Physics

1. A capacitor mai layi ɗaya yana farko ƙarfin C, izinin sarari kyauta is ε_o, yankin farantin is A, tazara tsakanin faranti is d. Idan the Faɗin farantin ya ƙaru sau 4, nisan da ke tsakanin faranti ya zama 2d kuma izinin sarari kyauta 5 neε_o, menene ƙarfin ƙarshe na capacitor-panel mai layi ɗaya.

An sani:

Ƙarfin capacitor = C

Izinin sarari kyauta = ε _o

Yankin P na ƙarshen = A

Nisa tsakanin faranti = d

Ana so: Ƙarfin capacitor (C)

Magani:

[irp]

2. Na'urar capacitor da ke da mafi girman ƙarfin aiki bisa ga hoton da ke ƙasa ita ce …

Magani:

Tsarin capacitor mai layi ɗaya:

Ƙarfin capacitors:

Capacitor wanda ke da mafi girman ƙarfin y shine capacitor C3_.

[irp]

3.

Ka yi la'akari da waɗannan abubuwan!
(1) Dielectric constant
(2) A bambancin yuwuwar tsakanin faranti
(3) Faranti thick
(4) Faɗin saman farantin
(5) Nisa tsakanin faranti
(6) Adadin cajin lantarki
Abubuwan da ke shafar ƙarfin faranti masu layi ɗaya capacitors sune…

Magani:

(1), (4) da (5)

4. Kwatanta ƙarfin capacitor 1 da 2 shine…

An sani:

Capacitor 1:

Faɗin saman = 2A

Nisa tsakanin faranti = d ₁

Capacitor 2:

Faɗin saman = A

Nisa tsakanin faranti = 2 d ₁

Ana so: Kwatanta ƙarfin capacitors 1 da 2

Magani:

[irp]

5.

Kayyade kwatancen ƙarfin capacitor na Parallel-plate I da II.

Magani:

Kwatanta ƙarfin capacitors na Parallel-plate I da II:

[irp]

6. Kapasito guda biyu masu layi ɗaya da aka nuna a cikin hoton da ke ƙasa.

Idan A ₁ = ½ A ₂ da d ₂ = 3 d ₁ to a tantance rabon ƙarfin capacitor mai layi ɗaya tsakanin hoton 2 da hoton 1.

An sani:

Mai haɗa farantin layi ɗaya I:

A ₁ = 1

d ₁ = 1

Layin capacitor II:

A ₂ = 2

d ₂ = 3

Ana so: rabon ƙarfin capacitor mai layi ɗaya tsakanin hoton 2 da hoton 1

Magani:

Mai haɗa farantin layi ɗaya I:

Layin capacitor II:

Rabon ƙarfin capacitors na parallel- plat II da I:

Karin bayani

1. Ƙayyade ƙarfin wutar lantarki a wani wuri da ke da nisan 1 cm daga caji 5.0 μ C. Madaidaitan Coulomb (k) = 9 x 10 ⁹ Nm ² C ⁻² , 1 μC = 10 ⁻⁶ C.

An sani:

Nisa daga caji (r) = 1 cm = 1/100 m = 0.01 m = 10 ^-2 m

Cajin (q) = 5.0 μ C = 5.0 x 10 ^-6 C

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

Ana nema: Ƙarfin wutar lantarki (V)

Magani:

Ƙarfin wutar lantarki:

Ƙarfin wutar lantarki shine 4.5 x 10 ⁶ Volts

[irp]

2. cajin Q₁ = 5.0 μC da caji Q₂ = 6.0 μC. Coulomb na yau da kullun (k) = 9 x 10⁹ Nm²C^-2, 1 μC = 10^-6 C. Maki na A yana tsakanin cajin. Ƙayyade ƙarfin wutar lantarki a wurin A.

An sani:

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

Nisa daga maki A daga Q ₁ = 10 cm = 0.1 m = 10 ^-1 m

Cajin Q ₂ = 6.0 μ C = 6.0 x 10 ^-6 C

Nisa daga maki a daga Q ₂ = 10 cm = 0.1 m = 10 ^-1 m

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

Ana so: Ƙarfin wutar lantarki a wuri na A

Magani:

Ƙarfin wutar lantarki 1 :

Ƙarfin wutar lantarki 2 :

Ƙarfin wutar lantarki a wuri na A :

V = V ₂ – V ₁

V = (54 – 45) x 10 ⁴

V = 9 x 10 ⁴

[irp]

3. cajin q₁ = 5.0 μC da caji q₂ = 6.0 μC. Coulomb na yau da kullun (k) = 9 x 10⁹ Nm²C^-2, 1 μC = 10^-6 C. Ƙayyade ƙarfin wutar lantarki a wurin A.

An sani:

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

Nisa daga maki A daga Q ₁ = 40 cm = 0.4 m = 4 x 10 ^-1 m

Cajin Q ₂ = 6.0 μ C = 6.0 x 10 ^-6 C

Nisa daga maki A daga Q ₂ = 50 cm = 0.5 m = 5 x 10 ^-1 m

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

Ana so: Ƙarfin wutar lantarki a wuri na A

Magani:

Ƙarfin wutar lantarki 1 :

Ƙarfin wutar lantarki 2 :

Ƙarfin wutar lantarki a wuri na 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

Ƙarfin wutar lantarki a wurin A shine -4.5 x 10 ³ Volts

Karin bayani

1. Ana hanzarta wutar lantarki daga hutawa ta hanyar bambancin V na 12. Menene canjin makamashin wutar lantarki na wutar lantarki?

An sani:

Cajin da ke kan na'urar lantarki (e) = -1.60 x 10 ^-19 Coulomb

Ƙarfin wutar lantarki = ƙarfin lantarki (V) = Volts 12

Ana so: Canjin makamashin lantarki na lantarki (ΔPE)

Magani:

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

Alamar ragewa tana nuna cewa ƙarfin kuzarin da ake da shi yana raguwa.

[irp]

2. Ana cajin faranti biyu masu layi daya. Raba tsakanin faranti shine 2 cm kuma girman filin lantarki tsakanin faranti shine 500 Volt/mita. Menene canjin kuzarin da ake samu na proton idan aka hanzarta daga farantin da aka caji mai kyau zuwa farantin da aka caji mara kyau.

An sani:

Girman filin lantarki tsakanin faranti (E) = Volt 500/mita

Nisa tsakanin faranti (s) = 2 cm = 0,02 m

Cajin da ke kan proton = +1.60 x 10 ^-19 Coulombs

Ana so: Canjin makamashin wutar lantarki (ΔPE)

Magani:

Ƙarfin wutar lantarki:

V = Es

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

V = Volt 10

Canjin makamashin lantarki:

ΔPE = qV

ΔPE = (1,60 x 10 ^-19 C)(10 V)

ΔPE = Joule 16 x 10 ^-19

ΔPE = 1.6 x 10 ^{-1 8} Joule

[irp]

3. An raba cajin maki biyu da nisan santimita 10. Caji a kan maki A = +9 μC da caji a kan maki B = -4 μC. k = 9 x 10 ⁹ Nm ² C ⁻² , 1 μC = 10 ⁻⁶ C. Menene canjin kuzarin wutar lantarki na caji a wurin B idan aka hanzarta zuwa wurin A?

An sani:

Cajin A (q ₁ ) = +9 μC = +9 x 10 ⁻⁶ C

Cajin B (q ₁ ) = -4 μC = -4 x 10 ⁻⁶ C

k = 9 x 10 ⁹ Nm ² C ⁻²

Nisa tsakanin caji A da B (r) = 10 cm = 0.1 m = 10 ^-1 m

Ana so: Canjin makamashin wutar lantarki (ΔEP)

Magani:

Karin bayani

1. Filin lantarki iri ɗaya E = 8000 N/C yana ratsawa ta cikin murabba'in fili mai faɗi A = 10 m2 ^. Kayyade kwararar wutar lantarki.

An sani:

Girman filin lantarki (E) = 8000 N/C

Yanki (A) = 10 m ²

θ = 0 ^o (kusurwar da ke tsakanin alkiblar filin lantarki da layin da aka zana a tsaye zuwa yankin)

Ana so: Fitar wutar lantarki ( Φ)

Magani:

Tsarin kwararar lantarki:

Φ = EA cos q

Φ = kwararar lantarki ( Nm ² /C) , E = filin lantarki (N/C), A = yanki (m ² ), q = kusurwa tsakanin layin filin lantarki da layin al'ada.

Gudun wutar lantarki:

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

[irp]

2. Filin lantarki iri ɗaya E = 5000 N/C yana ratsawa ta cikin murabba'in fili mai faɗi A = 2 m2 ^. Kayyade kwararar wutar lantarki.

An sani:

Filin wutar lantarki (E) = 5000 N/C

Yanki (A) = 2 m ²

θ = 60 ^o (kusurwar da ke tsakanin alkiblar filin lantarki da layin da aka zana a tsaye zuwa yankin)

Ana so: Fitar wutar lantarki ( Φ)

Magani:

Gudun wutar lantarki:

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

[irp]

3. Ƙwallon da ta yi ƙarfi mai radius na mita 0.5 tana da wutar lantarki mai ƙarfin μC 10 a tsakiyarta. Ka ƙayyade kwararar wutar lantarki da ke ratsa ƙwallon mai ƙarfi.

An sani:

Radius na ƙwallon (r) = 0.5 m

Cajin wutar lantarki (Q) = 10 μC = 10 x 10 ^-6 C

Ana so: Fitar wutar lantarki ( Φ)

Magani:

Filin lantarki:

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

Faɗin saman:

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

Gudun wutar lantarki:

Layukan filin lantarki suna daidaita da yankin, don haka kusurwar da ke tsakanin alkiblar filin lantarki da layin da aka zana a tsaye zuwa yankin, shine 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

Karin bayani

1. Lissafa girma da alkiblar filin lantarki a wurin A da ke da nisan 5 cm daga cajin maki Q = +10 μC. k = 9 x 10 ⁹ Nm ² C ⁻² , 1 μC = 10 ⁻⁶ C)

An sani:

Cajin wutar lantarki (Q) = +10 μC = +10 x 10 ^-6 C

Nisa tsakanin maki A da cajin maki Q (r _A ) = 5 cm = 0.05 m = 5 x 10 ^-2 m

k = 9 x 10 ⁹ Nm ² C ⁻²

Ana so: Girma da alkiblar filin lantarki a wuri na A

Magani:

Alkiblar filin lantarki a wuri na A:

Cajin wutar lantarki yana da kyau, don haka alkiblar filin wutar lantarki tana nesa da cajin wutar lantarki da maki A.

[irp]

2. Lissafa girma da alkiblar filin lantarki a wurin P da ke da nisan santimita 10 daga cajin maki Q = -2 0 μC. k = 9 x 10 ⁹ Nm ² C ⁻² , 1 μC = 10 ⁻⁶ C.

An sani:

Cajin wutar lantarki (q) = -20 μC = -20 x 10 ^-6 C

Nisa tsakanin wurin P da cajin lantarki (r _P ) = 10 cm = 0.1 m = 1 x 10 ^-1 m

k = 9 x 10 ⁹ Nm ² C ⁻²

Ana so: Girma da alkiblar filin lantarki a wuri na P

Magani:

Alkiblar filin lantarki a wuri na A:

Cajin wutar lantarki yana da rashin tabbas, don haka alkiblar filin wutar lantarki zuwa cajin wutar lantarki.

[irp]

3. An raba cajin maki biyu da nisan santimita 40. Menene girma da alkiblar filin lantarki a wurin P tsakanin cajin guda biyu, wato santimita 20 daga wurin A?

An sani:

Cajin A (qA ₎ = -2 μC = -2 x 10 ^-6 C

Cajin B (qB ₎ = + 4 μC = +4 x 10 ^-6 C

Nisa tsakanin caji A da maki P (r _AP ) = 20 cm = 0.2 m = 2 x 10 ^-1 m

Nisa tsakanin caji B da maki P (r _BP ) = 20 cm = 0.2 m = 2 x 10 ^-1 m

Ana so: Girma da alkiblar filin lantarki a wuri na P.

Magani:

Cajin A yana da korau don haka alkiblar wuraren filin lantarki tana zuwa ga Q _A (zuwa hagu).

Caji B yana da kyau don haka alkiblar wuraren filin lantarki tana nesa da Q _B (zuwa hagu).

Jimlar filin lantarki a aya ta A:

E = E _A + E _B

E = (4.5 x 10 ⁵ ) + (9 x 10 ⁵ )

E = 13.5 x 10 ⁵ N/C

Alkiblar filin lantarki tana nuni zuwa ga Q _A (zuwa hagu).

[irp]

4. Girman filin lantarki sifili ne a…

Cajin A yana da kyau kuma cajin B yana da kyau don haka girman filin lantarki shine sifili a wurin P, tsakanin caji biyu.

An sani:

Cajin A (qA ₎ = + 20 μC = +20 x 10 ⁻⁶ C

Cajin B (qB ₎ = +40 μC = +40 x 10 ⁻⁶ C

k = 9 x 10 ⁹ Nm ² C ⁻²

Nisa tsakanin caji A da caji B = 20 cm

Cajin tsakanin caji A da maki P (r _AP ) = a

Nisa tsakanin caji B da maki P (r _BP ) = 20 – a

Ana so: Girman filin lantarki sifili ne a….

Magani:

Girman filin lantarki da aka samar ta hanyar caji A a wurin P

Caji A yana da kyau don haka alkiblar wuraren filin lantarki tana nesa da caji A (zuwa dama).

Girman filin lantarki da aka samar ta hanyar caji B a wurin P:

Caji B yana da kyau don haka alkiblar wuraren filin lantarki tana nesa da caji B (zuwa hagu).

Jimlar filin lantarki a wurin P = 0:

Muna amfani da dabarar quadratic don tantance a.

Girman filin lantarki sifili ne wanda yake a 8 cm daga caji A ko 12 cm daga caji B.

[irp]

5. Dangane da hoton da ke ƙasa, w ga ma'anar P don haka filin lantarki a wurin P sifili ne? (k = 9 x 10 ⁹ Nm ² C ⁻² , 1 μC = 10 ⁻⁶ C)

Magani

To ƙididdige ƙarfin filin lantarki a wurin P, ana ɗauka a wurin P akwai cajin gwaji mai kyau. Q₁ yana da kyau kuma Q₂ yana da korau, don haka dole ne aya ta P ta kasance a gefen dama na Q₂ ko hagu na Q₁Idan batu na P yana gefen hagu na Q₁; filin lantarki da Q ya samar₁ a wurin P yana hagu (daga Q)₁) da kuma filin lantarki da Q ya samar₂ a wurin P zuwa dama (zuwa Q₁Alkiblar filin lantarki ta kasance akasin haka ta yadda filayen lantarki guda biyu suka kawar da juna ta yadda ƙarfin filin lantarki a wurin P ya zama sifili.

An sani:

Q1 = +9 _μC = +9 x 10 ⁻⁶ C

Q2 = -4 _μC = -4 x 10 ⁻⁶ C

k = 9 x 10 ⁹ Nm ² C ⁻²

Nisa tsakanin caji 1 da caji 2 = 3 cm

Nisa tsakanin Q ₁ da maki P (r _1P ) = a

Nisa tsakanin Q ₂ da maki P (r _2P ) = 3 + a

Ana so: wurin da wurin P yake domin filin lantarki a wurin P ya zama sifili

Magani:

Maki na P yana gefen hagu na _Q1.

Filin wutar lantarki da Q1 ya samar _a wuri na P:

Cajin gwajin yana da kyau kuma Q1 _yana da kyau don haka alkiblar filin lantarki tana zuwa hagu.

Filin wutar lantarki da Q2 ya samar _a wuri na P:

Cajin gwajin yana da kyau kuma Q2 _yana da rashin tabbas don haka alkiblar filin lantarki tana zuwa dama.

Filin wutar lantarki mai cikakken ƙarfi a wuri na A:

Yi amfani da dabarar quadratic don tantance:

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

Nisa tsakanin Q ₂ da maki P (r _2P ) = 3 + a = 3 – 1.8 = 1.2 cm.

Maki na P yana kan 1.2 cm dama na _Q1.

[irp]

Karin bayani

1. Cajin maki biyu, Q_A = +8 μC da Q_B = -5 μC, an raba su da nisa r = 10 cm. Menene girman ƙarfin lantarki. Daidaiton k = 8.988 x 10⁹ Nm²C^-2 = 9 x10⁹ Nm²C^-2.

An sani:

Cajin A (qA ₎ = +8 μC = +8 x 10 ^-6 C

Cajin B (qB ₎ = -5 μC = -5 x 10 ^-6 C

k = 9 x 10 ⁹ Nm ² C ⁻²

Nisa tsakanin caji A da B (r _AB ) = 10 cm = 0.1 m

Ana so : Girman ƙarfin wutar lantarki

Magani:

Tsarin dokar Coulomb :

Girman ƙarfin wutar lantarki:

[irp]

2. Ƙwayoyin cuta guda biyu da aka caji kamar yadda aka nuna a cikin hoton da ke ƙasa. Q _P = +10 μC da Q _q = +20 μC an raba su da nisa r = 10 cm. Menene girman ƙarfin lantarki.

An sani:

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

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

k = 9 x 10 ⁹ Nm ² C ⁻²

Nisa tsakanin caji P da Q (r _PQ ) = 12 cm = 0.12 m = 12 x 10 ^-2 m

Ana so: Girman ƙarfin wutar lantarki

Magani:

[irp]

3. An shirya ƙwayoyin cuta guda uku da aka caji a layi kamar yadda aka nuna a cikin hoton da ke ƙasa. Caji A = -5 μC, caji B = +10 μC da caji C = -12 μC. Lissafa ƙarfin lantarki mai ƙarfi akan ƙwayoyin B saboda sauran caji guda biyu.

An sani:

Cajin A (qA ₎ = -5 μC = -5 x 10 ^-6 C

Cajin B (qB ₎ = +10 μC = +10 x 10 ^-6 C

Cajin C (qC ₎ = -12 μC = -12 x 10 ^-6 C

k = 9 x 10 ⁹ Nm ² C ⁻²

Nisa tsakanin barbashi A da B (r _AB ) = 6 cm = 0.06 m = 6 x 10 ^-2 m

Nisa tsakanin barbashi B da C (r _BC ) = 4 cm = 0.04 m = 4 x 10 ^-2 m

Ana so: Girma da kuma alkiblar ƙarfin lantarki mai ƙarfi akan barbashi B

Magani:

Ƙarfin da ke kan barbashi B shine jimlar ƙarfin da F _BA ke yi akan barbashi B ta hanyar barbashi A da kuma ƙarfin da F _BC ke yi akan barbashi B ta hanyar barbashi C.

Ƙarfin F _BA da aka yi amfani da shi akan barbashi B ta hanyar barbashi A:

 

Alkiblar ƙarfin lantarki yana nuna barbashi A (yana nuna hagu).

Ƙarfin F _BC da aka yi amfani da shi akan barbashi B ta hanyar barbashi A:

 

Alkiblar ƙarfin lantarki yana nuna barbashi C (yana nuna dama).

Ƙarfin lantarki mai ƙarfi akan barbashi B :

F _B = F _AB – F _BC = 675 N – 125 N = 550 Newtons.

Alkiblar ƙarfin lantarki mai ƙarfi akan barbashi B yana nuna barbashi C (yana nuna dama).

[irp]

4. +Q ₁ = 10 μC, +Q ₂ = 50 μC da Q ₃ an raba su kamar yadda aka nuna a cikin hoton da ke ƙasa. Menene cajin lantarki akan barbashi 3 idan ƙarfin lantarki na lantarki akan barbashi 2 sifili ne.

An sani:

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

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

Nisa tsakanin caji 1 da 2 (r ₁₂ ) = 2 cm = 0.02 m = 2 x 10 ^-2 m

Nisa tsakanin caji 2 da caji 3 (r ₂₃ ) = 6 cm = 0.06 m = 6 x 10 ^-2 m

Ƙarfin lantarki mai ƙarfi akan barbashi 2 (F2 ₎ = 0

Ana nema : caji 3 (q ₃ )

Magani:

Ƙarfin da ke kan barbashi 2 shine jimillar ƙarfin F ₂₁ da aka yi wa barbashi 2 ta hanyar barbashi 1 da kuma ƙarfin F ₂₃ da aka yi wa barbashi 2 ta hanyar barbashi 3.

Ƙarfin F ₂₁ da aka yi amfani da shi akan barbashi 2 ta hanyar barbashi 1:

Alkiblar ƙarfin lantarki yana nuna barbashi na 3 (nuna zuwa dama).

Ƙarfin F₂₃ an yi amfani da barbashi 2 ta hanyar barbashi 3: 

Alkiblar ƙarfin lantarki yana nuna barbashi 1 (yana nuna hagu).

Ƙarfin lantarki mai ƙarfi akan barbashi 2 = 0:

Karin bayani

1. Abu mai lokacin inertia na 2 kg m ² yana juyawa a 1 rad/s. Menene kusurwar motsin abu?

An sani:

Lokacin inertia (I) = 2 kg m ²

Gudun kusurwa ( ω ) = 1 rad/s

Ana so: Motsin kusurwa (L)

Magani:

Tsarin motsin kusurwa:

L = I ω

L = ƙarfin kusurwa (kg m ² / s), I = lokacin inertia (kg m ² ), ω = saurin kusurwa (rad/s)

Motsin kusurwa:

L = I ω = (2)(1) = 2 kg m ² / s

[irp]

2. Pulley na silinda mai nauyin kilogiram 2 tare da radius na mita 0.1 yana juyawa a daidai lokacin da yake gudu 2 rad/s. Menene saurin kusurwar pulley ɗin?

An sani:

Nauyin kura (m) = 2 kg g

Radius na kura (r) = 0.1 m

Gudun kusurwa (ω) = 2 r ad/s

Ana so: Motsin kusurwa

Magani:

Tsarin lokacin inertia don silinda mai ƙarfi:

I = 1/2 mr ²

I = lokacin inertia (kg m ² ), m = nauyi (kg), r = radius (m)

Lokacin inertia:

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

Gudun kusurwa:

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

[irp]

3. Wani yanki mai nauyin kilogiram 2 wanda ke da radius na mita 0.2 yana juyawa a 4 rad/s. Menene ma'aunin kusurwa na ƙwallon.

An sani:

Nauyin ƙwallon (m) = 2 kg

Radius na ƙwallon (r) = 0.2 m

Gudun kusurwa (ω) = 4 ra d/s

Ana so: Motsin kusurwa

Magani:

Tsarin lokacin inertia don siffa mai kama da juna:

I = (2/5) mr ²

I = lokacin t na inertia (kg m ² ), m = nauyi (kg), r = radius (m)

Lokacin inertia don ƙwallo mai tsari ɗaya:

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

Motsin kusurwa na ƙwallo:

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

[irp]

4. Ƙwayar da ke da nauyin kilogiram 1 tana juyawa a daidai gudun kusurwa na 2 rad/s. Menene saurin kusurwa idan radius na da'irar ya kai santimita 10.

An sani:

Nauyin abu (m) = 1 kg g

Radius na da'ira (r) = 10 cm = 10/100 = 0.1 m

Gudun kusurwa (ω) = 2 rad/ s

Ana so: Motsin kusurwa

Magani:

Tsarin lokacin inertia ga barbashi:

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

Motsin kusurwa:

L = I ω = (0.01)(2) = 0.02 kg m ² / s

Karin bayani

1. Abu yana da lokacin inertia na 1 kg m ² yana juyawa a daidai gudun kusurwa na 2 rad/s. Menene kuzarin motsi na abin?

An sani:

Lokacin inertia (I) = 1 kg m ²

Gudun kusurwa ( ω) = 2 rad/s

Ana so: Ƙarfin motsi na juyawa (K E )

Magani:

Tsarin makamashin motsi na juyawa:

KE = 1/2 I ω ²

KE = makamashin motsi na juyawa (kg m²/s²), Ina = lokacin inertia (kg m²), ω = saurin kusurwa (ra/s)

Makamashin motsi na juyawa:

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

[irp]

2. Pulley na silinda mai nauyin kilogiram 20 tare da radius na mita 0.2 yana juyawa a daidai lokacin da yake gudu na 4 rad/s. Menene kuzarin motsi na juyawa na pulley?

An sani:

Nauyin silinda mai kura (m) = 20 kg g

Radius na silinda (r) = 0.2 m

Gudun kusurwa (ω) = 4 ra d/s

Ana so: Menene makamashin motsi na juyawa

Mafita;

Tsarin lokacin inertia na silinda:

I = 1/2 mr ²

I = lokacin inertia ( kg m ² ), m = nauyi (kg), r = radius (mita)

Lokacin inertia na silinda pulley:

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

Ƙarfin motsi na juyawa na kura:

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

[irp]

3. A- ƙwallon kilogiram 10 mai radius na mita 0.1 yana juyawa a daidai gwargwado na 10 rad/s. Menene kuzarin motsi na ƙwallon.

An sani:

Nauyin ƙwallon (m) = 10 kg

Radius na ƙwallon (r) = 0.1 m

Gudun kusurwa (ω) = 10 r ad/s

Ana so: Ƙarfin motsi na juyawa

Magani:

Tsarin lokacin inertia:

I = (2/5) mr ²

I = lokacin inertia (kg m ² ), m = nauyi (kg), r = radius (m)

Lokacin rashin kuzari na ƙwallon:

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

Ƙarfin motsi na juyawa na ƙwallon:

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

[irp]

4. Ƙwayar ƙwaya mai nauyin kilogiram 0.5 tana juyawa a daidai gudun kusurwa na 2 rad/s. Menene kuzarin motsi na jujjuyawar ƙwayar idan radius na da'irar ya kai santimita 10.

An sani:

Nauyin barbashi (m) = 0.5 kg g

Radius na ƙwallon (r) = 10 cm = 10/100 = 0.1 m

Gudun kusurwa (ω) = 2 rad/ s

Ana so: Ƙarfin motsi na juyawa

Magani:

Lokacin inertia ga ƙwayoyin cuta:

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

Makamashin motsi na juyawa:

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

Karin bayani

1. Ana amfani da ƙarfin F a kan igiya da aka naɗe a kusa da pulley na silinda. Ƙarfin juyi shine 2 N m kuma lokacin inertia shine 1 kg m ² , menene saurin kusurwa na silinda.

An sani:

Juyin juyi ( τ ) = 2 N m

Lokacin inertia (I) = 1 kg m ²

Ana so: Saurin kusurwa na silinda

Magani:

Στ = I α

Στ = karfin juyi, I = lokacin inertia, α = hanzarin kusurwa

Haɓaka silinda a kusurwa:

α = Στ / I = 2/1 = 2 rad/s ²

[irp]

2. Ana amfani da ƙarfin F a kan igiya da aka naɗe a kusa da pulley na silinda. Girman ƙarfin shine 10 N, radius na silinda shine 0.2 m kuma lokacin inertia shine 1 kg m 2, Menene saurin kusurwa na silinda?

An sani:

Ƙarfi (F) = 10 N

Radius na silinda (R) = 0.2 m

Lokacin inertia (I) = 1 kg m ²

Ana so: Saurin kusurwa na silinda.

Magani:

τ = FR

τ = ƙarfin juyi, F = ƙarfi, R = radius na silinda

Karfin juyi:

τ = FR = (1 0 N) (0.2 m) = 2 N m

Στ = I α

Στ = karfin juyi, I = lokacin inertia, α = hanzarin kusurwa

Haɓaka silinda a kusurwa:

α = Στ / I = 2/1 = 2 rad/s ²

[irp]

3. Ana amfani da ƙarfin F a kan igiya da aka naɗe a kusa da pulley na silinda. Girman ƙarfin shine 10 N, radius na silinda shine 0.2 m kuma nauyin silinda shine 20 kg m ^2, Menene saurin kusurwa na silinda.

An sani:

Ƙarfi (F) = 10 N

Radius na silinda (R) = 0.2 m

Nauyin silinda (M) = 20 kg

Ana so: Hanzarin kusurwa na silinda

Magani:

τ = FR = (1 0 N) (0.2 m) = 2 N m

Lokacin inertia:

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

Haɓaka silinda a kusurwa:

α = Στ / I = 2 / 0.4 = 5 rad/s ²

[irp]

4. Toshe mai nauyin kilogiram 1 da aka rataye a kan igiya da aka naɗe a kan injin silinda. Lokacin inertia na injin shine kilogiram 1 m ² kuma radius na injin shine mita 0.2. Menene saurin kusurwa na injin. Saurin nauyi saboda nauyi shine mita 10/s ².

An sani:

Lokacin inertia na kura (I) = 1 kg m ²

Nauyin tubali (m) = 1 kg

Saurin gudu saboda nauyi (g) = 10 m/s ²

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

Radius na kura (R) = 0.2 m

Ana so: Haɓaka kusurwa

Magani:

Karfin juyi:

τ = FR = w R = (1 0 N) (0.2 m) = 2 N m

Lokacin inertia:

I = 1 kg m ²

Haɓaka kusurwa:

α = Στ / I = 2/1 = 2 rad/s ²

[irp]

5. Toshe mai nauyin kilogiram 1 da aka rataye a kan igiya da aka naɗe a kan injin silinda. Nauyin injin pulley ɗin shine kilogiram 20 kuma radius na injin pulley ɗin shine mita 0,2. Menene saurin kusurwa na injin pulley ɗin da kuma saurin faɗuwar bututun. Saurin nauyi saboda nauyi shine mita 10/s ².

An sani:

Nauyin kura (M) = 20 kg

Radius na kura (R) = 0,2 m

Nauyin tubali (m) = 1 kg

Saurin gudu saboda nauyi (g) = 10 m/s ²

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

Ana so: saurin kusurwa na pulley da kuma saurin faɗuwar tubalan kyauta.

Magani:

Ƙarfin juyi:

τ = FR = w R = (1 0 N) (0.2 m) = 2 N m

Lokacin inertia na silinda pulley:

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

Saurin kusurwa na kusurwoyi:

α = Στ / I = 2 / 0.4 = 5 rad/s ²

Haɓaka faɗuwar free fall na tonon:

a = R α = (0.2)(5) = 1 m/s ²

Karin bayani

Lokacin inertia na barbashi

1. Kwallo mai nauyin gram 100 da aka haɗa da ƙarshen igiya mai tsawon santimita 30. Menene lokacin inertia na ƙwallon a kusa da axis na juyawa AB? Yi watsi da nauyin igiya.

An sani:

Axis na juyawa a AB

Kwallo mai nauyi (m) = gram 100 = 100/1000 = 0.1 kg

Nisa tsakanin ƙwallon da kuma axis na juyawa (r) = 30 cm = 0.3 m

Ana so: Lokacin rashin kuzari na ƙwallo (I)

Magani:

I = mr ² = (0.1 kg)(0.3 m) ²

I = (0.1 kg)(0.09 m ² )

I = 0.009 kg m ²

[irp]

2. Kwallo mai nauyin gram 100, m ₁ , da kuma ƙwallon gram 200, m ₂ , wanda aka haɗa ta da sanda mai tsawon santimita 60. Ba a yi la'akari da nauyin sandar ba. Axis na juyawa yana tsakiyar sandar. Menene lokacin inertia na ƙwallon game da axis na juyawa?

An sani:

Nauyin ƙwallon 1 (m ₁ ) = gram 100 = 100/1000 = 0.1 kg

Nisa tsakanin ƙwallon 1 da kuma axis na juyawa (r1 ₎ = 30 cm = 30/100 = 0.3 m

Nauyin ƙwallon (m2 ₎ = gram 200 = 200/1000 = 0.2 kg

Nisa tsakanin ƙwallon 2 da kuma axis na juyawa (r2 ₎ = 30 cm = 30/100 = 0.3 m

Ana so: lokacin rashin ƙarfin ƙwallo

Amsa:

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

[irp]

3. Kwallo mai nauyin gram 200, m ₁ da ƙwallon gram 100, m _2, an haɗa ta da sanda mai tsawon santimita 60. Yi watsi da nauyin sandar. Axis na juyawa yana wurin ƙwallon m _2. Menene lokacin inertia na ƙwallon. Yi watsi da nauyin sandar.

An sani:

Nauyin ƙwallon 1 (m ₁ ) = gram 200 = 200/1000 = 0.2 kg

Nisa tsakanin ƙwallon 1 da kuma axis na juyawa (r ₁ ) = 60 cm = 60/100 = 0.6 m

Nauyin ƙwallon 2 (m2 ₎ = gram 100 = 100/1000 = 0.1 kg

Nisa tsakanin ƙwallon 2 da kuma axis na juyawa (r2 ₎ = 0 m

Ana so: Lokacin rashin kuzarin ƙwallo

Magani:

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

[irp]

4. Nauyin kowanne ƙwallon gram 100 ne, wanda aka haɗa ta hanyar igiya. Tsawon igiyar shine cm 60 kuma faɗin igiyar shine cm 30. Menene lokacin inertia na ƙwallon a kusa da axis na juyawa. Yi watsi da nauyin igiya.

An sani:

Nauyin ƙwallon = m ₁ = m ₂ = m ₃ = m ₄ = 1 00 grams = 100/1000 = 0.1 kg

Nisa tsakanin ƙwallon da kuma axis na juyawa (r ₁ ) = 30 cm = 30/100 = 0.3 m

Nisa tsakanin ƙwallon 2 da kuma axis na juyawa (r ₂ ) = 30 cm = 30/100 = 0.3 m

Nisa tsakanin ƙwallon 3 da kuma axis na juyawa (r ₃ ) = 30 cm = 30/100 = 0.3 m

Nisa tsakanin ƙwallon 4 da kuma axis na juyawa (r ₄ ) = 30 cm = 30/100 = 0.3 m

Sananne: Lokacin rashin kuzari

Magani:

Ni = 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 ) + (0.1 kg)(0.09 m ² )

I = 0.036 kg m ²

[irp]

Lokacin inertia na abu mai tauri

5. Menene lokacin rashin ƙarfin sanda mai tsayin kilogiram 2 mai tsawon mita 2. Axis na juyawa yana tsakiyar sandar.

An sani:

Nauyin sanda (M) = 2 kg

Tsawon sandar (L) = 2 m

Ana so: Lokacin rashin kuzari

Magani:

Tsarin lokacin inertia lokacin da axis na juyawa yake a tsakiyar dogon sandar iri ɗaya:

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 ^m2

I = 2/3 kg ^m2

[irp]

6. Menene lokacin rashin ƙarfin sanda mai tsayin kilogiram 2 mai tsawon mita 2? Axis na juyawa yana nan a ƙarshen sandar.

An sani:

Nauyin sanda (M) = 2 kg

Tsawon sanda mai tauri (L) = 2 m

Ana so: Lokacin rashin kuzari

Magani:

Tsarin lokacin inertia lokacin da axis na juyawa yake a ƙarshen sandar:

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 ^m2

[irp]

7. Silinda mai ƙarfi mai nauyin kilogiram 10 tare da radius na mita 0.1. Axis na juyawa yana tsakiyar silinda mai ƙarfi, wanda aka nuna a cikin hoton da ke ƙasa. Menene lokacin inertia na silinda?

An sani:

Nauyin silinda mai ƙarfi (M) = 10 kg

Radius na silinda (L) = 0.1 m

Ana so: Lokacin inertia

Ana so: Lokacin inertia

Magani:

Tsarin lokacin inertia lokacin da axis na juyawa yake a tsakiyar silinda:

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. Wani yanki mai nauyin kilogiram 20 mai tsawon mita 0.1. An nuna axis na juyawa da ke tsakiyar yankin a cikin hoton da ke ƙasa.

An sani:

Nauyin ƙwallo (M) = 20 kg

Radius na ƙwallo (L) = 0.1 m

Ana so: lokacin rashin kuzari

Magani:

Tsarin lokacin inertia lokacin da axis na juyawa yake a tsakiyar ƙwallo:

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 ^m2

I = 0.08 kg m ²

[irp]

9. Faranti mai sirara mai nauyin kilogiram 2, mai tsawon mita 0.5 da faɗin mita 0.2. Axis na juyawa yana tsakiyar farantin mai kusurwa huɗu da aka nuna a cikin hoton da ke ƙasa. Menene lokacin inertia na murabba'in?

An sani:

Nauyin farantin murabba'i mai kusurwa huɗu (M) = 2 kg

Tsawon farantin (a) = 0.5 m

Faɗin farantin (b) = 0.2 m

Ana so: Lokacin Rashin Inertia

Magani:

Tsarin lokacin inertia lokacin da axis na juyawa yake a tsakiyar farantin:

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 ^m2

Karin bayani