad 100-garaam ah oo ku xiran hal dhinac oo xarig ah oo dhererkeedu yahay 30 cm. Waa maxay daqiiqadda firfircoonida kubbadda ee ku saabsan dhidibka wareegga AB? Iska ilow cufka xarigga.

La yaqaan:

dhidibka wareegga ee AB

Mass kubad (m) = 100 garaam = 100/1000 = 0.1 kg

Masaafada u dhaxaysa kubadda iyo dhidibka wareegga (r) = 30 cm = 0.3 m

SE buska: Daqiiqad firfircooni ah oo kubadda ah (I)

Xalka:

Aniga = Mudane² = (0.1 kg)(0.3 m)²

I = (0.1 kg)(0.09 m²)

I = 0.009 kg m²

2. Kubad 100-garaam ah, m₁iyo kubad 200-gram ah, m₂, oo ay ku xiran tahay ul dhererkeedu yahay 60 cm. Cufka usha waa la iska indho tiray. dhidibka wareegga wuxuu ku yaal bartamaha usha. Waa maxay daqiiqadda firfircoonida kubbadda ee ku saabsan dhidibka wareegga?

La yaqaan:

Cufka kubadda 1 (m₁) = 100 garaam = 100/1000 = 0.1 kg

Masaafada kubbadda 1 iyo dhidibka wareegga (r)₁) = 30 cm = 30/100 = 0.3 m

Cufka kubadda (m)₂) = 200 garaam = 200/1000 = 0.2 kg

The masaafada kubadda 2aad iyo dhidibka wareegga (r)₂) = 30 cm = 30/100 = 0.3 m

La doonayo: daqiiqad firfircoon oo kubadaha ah

Jawaab:

Aniga = m₁ r₁² +m₂ r₂²

I = (0.1 kg)(0.3 m)² + (0.2 kg)0.3 m)²

I = (0.1 kg)(0.09 m²) + (0.2 kg)0.09 m²)

I = 0.009 kg m² + 0.018 kg m²

I = 0.027 kg m²

3. Kubad 200-garaam ah, m₁ iyo kubad 100-garaam ah, m_2, Ku xiran ul dhererkeedu yahay 60 cm. Iska ilow cufka ul. Dhidibka wareegga wuxuu ku yaal kubbadda m₂Waa maxay daqiiqadda firfircoonida kubbadaha. Iska ilow cufnaanta usha.

La yaqaan:

Cufka kubadda 1 (m₁= 200 garaam = 200/1000 = 0.2 kg

Masaafada u dhaxaysa kubbadda 1 iyo dhidibka wareegga (r)₁) = 60 cm = 60/100 = 0.6 m

Cufka kubadda 2 (m₂) = 100 garaam = 100/1000 = 0.1 kg

Masaafada u dhaxaysa kubbadda 2 iyo dhidibka wareegga (r)₂) = 0 m

La doonayo: Daqiiqad firfircoon oo kubadaha ah

Xalka:

Aniga = m₁ r₁² +m₂ r₂²

I = (0.2 kg)(0,6 m)² + (0.2 kg)(0))²

I = (0.2 kg)(0.36 m²) + 0

I = 0.072 kg m²

4. Cufnaanta kubbad kasta waa 100 garaam, oo lagu xiray xarig. Dhererka xariggu waa 60 cm, ballaca xarigguna waa 30 cm. Waa maxay daqiiqadda firfircoonida kubbadda ee ku saabsan dhidibka wareegga. Iska ilow cufka xarigga.

La yaqaan:

Cufka kubadda = m₁ = m₂ = m₃ = m₄ = 100 garaam = 100/1000 = 0.1 kg

Masaafada u dhaxaysa kubadda iyo dhidibka wareegga (r)₁) = 30 cm = 30/100 = 0.3 m

Masaafada u dhaxaysa kubbadda 2 iyo dhidibka wareegga (r)₂) = 30 cm = 30/100 = 0.3 m

Masaafada u dhaxaysa kubbadda 3 iyo dhidibka wareegga (r)₃) = 30 cm = 30/100 = 0.3 m

Masaafada u dhaxaysa kubbadda 4 iyo dhidibka wareegga (r)₄) = 30 cm = 30/100 = 0.3 m

La yaqaan: Waqtiga inertia

Xalka:

Aniga = m₁ r₁² +m₂ r₂² +m₃ r₃² +m₄ r₄²

I = (0.1 kg)(0.3 m)² + (0.1 kg)0.3 m)² + (0.1 kg)(0.3 m)² + (0.1 kg)(0.3 m)²

I = (0.1 kg)(0.09 m²) + (0.1 kg) (0.09 m²) + (0.1 kg) (0.09 m²) + (0.1 kg) (0.09 m²)

I = 0.036 kg m²

Daqiiqadda firfircoonida shay adag

5. Waa maxay daqiiqadda firfircoonida usha isku midka ah ee 2-kg ah oo dhererkeedu yahay 2 m. Dhidibka wareegga wuxuu ku yaal bartamaha usha.

La yaqaan:

Cufka usha (M) = 2 kg

Dhererka usha (L) = 2 m

SE buska: Waqtiga inertia

Xalka:

Qaacidada daqiiqadda inertia marka dhidibka wareegga uu ku yaal bartamaha usha dheer ee isku midka ah:

I = (1/12) ML²

I = (1/12) (2 kg)(2 m)²

I = (1/12) (2 kg) (4 m)²)

I = (1/12)(8 kg m²)

I = 8/12 kg m²

I = 2/3 kg m²

6. Waa maxay daqiiqadda firfircoonida usha isku midka ah ee 2-kg ah oo dhererkeedu yahay 2 m? Dhidibka wareegga wuxuu ku yaal hal dhinac oo usha ah.

La yaqaan:

Cufka usha (M) = 2 kg

Dhererka usha adag (L) = 2 m

SE buska: Waqtiga inertia

Xalka:

Qaacidada daqiiqadda inertia marka dhidibka wareegga uu ku yaal hal dhammaad oo usha ah:

I = (1/3) ML²

I = (1/3) (2 kg)(2 m)²

I = (1/3) (2 kg) (4 m)²)

I = (1/3)(8 kg m²)

I = 8/3 kg m²

7. Dhululubo adag oo 10-kg ah oo leh gacan 0.1 m. Dhidibka wareegga wuxuu ku yaal bartamaha dhululubada adag, oo lagu muujiyey sawirka hoose. Waa maxay daqiiqadda firfircoonida dhululubada?

La yaqaan:

Cufka dhululubo adag (M) = 10 kg

Radius-ka dhululubada (L) = 0.1 m

SE buska: Daqiiqadda firfircoonida

SE buska: Daqiiqadda firfircoonida

Xalka:

Qaacidada daqiiqadda inertia marka dhidibka wareeggu uu ku yaal bartamaha dhululubada:

I = (1/2) MR²

I = (1/2) (10 kg) (0.1 m)²

I = (1/2) (10 kg) (0.01 m)²)

I = (1/2)(0.1 kg m²)

I = 0.05 kg m²

8. Goob wareegsan oo isku mid ah oo 20-kg ah oo dhererkeedu yahay 0.1 m. Dhidibka wareegga ee ku yaal bartamaha wareegga waxaa lagu muujiyay sawirka hoose.

La yaqaan:

Cufka wareegga (M) = 20 kg

Gacanka kubbadda (L) = 0.1 m

SE buska: daqiiqad firfircooni ah

Xalka:

Qaacidada daqiiqadda inertia marka dhidibka wareegga uu ku yaal bartamaha wareegga:

I = (2/5) MR²

I = (2/5)(20 kg)(0.1 m)²

I = (2/5)(20 kg)(0.01 m²)

I = (2/5)(0.2 kg m²)

I = 0.4/5 kg m²

I = 0.08 kg m²

9. Saxan khafiif ah oo leydi ah oo 2-kg ah oo dhererkiisu yahay 0.5 m iyo ballaciisuna yahay 0.2 m. Dhidibka wareegga wuxuu ku yaal bartamaha saxanka leydi ee lagu muujiyay sawirka hoose. Waa maxay daqiiqadda firfircoonida leydiga?

La yaqaan:

Cufka saxanka leydi ah (M) = 2 kg

Dhererka saxanka (a) = 0.5 m

Ballaca saxanka (b) = 0.2 m

La doonayo: Waqtiga inertia

Xalka:

Qaaciddada daqiiqadda inertia marka dhidibka wareegga uu ku yaal bartamaha saxanka:

I = (1/12) M (a² + b²)

I = (1/12)(2)(0.5)² + 0.2²)

I = (2/12)(0.25 + 0.04)

I = (1/6)(0.29)

I = 0.29/6 kg m²

Akhri wax dheeraad ah