Physics

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 = C

Ogolaansho la'aanta booska bannaan = ε _o

Aagga dambe ee P = A

Masaafada u dhaxaysa saxamada = d

La Rabay: Awoodda kaabsootarka (C)

Xalka:

2. Kaadhka leh awoodda ugu weyn iyadoo lagu saleynayo sawirka hoose waa …

Xalka:

Qaacidada kaabsoosha-saxanka barbar socda:

Awoodda capacitors:

Kaaliyaha leh awoodda ugu weyn ee y 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:

Kondenser 1:

Bedka dusha sare = 2A

Masaafada u dhaxaysa saxamada = d ₁

Kondenser 2:

Bedka dusha sare = A

Masaafada u dhaxaysa saxamada = 2 d ₁

La Doonayo: Isbarbardhigga awoodda ee kaabsoodhada 1 iyo 2

Xalka:

5.

Go'aami isbarbardhigga awoodda capacitor-ka Parallel-plate I iyo II.

Xalka:

Isbarbardhigga awoodda capacitors-ka Parallel-plate I iyo II:

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

Haddii A ₁ = ½ A ₂ iyo d ₂ = 3 d ₁ markaa go'aami saamiga awoodaha capacitor-ka saxanka is barbar socda ee u dhexeeya sawirka 2 iyo sawirka 1.

La yaqaan:

Kaaliyaha saxanka is barbar socda I:

A ₁ = 1

d ₁ = 1

Kondensarka is barbar socda II:

A ₂ = 2

d ₂ = 3

La rabay: 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:

Saamiga awoodaha ee capacitors-ka barbar socda II iyo I:

Akhri wax dheeraad ah

1. Go'aami awoodda korantada meel ku taal 1 cm laga bilaabo dallacaadda 5.0 μ C. Joogtada Coulomb (k) = 9 x 10 ⁹ Nm ² C ⁻² , 1 μC = 10 ⁻⁶ C.

La yaqaan:

Masaafada u dhaxaysa dallacaadda (r) = 1 cm = 1/100 m = 0.01 m = 10 ^-2 m

Dalac (q) = 5.0 μ C = 5.0 x 10 ^-6 C

Joogtada Coulomb (k) = 9 x 10 ⁹ Nm ² C ⁻²

La Doonayo: Awoodda korantada (V)

Xalka:

Awoodda korontada:

Awoodda korontadu waa 4.5 x 10 ⁶ Volts

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:

Dalacaadda Q ₁ = -5.0 μ C = -5.0 x 10 ^-6 C

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

Dalacaadda Q ₂ = 6.0 μ C = 6.0 x 10 ^-6 C

Masaafada barta a laga bilaabo Q ₂ = 10 cm = 0.1 m = 10 ^-1 m

Joogtada Coulomb (k) = 9 x 10 ⁹ Nm ² C ⁻²

La Doonayo: Awoodda korantada ee barta A

Xalka:

Awoodda korontada 1 :

Awoodda korontada 2 :

Awoodda korantada ee barta A :

V = V ₂ – V ₁

V = (54 – 45) x 10 ⁴

V = 9 x 10 ⁴

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:

Dalacaadda Q ₁ = -5.0 μ C = -5.0 x 10 ^-6 C

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

Dalacaadda Q ₂ = 6.0 μ C = 6.0 x 10 ^-6 C

Masaafada barta A laga bilaabo Q ₂ = 50 cm = 0.5 m = 5 x 10 ^-1 m

Joogtada Coulomb (k) = 9 x 10 ⁹ Nm ² C ⁻²

La Doonayo: Awoodda korantada ee barta A

Xalka:

Awoodda korontada 1 :

Awoodda korontada 2 :

Awoodda korantada ee barta A :

V = V ₁ + V ₂

V = (-11.25 + 10.8) x 10 ⁴

V = -0.45 x 10 ⁴

V = -4.5 x 10 ³

Awoodda korantada ee barta A waa -4.5 x 10 ³ Volts

Akhri wax dheeraad ah

1. Elektaroonku wuxuu ka soo kacaa nasashada iyadoo loo marayo farqi suurtagal ah oo ah 12 V. Waa maxay isbeddelka tamarta awoodda korantada ee elektaroonku?

La yaqaan:

Xaraashka elektaroonada (e) = -1.60 x 10 ^-19 Coulomb

Awoodda korantada = danab (V) = 12 Volts

La Doonayo: Isbeddelka tamarta korontada ee elektarooniga (ΔPE)

Xalka:

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

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, baaxadda goobta korontada ee u dhaxaysa saxannadana waa 500 Volt/mitir. Waa maxay isbeddelka tamarta suurtagalka ah ee proton-ka marka laga dardargeliyo saxanka togan ee loo dallacsiiyay saxanka taban.

La yaqaan:

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

Masaafada u dhaxaysa saxanadaha (yada) = 2 cm = 0,02 m

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

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 ^{-1 8} Joule

3. Laba dalac oo dhibic ah ayaa kala fogaada masaafada 10 cm. Dalac oo dhibicda A = +9 μC ah iyo dalac oo dhibicda B = -4 μC ah. k = 9 x 10 ⁹ Nm ² C ⁻² , 1 μC = 10 ⁻⁶ C. Waa maxay isbeddelka tamarta korontada ee awoodda dallacaadda ee dhibicda B haddii loo dedejiyo dhibicda A?

La yaqaan:

Dalac A (q ₁ ) = +9 μC = +9 x 10 ⁻⁶ C

Dalacaadda B (q ₁ ) = -4 μC = -4 x 10 ⁻⁶ C

k = 9 x 10 ⁹ Nm ² C ⁻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. Goob koronto oo isku mid ah E = 8000 N/C waxay dhex martaa dhul siman oo laba jibbaaran A = 10 m 2. ^Go'aami qulqulka korantada.

La yaqaan:

Baaxadda goobta korantada (E) = 8000 N/C

Bedka (A) = 10 m ²

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

La Rabay: Dareeraha Korontada ( Φ)

Xalka:

Qaacidada qulqulka korantada:

Φ = EA cos q

Φ = qulqulka korantada ( Nm ² /C) , E = goobta korantada (N/C), A = aagga (m ² ), q = xagal u dhexeeya xariiqda goobta korantada iyo xariiqda caadiga ah.

Dareeraha korantada:

Φ = EA cos q = (8000) (10) ( cos 0) = (8000) (10) (1 ) = 80,000 = 8 x 10 ⁴ Nm ² /C

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

La yaqaan:

Goobta korontada (E) = 5000 N/C

Bedka (A) = 2 m ²

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

La Rabay: Dareeraha Korontada ( Φ)

Xalka:

Dareeraha korantada:

Φ = EA cos q = (5000) (2) ( cos 60) = (5000) (2) (0.5 ) = 5000 = 5 x 10 ³ Nm ² /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

Dakhli koronto (Q) = 10 μC = 10 x 10 ^-6 C

La Rabay: Dareeraha Korontada ( Φ)

Xalka:

Goobta korontada:

E = kq/r ²

E = (9 x 10 ⁹ Nm ² /C ² )( 10 x 10 ^-6 C) / 0.5 ²

E = (90 x 10 ³ ) / 0,25

E = 360 x 10 ³

E = 3.60 x 10 ⁵ N/C

Bedka dusha sare:

A = 4 π r ² = 4 (3.14) (0.5) ² = (12.56) (0.25) = 3.14 m ²

Dareeraha korantada:

Xariiqyada goobta korantada ee ku toosan aagga, si xagasha u dhaxaysa jihada goobta korontada iyo xariiqda u janjeerta dhinaca aagga, ay u noqoto 0 ^o.

Φ = EA cos q

Φ = ( 3.60 x 10 ⁵ ) ( 3.14 ) ( cos 0)

Φ = (11.304 x 10 ⁵ )(1)

Φ = 11.304 x 10 ⁵

Φ = 1.13 x 10 ⁶ Nm ² /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 10 ⁹ Nm ² C ⁻² , 1 μC = 10 ⁻⁶ C)

La yaqaan:

Dakhli koronto (Q) = +10 μC = +10 x 10 ^-6 C

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

k = 9 x 10 ⁹ Nm ² C ⁻²

La Doonayo: Cabbirka iyo jihada garoonka korantada ee 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 korantada ee barta P oo ku taal 10 cm laga bilaabo dallacaadda dhibicda Q = -2 0 μC. k = 9 x 10 ⁹ Nm ² C ⁻² , 1 μC = 10 ⁻⁶ C.

La yaqaan:

Dakhli koronto (q) = -20 μC = -20 x 10 ^-6 C

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

k = 9 x 10 ⁹ Nm ² C ⁻²

La Doonayo: Cabbirka iyo jihada garoonka 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

Dalacaadda 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 si jihada meelaha korontada ku shaqeeya ay u jeeddo Q _A (bidixda).

Dalacaadda B waa mid togan si jihada meelaha korontada laga helo ay uga fogaato Q _B (bidixda).

Wadarta garoon koronto ee barta A:

E = E _A + E _B

E = (4.5 x 10 ⁵ ) + (9 x 10 ⁵ )

E = 13.5 x 10 ⁵ N/C

Jihada goobta korontada waxay u jeeddaa Q _A (bidixda).

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

Dalacaadda A waa togan, dalacaadda B-na waa togan, sidaa darteed baaxadda goobta korontada waa eber oo ku taal barta P, labada dallacba u dhexeeya.

La yaqaan:

Dalac A (q _A ) = + 20 μC = +20 x 10 ⁻⁶ C

Dalac B (q _B ) = +40 μC = +40 x 10 ⁻⁶ C

k = 9 x 10 ⁹ Nm ² C ⁻²

Masaafada u dhaxaysa dallacaadda A iyo dallacaadda 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: Cabbirka garoonka korontadu waa eber oo ku yaal….

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.

5. Iyada oo ku saleysan sawirka hoose, w halkan waa barta P si goobta korantada ee barta P ay eber u noqoto? (k = 9 x 10 ⁹ Nm ² C ⁻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:

Q ₁ = +9 μC = +9 x 10 ⁻⁶ C

Q ₂ = -4 μC = -4 x 10 ⁻⁶ C

k = 9 x 10 ⁹ Nm ² C ⁻²

Masaafada u dhaxaysa dallacaadda 1 iyo dallacaadda 2 = 3 cm

Masaafada u dhaxaysa Q ₁ iyo barta P (r _1P ) = a

Masaafada u dhaxaysa Q ₂ iyo barta P (r _2P ) = 3 + a

La Doonayo: goobta barta P si goobta korantada ee barta P ay eber u noqoto

Xalka:

Barta P waxay ku taal bidixda Q ₁.

Goobta korantada ee ay soo saartay Q1 _barta P:

Dakhligu waa togan yahay, Q1 _-na waa togan yahay, sidaa darteed jihada garoon koronto ayaa u jeedda bidixda.

Goobta korantada ee ay soo saartay Q2 _barta P:

Kharashka tijaabada waa togan yahay Q2 _-na waa taban yahay si jihada gaaska korontada uu u socdo dhanka 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 Q ₂ iyo barta P (r _2P ) = 3 + a = 3 – 1.8 = 1.2 cm.

Barta P waxay ku taal 1.2 cm midig ee Q ₁.

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:

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

Dalacaadda B (q _B ) = -5 μC = -5 x 10 ^-6 C

k = 9 x 10 ⁹ Nm ² C ⁻²

Masaafada u dhaxaysa khidmadaha A iyo B (r _AB ) = 10 cm = 0.1 m

La Doonayo : Cabbirka xoogga korontada

Xalka:

Qaacidda sharciga Coulomb :

Baaxadda xoogga korontada:

2. Laba walxood oo la dallacay sida ku cad jaantuska hoose. Q _P = +10 μC iyo Q _q = +20 μC waxaa kala soocaya masaafo r = 10 cm. Waa maxay baaxadda xoogga korontada ku shaqeeya.

La yaqaan:

Dalacsiinta P (QP ₎ = +10 μC = +10 x 10 ^-6 C

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

k = 9 x 10 ⁹ Nm ² C ⁻²

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

La Doonayo: Cabbirka 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

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

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

k = 9 x 10 ⁹ Nm ² C ⁻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: Cabbirka iyo jihada xoogga elektrostaatigga saafiga ah ee ku yaal walxaha B

Xalka:

Xoogga saafiga ah ee ku jira walxaha B waa wadarta vector-ka ee xoogga F _BA ee ku jira walxaha B ee ku jira walxaha A iyo xoogga F _BC ee ku jira walxaha B ee ku jira walxaha C.

Xoogga F _BA ee ku kacay walxaha B iyadoo loo marayo walxaha A:

 

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

Xoogga F _BC ee ku kacay 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. +Q ₁ = 10 μC, +Q ₂ = 50 μC iyo Q ₃ waa la kala saaraa sida ku cad sawirka hoose. Waa maxay dallacaadda elektrostaatigga ah ee ku jirta walaxda 3 haddii xoogga elektrostaatigga ah ee saafiga ah ee walaxda 2 uu eber yahay.

La yaqaan:

Dalacaadda 1 (q ₁ ) = +10 μC = +10 x 10 ^-6 C

Dalacaadda 2 (q ₂ ) = +50 μC = +50 x 10 ^-6 C

Masaafada u dhaxaysa kharashyada 1 iyo 2 (r ₁₂ ) = 2 cm = 0.02 m = 2 x 10 ^-2 m

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

Xoogga elektaroonigga saafiga ah ee ku jira walxaha 2 (F2 ₎ = 0

La Rabay : charge 3 (q ₃ )

Xalka:

Xoogga saafiga ah ee ku jira walaxda 2 waa wadarta vektor-ka ee xoogga F ₂₁ ee ku shaqeeya walaxda 2 iyadoo la adeegsanayo walaxda 1 iyo xoogga F ₂₃ ee ku shaqeeya walaxda 2 iyadoo la adeegsanayo walaxda 3.

Xoogga F ₂₁ ee ku shaqeeya walxaha 2 iyadoo loo marayo 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 daqiiqadda firfircoonida 2 kg m ² ayaa ku wareegaysa 1 rad/s. Waa maxay xawaaraha xagasha ee shayga?

La yaqaan:

Daqiiqadda firfircoonida (I) = 2 kg m ²

Xawaaraha xagasha ( ω ) = 1 rad/s

La Raadinayo: Xawaaraha xagasha (L)

Xalka:

Qaacidada dhaqdhaqaaqa xagasha:

L = I ω

L = dhaqdhaqaaqa xagasha (kg m ² / s), I = daqiiqadda firfircoonida (kg m ² ), ω = xawaaraha xagasha (rad/s)

Xawaaraha xagasha:

L = I ω = (2)(1) = 2 kg m ² / s

2. Barkin dhululubo ah oo 2- kg ah oo leh gacan 0.1 m ayaa ku wareegaysa xawaare xagal joogto ah oo ah 2 rad/s. Waa maxay xawaaraha xagasha ee barkinta?

La yaqaan:

Cufka roogga (m) = 2 kg g

Gacanka roogga (r) = 0.1 m

Xawaaraha xagasha (ω) = 2 r xayeysiis/s

La Doonayo: Xawaaraha xagasha

Xalka:

Qaacidada daqiiqadda firfircoonida ee dhululubada adag:

I = 1/2 mr ²

I = daqiiqadda firfircoonida (kg m ² ), m = cufnaanta (kg), r = gacanka (m)

Daqiiqadda firfircoonida:

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

Xawaaraha xagasha:

L = I ω = ( 0.01 )( 2 ) = 0.02 kg m ² / s

3. Goobo isku mid ah oo 2-kg ah oo leh gacan 0.2 m ayaa ku wareegaysa 4 rad/s. Waa maxay xawaaraha xagasha ee kubbadda.

La yaqaan:

Cufka kubadda (m) = 2 kg

Gacanka kubbadda (r) = 0.2 m

Xawaaraha xagasha (ω) = 4 ra d/s

La Doonayo: Xawaaraha xagasha

Xalka:

Qaacidada daqiiqadda firfircoonida ee wareegga isku midka ah:

I = (2/5) mr ²

I = daqiiqad t ee inertia (kg m ² ), 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 m ²

Dhaqdhaqaaqa xagasha ee kubbadda:

L = I ω = ( 0.032 )( 4 ) = 0.128 kg m ² / s

4. Walax 1-kg ah ayaa ku wareegta xawaare xagal joogto ah oo ah 2 rad/s. Waa maxay xawaaraha xagasha haddii gacanka goobadu yahay 10 cm.

La yaqaan:

Cufka walaxda (m) = 1 kg g

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

Xawaaraha xagasha (ω) = 2 rad/ s

La Doonayo: Xawaaraha xagasha

Xalka:

Qaacidada daqiiqadda firfircoonida ee walxaha:

I = mr ² = (1)(0.1) ² = (1)(0.01) = 0.01 kg m ²

Dhaqdhaqaaqa xagasha:

L = I ω = (0.01)(2) = 0.02 kg m ² / s

Akhri wax dheeraad ah

1. Shay wuxuu leeyahay daqiiqadda firfircoonida oo ah 1 kg m ² oo ku wareegaysa xawaare xagal joogto ah oo ah 2 rad/s. Waa maxay tamarta dhaqdhaqaaqa wareega ee shayga?

La yaqaan:

Daqiiqadda firfircoonida (I) = 1 kg m ²

Xawaaraha xagasha ( ω) = 2 rad/s

La Doonayo: Tamarta dhaqdhaqaaqa wareega (K E )

Xalka:

Qaacidada tamarta wareegga wareegga:

KE = 1/2 I ω ²

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

Tamarta dhaqdhaqaaqa wareega:

KE = 1/2 I ω ² = 1/2 (1) (2) ² = 1/2 (1) (4) = 2 Joules

2. Barkin dhululubo ah oo 20-kg ah oo leh gacan 0.2 m ayaa ku wareegaysa xawaare xagal joogto ah oo ah 4 rad/s. Waa maxay tamarta dhaqdhaqaaqa wareega ee barkinta?

La yaqaan:

Cufka boolal dhululubo ah (m) = 20 kg g

Gacanka dhululubada (r) = 0.2 m

Xawaaraha xagasha (ω) = 4 ra d/s

La Doonayo: Waa maxay tamarta dhaqdhaqaaqa wareega

Xalka;

Qaacidada daqiiqadda inertia ee silinda:

I = 1/2 mr ²

I = daqiiqadda firfircoonida ( kg m ² ), m = cufnaanta (kg), r = gacanka (mitir)

Daqiiqada firfircoonida ee boolal silinda ah:

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

Tamarta dhaqdhaqaaqa wareega ee boolalka:

KE = 1/2 I ω ² = 1/2 (0.4) (4) ² = (0.2) (16) = 3.2 Joules

3. A- Kubbad 10 kg ah oo leh gacan 0.1 m ah ayaa ku wareegta si joogto ah 10 rad/s. Waa maxay tamarta dhaqdhaqaaqa ee kubbadda.

La yaqaan:

Cufka kubadda (m) = 10 kg

Gacanka kubbadda (r) = 0.1 m

Xawaaraha xagasha (ω) = 10 r xayeysiis/s

La Doonayo: Tamarta dhaqdhaqaaqa wareega

Xalka:

Qaacidada daqiiqadda inertia:

I = (2/5) mr ²

I = daqiiqadda firfircoonida (kg m ² ), m = cufnaanta (kg), r = gacanka (m)

Daqiiqadda firfircoonida kubadda:

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

Tamarta dhaqdhaqaaqa wareega ee kubbadda:

K E = 1/2 I ω ² = 1/2 (0.04) (10) ² = (0.02) (100) = 2 Joules

4. Walax 0.5- kg ah ayaa ku wareegta xawaare xagal joogto ah oo ah 2 rad/s. Waa maxay tamarta dhaqdhaqaaqa wareega ee walaxda haddii gacanka goobada uu yahay 10 cm.

La yaqaan:

Cufka walxaha (m) = 0.5 kg g

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

Xawaaraha xagasha (ω) = 2 rad/ s

La Doonayo: Tamarta dhaqdhaqaaqa wareega

Xalka:

Daqiiqadda firfircoonida walxaha:

I = mr ² = (0.5)(0.1) ² = (0.5)(0.01) = 0.005 kg m ²

Tamarta dhaqdhaqaaqa wareega:

K E = 1/2 I ω ² = 1/2 (0.005) (2) ² = 1/2 (0.005) (4) = (0.005) (2) = 0.01 Joule

Akhri wax dheeraad ah

1. Xoog F ah ayaa lagu dabaqaa xadhig ku duuban jiid dhululubo ah. Xuddunta waa 2 N m daqiiqadda inertia- na waa 1 kg m ² , waa maxay dardargelinta xagasha dhululubada.

La yaqaan:

Wareegga ( τ ) = 2 N m

Daqiiqadda firfircoonida (I) = 1 kg m ²

La Doonayo: Xawaaraha xagasha ee silinda

Xalka:

Στ = I α

Στ = qulqulka saafiga ah, I = daqiiqad firfircooni, α = dardargelinta xaglaha

Dardargelinta xagasha ee dhululubada:

α = Στ / I = 2/1 = 2 rad/s ²

2. Xoog F ah ayaa lagu dabaqaa xadhig ku duuban jiid dhululubo ah. Cabbirka xoogga waa 10 N, gacanka dhululubo waa 0.2 m daqiiqadda inertia-na waa 1 kg m 2, Waa maxay dardargelinta xagasha dhululubo?

La yaqaan:

Xoogga (F) = 10 N

Radius-ka dhululubada (R) = 0.2 m

Daqiiqadda firfircoonida (I) = 1 kg m ²

La Doonayo: Xawaaraha xagasha ee silinda.

Xalka:

τ = FR

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

Xuddunta:

τ = FR = (1 0 N) (0.2 m) = 2 N m

Στ = I α

Στ = qulqulka saafiga ah, I = daqiiqad firfircooni, α = dardargelinta xaglaha

Dardargelinta xagasha ee dhululubada:

α = Στ / I = 2/1 = 2 rad/s ²

3. Xoog F ah ayaa lagu dabaqaa xadhig ku duuban jiid dhululubo ah. Cabbirka xooggu waa 10 N, dhexroorka dhululubada waa 0.2 m, miisaanka dhululubadana waa 20 kg m ^2,. Waa maxay dardargelinta xagasha dhululubada.

La yaqaan:

Xoogga (F) = 10 N

Radius-ka dhululubada (R) = 0.2 m

Cufka dhululubada (M) = 20 kg

La Doonayo: Xawaaraha xagal ee dhululubada

Xalka:

τ = FR = (1 0 N) (0.2 m) = 2 N m

Daqiiqadda firfircoonida:

I = 1⁄2 MR ² = 1⁄2 (20)(0.2) ² = 1⁄2 (20)(0.04) = 0.4 kg m ²

Dardargelinta xagasha ee dhululubada:

α = Στ / I = 2 / 0.4 = 5 rad/s ²

4. Baloog 1-kg ah oo ka laalaada xadhig ku duuban baloog dhululubo ah. Daqiiqadda firfircoonida ee baloogga waa 1 kg m ² iyo gacanka baloogga waa 0.2 m. Waa maxay dardargelinta xagasha ee baloogga. Dardargelinta cufisjiidadka awgeed waa 10 m/s ².

La yaqaan:

Daqiiqadda firfircoonida ee roogga (I) = 1 kg m ²

Cufka baloogga (m) = 1 kg

Dardargelinta cufisjiidadka awgeed (g) = 10 m/s ²

Miisaanka (w) = mg = (1 kg)(10 m/s ² ) = 10 kg m/s ² = 10 N

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 m ²

Dardargelinta xagasha:

α = Στ / I = 2/1 = 2 rad/s ²

5. Baloog 1-kg ah oo ka laalaada xadhig ku duuban baloog dhululubo ah. Cufnaanta baloogga waa 20 kg halka gacanka baloogga uu yahay 0,2 m. Waa maxay dardargelinta xagasha ee baloogga iyo dardargelinta xorta ah ee baloogga. Dardargelinta cufisjiidadka awgeed waa 10 m/s ².

La yaqaan:

Cufka roogga (M) = 20 kg

Gacanka roogga (R) = 0,2 m

Cufka baloogga (m) = 1 kg

Dardargelinta cufisjiidadka awgeed (g) = 10 m/s ²

Miisaanka (w) = mg = (1 kg)(10 m/s ² ) = 10 kg m/s ² = 10 N

La rabay: 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 MR ² = 1⁄2 (20)(0.2) ² = (10)(0.04) = 0.4 kg m ²

Dardargelinta xagasha ee boolalku:

α = Στ / I = 2 / 0.4 = 5 rad/s ²

Dardargelinta dhicitaanka xorta ah ee baloogga:

a = R α = (0.2)(5) = 1 m/s ²

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

Kubbad cufnaan (m) = 100 garaam = 100/1000 = 0.1 kg

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

La Doonayo: Daqiiqada firfircoonida kubadda (I)

Xalka:

I = mr ² = (0.1 kg)(0.3 m) ²

I = (0.1 kg)(0.09 m ² )

I = 0.009 kg m ²

2. Kubbad 100-garaam ah, m ₁ , iyo kubad 200-garaam ah, m ₂ , oo ay ku xiran yihiin 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 kubbadda 1 (m ₁ ) = 100 garaam = 100/1000 = 0.1 kg

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

Cufka kubbadda (m2 ₎ = 200 garaam = 200/1000 = 0.2 kg

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

La doonayay: daqiiqad firfircoon oo kubadaha ah

Jawaab:

I = m ₁ r ₁ ² + m ₂ r ₂ ²

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

I = (0.1 kg)( 0.09 m ² ) + (0.2 kg)( 0.09 m ² )

I = 0.009 kg m ² + 0.018 kg m ²

I = 0.027 kg m ²

3. Kubbad 200-garaam ah, m ₁ iyo kubad 100-garaam ah, m _2, oo ay ku xiran yihiin ul dhererkeedu yahay 60 cm. Iska ilow cufka usha. Dhidibka wareegga wuxuu ku yaal kubbadda m _2. Waa maxay daqiiqadda firfircoonida kubbadaha. Iska ilow cufka usha.

La yaqaan:

Cufka kubbadda 1 (m ₁ ) = 2 00 garaam = 200/1000 = 0.2 kg

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

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

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

La Doonayay: Daqiiqada firfircoonida kubadaha

Xalka:

I = m ₁ r ₁ ² + m ₂ r ₂ ²

I = (0.2 kg)( 0,6 m) ² + (0.2 kg)(0 ) ²

I = (0.2 kg)( 0.36 m ² ) + 0

I = 0.072 kg m ²

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 = m ₁ = m ₂ = m ₃ = m ₄ = 1 00 garaam = 100/1000 = 0.1 kg

Masaafada u dhaxaysa kubbadda 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: Daqiiqada firfircoonida

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 m ² ) + (0.1 kg)(0.09 m ^{2 )} + (0.1 kg)(0.09 m ² )

I = 0.036 kg m ²

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

La Doonayo: Daqiiqada firfircoonida

Xalka:

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

I = (1/12) ML ²

I = (1/12) (2 kg)( 2 m) ²

I = (1/12) (2 kg)(4 m ² )

I = (1/12)(8 kg m ² )

I = 8/12 kg m ²

I = 2/3 kg m ²

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

La Doonayo: Daqiiqada firfircoonida

Xalka:

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

I = (1/3) ML ²

I = (1/3) (2 kg)( 2 m) ²

I = (1/3) (2 kg)(4 m ² )

I = (1/3)(8 kg m ² )

I = 8/3 kg m ²

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

La Doonayo: Daqiiqada firfircoonida

La Doonayo: Daqiiqada firfircoonida

Xalka:

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

I = (1/2) MR ²

I = (1/2) (10 kg)(0.1 m) ²

I = (1/2) (10 kg)(0.01 m ² )

I = (1/2)(0.1 kg m ² )

I = 0.05 kg m ²

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

La rabay: daqiiqad firfircooni ah

Xalka:

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

I = (2/5) MR ²

I = (2/5)(20 kg)(0.1 m) ²

I = (2/5)(20 kg)(0.01 m ² )

I = (2/5)(0.2 kg m ² )

I = 0.4/5 kg m ²

I = 0.08 kg m ²

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: Daqiiqada Inertia

Xalka:

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

I = (1/ 12 ) M (a ² + b ² )

I = (1/12)(2)(0.5 ² + 0.2 ² )

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

I = (1/6)(0.29)

I = 0.29/6 kg m ²

Akhri wax dheeraad ah