fisiks

1. A capacitor ea poleiti e bapileng na lekgetlo la ntlha bokgoni C, tumello ya sebaka se lokolohileng is ε_o, sebaka sa poleiti is A, sebaka se pakeng tsa lipoleiti is d. Haebae sebaka sa poleiti se eketsehile ka makhetlo a 4, sebaka se pakeng tsa lipoleiti e fetoha 2d mme tumello ea sebaka se lokolohileng ke 5ε_o, bokgoni ba ho qetela ba capacitor ya parallel-plate ke bofe.

Tse tsejoang:

Bokgoni ba capacitor = C

T he tumello ea sebaka se lokolohileng = ε _o

Sebaka sa morao sa P = A

Sebaka se pakeng tsa lipoleiti = d

Ho batloa: Bokhoni ba capacitor (C)

Tharollo:

2. Capacitor e nang le bokgoni bo boholo ka ho fetisisa ho latela setshwantsho se ka tlase ke …

Tharollo:

Foromo ea capacitor ea poleiti e tšoanang:

Bokhoni ba li-capacitor:

Capacitor e nang le bokgoni bo boholo ka ho fetisisa ba y ke capacitor C3_.

3.

Nahana ka lintlha tse latelang!
(1) Phetoho ea dielektri
(2) The phapang e ka bang teng pakeng tsa lipoleiti
(3) Poleiti thick
(4) Sebaka se ka holimo sa poleiti
(5) Sebaka se pakeng tsa lipoleiti
(6) Palo ea tefiso ea motlakase
Mabaka a amang bokgoni ba dipoleiti tse bapileng li-capacitor ke…

Tharollo:

(1), (4) le (5)

4. Papiso ea bokhoni ba capacitor 1 le 2 ke…

Tse tsejoang:

Motlakase oa 1:

Sebaka sa bokaholimo = 2A

Sebaka se pakeng tsa lipoleiti = d ₁

Motlakase oa 2:

Sebaka sa bokaholimo = A

Sebaka se pakeng tsa lipoleiti = 2 d ₁

Ho batloa: Papiso ea bokhoni ba li-capacitor 1 le 2

Tharollo:

5.

Fumana papiso ea bokhoni ba capacitor ea Parallel-plate I le II.

Tharollo:

Papiso ea bokhoni ba li-capacitor tsa Parallel-plate I le II:

6. Li-capacitor tse peli tsa poleiti e bapileng tse bontšitsoeng setšoantšong se ka tlase.

Haeba A ₁ = ½ A ₂ le d ₂ = 3 d ₁ , fumana karolelano ea bokhoni ba capacitor ea parallel -plate pakeng tsa setšoantšo sa 2 le setšoantšo sa 1.

Tse tsejoang:

Capacitor ea poleiti e bapileng I:

A ₁ = 1

d ₁ = 1

Capacitor ea poleiti e bapileng II:

A ₂ = 2

d ₂ = 3

Ho batlwa: karolelano ea bokhoni ba capacitor ea poleiti e bapileng pakeng tsa setšoantšo sa 2 le setšoantšo sa 1

Tharollo:

Capacitor ea poleiti e bapileng I:

Capacitor ea poleiti e bapileng II:

Karolelano ea bokhoni ba li-capacitor tsa parallel- plate II le I:

Bala haholoanyane

1. Fumana matla a motlakase ntlheng e fumanehang ho 1 cm ho tloha tefisong 5.0 μ C. Coulomb constant (k) = 9 x 10 ⁹ Nm ² C ⁻² , 1 μC = 10 ⁻⁶ C.

Tse tsejoang:

Sebaka ho tloha ho tjhaja (r) = 1 cm = 1/100 m = 0.01 m = 10 ^-2 m

Tefiso (q) = 5.0 μ C = 5.0 x 10 ^-6 C

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

Ho batloa: Bokhoni ba motlakase (V)

Tharollo:

Bokhoni ba motlakase:

Bokgoni ba motlakase ke 4.5 x 10 ⁶ Volts

2. lefisa Q₁ = 5.0 μC le tefello Q₂ = 6.0 μC. Kamehla ea Coulomb (k) = 9 x 10⁹ Nm²C^-2, 1 μC = 10^-6 C. Ntlha ea A e pakeng tsa liqoso. Fumana matla a motlakase ntlheng ea A.

Tse tsejoang:

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

Sebaka sa ntlha A ho tloha ho Q ₁ = 10 cm = 0.1 m = 10 ^-1 m

Tefiso Q ₂ = 6.0 μ C = 6.0 x 10 ^-6 C

Sebaka sa ntlha a ho tloha Q ₂ = 10 cm = 0.1 m = 10 ^-1 m

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

Ho batloa: Matla a motlakase ntlheng ea A

Tharollo:

Bokhoni ba motlakase 1 :

Bokhoni ba motlakase 2 :

Bokhoni ba motlakase ntlheng ea A :

V = V ₂ – V ₁

V = (54 – 45) x 10 ⁴

V = 9 x 10 ⁴

3. lefisa q₁ = 5.0 μC le tefello q₂ = 6.0 μC. Kamehla ea Coulomb (k) = 9 x 10⁹ Nm²C^-2, 1 μC = 10^-6 C. Fumana matla a motlakase ntlheng ea A.

Tse tsejoang:

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

Sebaka sa ntlha A ho tloha ho Q ₁ = 40 cm = 0.4 m = 4 x 10 ^-1 m

Tefiso Q ₂ = 6.0 μ C = 6.0 x 10 ^-6 C

Sebaka sa ntlha A ho tloha ho Q ₂ = 50 cm = 0.5 m = 5 x 10 ^-1 m

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

Ho batloa: Matla a motlakase ntlheng ea A

Tharollo:

Bokhoni ba motlakase 1 :

Bokhoni ba motlakase 2 :

Bokhoni ba motlakase ntlheng ea 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

Bokgoni ba motlakase ntlheng ya A ke -4.5 x 10 ³ Volts

Bala haholoanyane

1. Electron e potlakiswa ho tloha phomolong ka phapang ya bokgoni ba 12 V. Phetoho ya matla a bokgoni ba motlakase a electron ke efe?

Tse tsejoang:

Tefiso ho elektrone (e) = -1.60 x 10 ^-19 Coulomb

Bokhoni ba motlakase = motlakase (V) = li-volt tse 12

Ho Batloa: Phetoho ea matla a motlakase a elektrone (ΔPE)

Tharollo:

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

Letšoao la minus le bontša hore matla a ka bang teng aa fokotseha.

2. Lipoleiti tse peli tse bapileng li tjhajwa. Karohano pakeng tsa lipoleiti ke 2 cm 'me boholo ba tšimo ea motlakase pakeng tsa lipoleiti ke 500 Volt/meter. Phetoho ea matla a ka bang teng a proton ke efe ha e potlakisoa ho tloha poleiting e tjhajwang hantle ho ea poleiting e tjhajwang hampe?

Tse tsejoang:

Boholo ba tšimo ea motlakase pakeng tsa lipoleiti (E) = 500 Volt/meter

Sebaka se pakeng tsa lipoleiti (li) = 2 cm = 0,02 m

Tefiso ho proton = +1.60 x 10 ^-19 Coulombs

Ho Batloa: Phetoho ea matla a motlakase (ΔPE)

Tharollo:

Bokhoni ba motlakase:

V = E s

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

V = 10 Volt

Phetoho ea matla a motlakase:

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. Litefiso tse peli tsa lintlha li arotsoe ka sebaka sa 10 cm. Tefiso ntlheng ea A = +9 μC 'me tefiso ntlheng ea B = -4 μC. k = 9 x 10 ⁹ Nm ² C ⁻² , 1 μC = 10 ⁻⁶ C. Phetoho ea matla a motlakase a tefiso ntlheng ea B ke efe haeba e potlakisoa ho ea ntlheng ea A?

Tse tsejoang:

Tefiso A (q ₁ ) = +9 μC = +9 x 10 ⁻⁶ C

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

k = 9 x 10 ⁹ Nm ² C ⁻²

Sebaka se pakeng tsa litefiso A le B (r) = 10 cm = 0.1 m = 10 ^-1 m

Ho Batloa: Phetoho ea matla a motlakase (ΔEP)

Tharollo:

Bala haholoanyane

1. Tšimo ea motlakase e tšoanang E = 8000 N/C e feta sebakeng se bataletseng sa sekwere A = 10 m 2. ^Fumana phallo ea motlakase.

Tse tsejoang:

Boholo ba tšimo ea motlakase (E) = 8000 N/C

Sebaka (A) = 10 m ²

θ = 0 ^o (sekhutlo se pakeng tsa tataiso ea tšimo ea motlakase le mola o huloang o otlolohileng sebakeng seo)

Ho batloa: Motlakase o phallang ( Φ)

Tharollo:

Mokhoa oa ho phalla ha motlakase:

Φ = EA cos q

Φ = phallo ea motlakase ( Nm ² /C) , E = tšimo ea motlakase (N/C), A = sebaka (m ² ), q = sekhutlo pakeng tsa mola oa tšimo ea motlakase le mola o tloaelehileng.

Phallo ea motlakase:

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

2. Tšimo ea motlakase e tšoanang E = 5000 N/C e feta sebakeng se bataletseng sa sekwere A = 2 m 2. ^Fumana phallo ea motlakase.

Tse tsejoang:

Tšimo ea motlakase (E) = 5000 N/C

Sebaka (A) = 2 m ²

θ = 60 ^o (sekhutlo se pakeng tsa tataiso ea tšimo ea motlakase le mola o huloang o otlolohileng sebakeng seo)

Ho batloa: Motlakase o phallang ( Φ)

Tharollo:

Phallo ea motlakase:

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

3. Bolo e tiileng e nang le radius ea limithara tse 0.5 e na le tjhaja ea motlakase ea 10 μC bohareng ba eona. Fumana phallo ea motlakase e fetang ka har'a bolo e tiileng.

Tse tsejoang:

Radius ea bolo (r) = 0.5 m

Tefiso ea motlakase (Q) = 10 μC = 10 x 10 ^-6 C

Ho batloa: Motlakase o phallang ( Φ)

Tharollo:

Tšimo ea motlakase:

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

Sebaka sa bokaholimo:

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

Phallo ea motlakase:

Mela ea tšimo ea motlakase e otlolohile sebakeng seo, hoo sekhutlo se pakeng tsa tataiso ea tšimo ea motlakase le mola o huloang o otlolohile sebakeng seo, e leng 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

Bala haholoanyane

1. Bala boholo le tataiso ea tšimo ea motlakase ntlheng ea A e fumanehang ho 5 cm ho tloha tefisong ea ntlha Q = +10 μC. k = 9 x 10 ⁹ Nm ² C ⁻² , 1 μC = 10 ⁻⁶ C)

Tse tsejoang:

Tefiso ea motlakase (Q) = +10 μC = +10 x 10 ^-6 C

Sebaka se pakeng tsa ntlha A le tjhaja ya ntlha Q (r _A ) = 5 cm = 0.05 m = 5 x 10 ^-2 m

k = 9 x 10 ⁹ Nm ² C ⁻²

Ho batloa: Boholo le tataiso ea tšimo ea motlakase ntlheng ea A

Tharollo:

Tsela eo tšimo ea motlakase e tsamaeang ka eona ntlheng ea A:

Tefiso ea motlakase e ntle kahoo tsela eo tšimo ea motlakase e tsamaeang ka eona e hole le tefiso ea motlakase le lintlha tsa A.

2. Bala boholo le tataiso ea tšimo ea motlakase ntlheng ea P e fumanehang ho 10 cm ho tloha tefisong ea ntlha Q = -2 0 μC. k = 9 x 10 ⁹ Nm ² C ⁻² , 1 μC = 10 ⁻⁶ C.

Tse tsejoang:

Tefiso ea motlakase (q) = -20 μC = -20 x 10 ^-6 C

Sebaka se pakeng tsa ntlha P le tjhaja ya motlakase (r _P ) = 10 cm = 0.1 m = 1 x 10 ^-1 m

k = 9 x 10 ⁹ Nm ² C ⁻²

Ho batloa: Boholo le tataiso ea tšimo ea motlakase ntlheng ea P

Tharollo:

Tsela eo tšimo ea motlakase e tsamaeang ka eona ntlheng ea A:

Tefiso ea motlakase e mpe ka hona e lebisa tsela eo tšimo ea motlakase e eang ho tjhaja ea motlakase ka eona.

3. Litefiso tse peli tsa lintlha li arotsoe ka sebaka sa 40 cm. Boholo le tataiso ea tšimo ea motlakase ke efe ntlheng ea P pakeng tsa litefiso tse peli, e leng 20 cm ho tloha ntlheng ea A?

Tse tsejoang:

Tefiso A (q _A ) = -2 μC = -2 x 10 ^-6 C

Tefiso B (q _B ) = + 4 μC = +4 x 10 ^-6 C

Sebaka se pakeng tsa tjhaja A le ntlha P (r _AP ) = 20 cm = 0.2 m = 2 x 10 ^-1 m

Sebaka se pakeng tsa tjhaja B le ntlha P (r _BP ) = 20 cm = 0.2 m = 2 x 10 ^-1 m

Ho batloa: Boholo le tataiso ea tšimo ea motlakase ntlheng ea P.

Tharollo:

Tefiso A e mpe hoo tataiso ea lintlha tsa tšimo ea motlakase e lebileng ho Q _A (ka letsohong le letšehali).

Tefiso ea B e ntle hoo tataiso ea lintlha tsa tšimo ea motlakase e leng hole le Q _B (ka letsohong le letšehali).

Tšimo eohle ea motlakase ntlheng ea A:

E = E _A + E _B

E = (4.5 x 10 ⁵ ) + (9 x 10 ⁵ )

E = 13.5 x 10 ⁵ N/C

Tsela eo tšimo ea motlakase e tsamaeang ka eona e supa Q _A (ka letsohong le letšehali).

4. Boholo ba tšimo ea motlakase ke lefela ho…

Tefiso A e ntle 'me tefiso B e ntle hoo boholo ba tšimo ea motlakase e leng lefela bo fumanehang ntlheng ea P, lipakeng tsa litefiso ka bobeli.

Tse tsejoang:

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

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

k = 9 x 10 ⁹ Nm ² C ⁻²

Sebaka se pakeng tsa tjhaja A le tjhaja B = 20 cm

Tefiso pakeng tsa tefiso A le ntlha P (r _AP ) = a

Sebaka se pakeng tsa tjhaja B le ntlha P (r _BP ) = 20 – a

Ho Batloa: Boholo ba tšimo ea motlakase ke lefela bo fumanehang ho….

Tharollo:

Boholo ba tšimo ea motlakase e hlahisoang ke tjhaja A ntlheng ea P

Tefiso A e ntle hoo tataiso ea lintlha tsa tšimo ea motlakase e leng hole le tefiso A (ka ho le letona).

Boholo ba tšimo ea motlakase e hlahisoang ke tjhaja B ntlheng ea P:

Tefiso ea B e ntle hoo tataiso ea lintlha tsa tšimo ea motlakase e leng hole le tefiso ea B (ka letsohong le letšehali).

Tšimo eohle ea motlakase ntlheng ea P = 0:

Re sebelisa foromo ea quadratic ho fumana a.

Boholo ba tšimo ea motlakase ke lefela bo fumanehang ho 8 cm ho tloha ho tjhaja A kapa 12 cm ho tloha ho tjhaja B.

5. Ho latela setshwantsho se ka tlase, ntlha ya P ke ena hore tshimo ya motlakase ntlheng ya P e be lefela? (k = 9 x 10 ⁹ Nm ² C ⁻² , 1 μC = 10 ⁻⁶ C)

tharollo

To bala matla a tšimo ea motlakase ntlheng ea P, ho nahanoa hore ntlheng ea P ho na le tefiso ea teko e ntle.₁ e ntle 'me Q₂ ke negative, ka hona ntlha P e tlameha ho ba ka letsohong le letona la Q₂ kapa ka letsohong le letšehali la Q₁Haeba ntlha P e le ka letsohong le letšehali la Q₁; tšimo ea motlakase e hlahisoang ke Q₁ ntlheng ea P e ka letsohong le letšehali (hole le Q₁) le tšimo ea motlakase e hlahisoang ke Q₂ ntlheng ea P ka ho le letona (ho ea ho Q₁). Tsela eo tšimo ea motlakase e tsamaeang ka eona e fapane hoo masimo a mabeli a motlakase a arohanang e le hore matla a tšimo ea motlakase ntlheng ea P e be lefela.

Tse tsejoang:

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

Sebaka se pakeng tsa tjhaja 1 le tjhaja 2 = 3 cm

Sebaka se pakeng tsa Q ₁ le ntlha P (r _1P ) = a

Sebaka se pakeng tsa Q ₂ le ntlha P (r _2P ) = 3 + a

Ho batloa: sebaka sa ntlha P e le hore tšimo ea motlakase ntlheng P e be lefela

Tharollo:

Ntlha ea P e ka letsohong le letšehali la Q ₁.

Tšimo ea motlakase e hlahisoang ke Q ₁ ntlheng ea P:

Tefiso ea teko e ntle 'me Q ₁ e ntle hoo tsela ea tšimo ea motlakase e leng ka letsohong le letšehali.

Tšimo ea motlakase e hlahisoang ke Q ₂ ntlheng ea P:

Tefiso ea teko ke e ntle 'me Q2 _ke e mpe e le hore tsela eo tšimo ea motlakase e tsamaeang ka eona e be ka ho le letona.

Tšimo ea motlakase ea net ntlheng ea A:

Sebelisa foromo ea quadratic ho fumana :

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

Sebaka se pakeng tsa Q ₂ le ntlha P (r _2P ) = 3 + a = 3 – 1.8 = 1.2 cm.

Ntlha ea P e ka letsohong le letona la Q _{1 ka 1.2 cm}.

Bala haholoanyane

1. Litefiso tsa lintlha tse peli, Q_A = +8 μC le Q_B = -5 μC, di arotswe ka sebaka r = 10 cm. Boholo ba matla a motlakase. K e sa fetoheng = 8.988 x 10⁹ Nm²C^-2 = 9x10⁹ Nm²C^-2.

Tse tsejoang:

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

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

k = 9 x 10 ⁹ Nm ² C ⁻²

Sebaka se pakeng tsa litefiso A le B (r _AB ) = 10 cm = 0.1 m

Ho batloa : Boholo ba matla a motlakase

Tharollo:

Foromo ea molao oa Coulomb :

Boholo ba matla a motlakase:

2. Dikaroloana tse pedi tse tjhajilweng jwalo ka ha ho bontshitswe setshwantshong se ka tlase. Q _P = +10 μC le Q _q = +20 μC di arotswe ka sebaka r = 10 cm. Boholo ba matla a motlakase ke bofe.

Tse tsejoang:

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

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

k = 9 x 10 ⁹ Nm ² C ⁻²

Sebaka se pakeng tsa litefiso P le Q (r _PQ ) = 12 cm = 0.12 m = 12 x 10 ^-2 m

Ho batloa: Boholo ba matla a motlakase

Tharollo:

3. Dikaroloana tse tharo tse tjhajilweng di hlophisitswe ka mola jwalo ka ha ho bontshitswe setshwantshong se ka tlase. Tefiso A = -5 μC, tefiso B = +10 μC mme tefiso C = -12 μC. Bala matla a motlakase a sa fetoheng hodima karoloana B ka lebaka la ditefiso tse ding tse pedi.

Tse tsejoang:

Tefiso A (q _A ) = -5 μC = -5 x 10 ^-6 C

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

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

k = 9 x 10 ⁹ Nm ² C ⁻²

Sebaka se pakeng tsa dikarolwana A le B (r _AB ) = 6 cm = 0.06 m = 6 x 10 ^-2 m

Sebaka se pakeng tsa dikarolwana tsa B le C (r _BC ) = 4 cm = 0.04 m = 4 x 10 ^-2 m

Ho batloa: Boholo le tataiso ea matla a motlakase a sa fetoheng holim'a karoloana ea B

Tharollo:

Matla a letlooa holim'a karoloana ea B ke kakaretso ea vector ea matla a F _BA a sebelisoang holim'a karoloana ea B ke karoloana ea A le matla a F _BC a sebelisoang holim'a karoloana ea B ke karoloana ea C.

Matla a F _BA a sebelisitsoeng holim'a karoloana ea B ke karoloana ea A:

 

Tsela eo matla a motlakase a kgutlang ka yona e supa karoloana ya A (ntlha ho ya ka letsohong le letshehadi).

Matla a F _BC a sebelisitsoeng holim'a karoloana B ke karoloana A:

 

Tsela eo matla a motlakase a kgutlang ka yona e supa karoloana ya C (ntlha e yang ka ho le letona).

Matla a motlakase a sa fetoheng holim'a karoloana ea B :

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

Tsela eo matla a motlakase a sa fetoheng ka yona hodima karoloana ya B e supang karoloana ya C (e supa ka ho le letona).

4. +Q ₁ = 10 μC, +Q ₂ = 50 μC le Q ₃ di arotswe jwalo ka ha ho bontshitswe setshwantshong se ka tlase. Tefiso ya motlakase hodima karoloana ya 3 ke efe haeba matla a motlakase a sa fetoheng hodima karoloana ya 2 e le lefela.

Tse tsejoang:

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

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

Sebaka se pakeng tsa litefiso 1 le 2 (r ₁₂ ) = 2 cm = 0.02 m = 2 x 10 ^-2 m

Sebaka se pakeng tsa tjhaja 2 le tjhaja 3 (r ₂₃ ) = 6 cm = 0.06 m = 6 x 10 ^-2 m

Matla a motlakase a sa fetoheng holim'a karoloana ea 2 (F ₂ ) = 0

Ho batloa : tjhaja 3 (q ₃ )

Tharollo:

Matla a letlooa holim'a karoloana ea 2 ke kakaretso ea vector ea matla a F ₂₁ a sebelisoang holim'a karoloana ea 2 ke karoloana ea 1 le matla a F ₂₃ a sebelisoang holim'a karoloana ea 2 ke karoloana ea 3.

Matla a F ₂₁ a sebelisitsoeng holim'a karoloana ea 2 ka karoloana ea 1:

Tsela eo matla a motlakase a kgutlang ka yona e supa karoloana ya 3 (supa ho ya ka ho le letona).

Matla a F₂₃ e sebediswa hodima karoloana ya 2 ke karoloana ya 3: 

Tsela eo matla a motlakase a kgutlelang ho yona e supa karoloana ya 1 (ntlha ho ya ka letsohong le letshehadi).

Matla a motlakase a sa fetoheng holim'a karoloana ea 2 = 0:

Bala haholoanyane

1. Ntho e nang le motsotso wa ho se inertia wa 2 kg m ² e potoloha ka 1 rad/s. Momentum ya ntho ke efe ?

Tse tsejoang:

Nako ea ho se inertia (I) = 2 kg m ²

Lebelo la Angular ( ω ) = 1 rad/s

Ho batloa: Momentum ea Angular (L)

Tharollo:

Foromo ea momentum ea angular:

L = Ke ω

L = motsamao o potolohang (kg m ² /s), I = motsotso wa ho se tsitse (kg m ² ), ω = lebelo le potolohang (rad/s)

Motsamao o potolohang:

L = I ω = (2)(1) = 2 kg m ² /s

2. Pulley ea silindara ea 2- kg e nang le radius ea 0.1 m e potoloha ka lebelo le sa fetoheng la 2 rad/s. Momentum ea pulley ea angular ke efe?

Tse tsejoang:

Boima ba pulley (m) = 2 kg g

Radius ea pulley (r) = 0.1 m

Lebelo la Angular (ω) = 2 r ad/s

Ho batloa: Momentum ea Angular

Tharollo:

Foromo ea motsotso oa inertia bakeng sa silindara e tiileng:

Ke = 1/2 monghali ²

I = motsotso oa ho se be le nako (kg m ² ), m = boima (kg), r = radius (m)

Motsotso oa ho se inertia:

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

Lebelo la angular:

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

3. Selikalikoe se lekanang sa 2-kg se nang le radius ea 0.2 m se potoloha ka 4 rad/s. Momentum ea bolo ke efe?

Tse tsejoang:

Boima ba bolo (m) = 2 kg

Radius ea bolo (r) = 0.2 m

Lebelo la Angular (ω) = 4 ra d/s

Ho batloa: Momentum ea Angular

Tharollo:

Foromo ea motsotso oa inertia bakeng sa selika-likoe se ts'oanang:

Ke = (2/5) monghadi ²

I = motsotso oa t oa inertia (kg m ² ), m = boima (kg), r = radius (m)

Motsotso oa inertia bakeng sa sekala se ts'oanang:

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

Motsamao o potolohang oa selika-likoe:

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

4. Karolwana e boima ba 1-kg e potoloha ka lebelo le sa fetoheng la 2 rad/s. Lebelo le ka lehlakoreng ke lefe haeba radius ya sedikadikwe e le 10 cm.

Tse tsejoang:

Boima ba ntho (m) = 1 kg g

Radius ea selikalikoe (r) = 10 cm = 10/100 = 0.1 m

Lebelo la angular (ω) = 2 rad/ s

Ho batloa: Momentum ea Angular

Tharollo:

Foromo ea motsotso oa inertia bakeng sa likaroloana:

Ke = monghadi ² = (1)(0.1) ² = (1)(0.01) = 0.01 kg m ²

Motsamao o potolohang:

L = I ω = (0.01)(2) = 0.02 kg m ² /s

Bala haholoanyane

1. Ntho e na le motsotso wa ho se inertia wa 1 kg m ² e potoloha ka lebelo le sa fetoheng la 2 rad/s. Matla a kinetic a ntho ke afe ?

Tse tsejoang:

Motsotso oa ho se inertia (I) = 1 kg m ²

Lebelo la angular ( ω) = 2 rad/s

Ho batloa: Matla a kinetic a potolohang (K E )

Tharollo:

Foromo ea matla a kinetic a potolohang:

KE = 1/2 I ω ²

KE = matla a kinetic a potolohang (kg m²/s²), Ke = motsotso oa ho se inertia (kg m²), ω = lebelo la angular (rad/s)

Matla a kinetic a potolohang:

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

2. Pulley ea silindara ea boima ba lik'hilograma tse 20 e nang le radius ea 0.2 m e potoloha ka lebelo le sa fetoheng la 4 rad/s. Matla a kinetic a potolohang a pulley ke afe?

Tse tsejoang:

Boima ba pulley ea silindara (m) = 20 kg g

Radius ea silindara (r) = 0.2 m

Lebelo la angular (ω) = 4 ra d/s

Ho batloa: Matla a kinetic a potolohang ke eng

Tharollo;

Foromo ea motsotso oa inertia ea silindara:

Ke = 1/2 monghali ²

I = motsotso oa ho se be le nako ( kg m2 ⁾ , m = boima (kg), r = radius (limithara)

Motsotso oa inertia ea pulley ea silindara:

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

Matla a kinetic a potolohang a pulley:

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

3. Bolo ea A- 10 kg e nang le radius ea 0.1 m e potoloha ka sekhahla sa 10 rad/s. Matla a kinetic a bolo ke afe?

Tse tsejoang:

Boima ba bolo (m) = 10 kg

Radius ea bolo (r) = 0.1 m

Lebelo la Angular (ω) = 10 r ad/s

Ho batloa: Matla a kinetic a potolohang

Tharollo:

Foromo ea motsotso oa inertia:

Ke = (2/5) monghadi ²

I = motsotso oa ho se be le nako (kg m ² ), m = boima (kg), r = radius (m)

Motsotso oa ho se tsitse ha bolo:

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

Matla a kinetic a potolohang a bolo:

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

4. Karolwana ya 0.5- kg e potoloha ka lebelo le sa fetoheng la 2 rad/s. Matla a kinetic a potolohang a karolwana ke afe haeba radius ya sedikadikwe e le 10 cm.

Tse tsejoang:

Boima ba karoloana (m) = 0.5 kg g

Radius ea bolo (r) = 10 cm = 10/100 = 0.1 m

Lebelo la angular (ω) = 2 rad/ s

Ho batloa: Matla a kinetic a potolohang

Tharollo:

Motsotso oa ho se sebetse hantle ha likaroloana:

Ke = monghadi ² = (0.5)(0.1) ² = (0.5)(0.01) = 0.005 kg m ²

Matla a kinetic a potolohang:

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

Bala haholoanyane

1. Matla F a sebediswa thapong e phuthetsweng ka pulley ya silindara. Torque ke 2 N m mme motsotso wa inertia ke 1 kg m ² , ho potlaka ha silindara ka angular ke eng .

Tse tsejoang:

Torque ( τ ) = 2 N m

Motsotso oa ho se inertia (I) = 1 kg m ²

Ho batloa: Ho potlaka ha silindara ka tsela e potolohang

Tharollo:

Στ = I α

Στ = torque ea net, I = motsotso oa inertia, α = ho potlakisa ha angular

Ho potlakisa silindara ka mokhoa o potolohang:

α = Στ / I = 2 / 1 = 2 rad/s ²

2. Matla F a sebediswa thapong e phuthetsweng ka pulley ya silindara. Boholo ba matla ke 10 N, radius ya silindara ke 0.2 m mme motsotso wa inertia ke 1 kg m 2, Potlakiso ya angular ya silindara ke eng ?

Tse tsejoang:

Matla (F) = 10 N

Radius ea silindara (R) = 0.2 m

Motsotso oa ho se inertia (I) = 1 kg m ²

Ho batloa: Ho potlaka ha silindara ka tsela e potolohang.

Tharollo:

τ = FR

τ = torque, F = matla, R = radius ea silindara

Torque:

τ = FR = (1 0 N) (0.2 m) = 2 N m

Στ = I α

Στ = torque ea net, I = motsotso oa inertia, α = ho potlakisa ha angular

Ho potlakisa silindara ka mokhoa o potolohang:

α = Στ / I = 2 / 1 = 2 rad/s ²

3. Matla F a sebediswa thapong e phuthetsweng ho potoloha pulley ya silindara. Boholo ba matla ke 10 N, radius ya silindara ke 0.2 m mme boima ba silindara ke 20 kg m ^2, . Ho potlaka ha silindara ka angular ke eng .

Tse tsejoang:

Matla (F) = 10 N

Radius ea silindara (R) = 0.2 m

Boima ba silindara (M) = 20 kg

Ho batloa: Ho potlakisa silindara ka tsela e potolohang

Tharollo:

τ = FR = (1 0 N) (0.2 m) = 2 N m

Motsotso oa ho se inertia:

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

Ho potlakisa silindara ka mokhoa o potolohang:

α = Στ / I = 2 / 0.4 = 5 rad/s ²

4. Sekotwana sa 1-kg se leketlileng thapong se phuthetsweng ka pulley ya silindara. Motsotso wa ho se sebetse hantle ha pulley ke 1 kg m ² mme radius ya pulley ke 0.2 m. Ho potlaka ha pulley ka lehlakoreng le leng ke eng? Ho potlaka ka lebaka la matla a khoheli ke 10 m/s ².

Tse tsejoang:

Nako ea ho se sebetse ha pulley (I) = 1 kg m ²

Boima ba boloko (m) = 1 kg

Ho potlaka ka lebaka la matla a khoheli (g) = 10 m/s ²

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

Radius ea pulley (R) = 0.2 m

Ho batloa: Ho potlaka ha Angular

Tharollo:

Torque:

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

Motsotso oa ho se inertia:

I = 1 kg m ²

Ho potlaka ha Angular:

α = Στ / I = 2 / 1 = 2 rad/s ²

5. Sekotwana sa 1-kg se leketlileng thapong se phuthetsweng ka pulley ya silindara. Boima ba pulley ke 20 kg mme radius ya pulley ke 0,2 m. Ho potlaka ha angular ha pulley le ho potlaka ha ho wa ha free block ke eng. Ho potlaka ka lebaka la matla a khoheli ke 10 m/s ².

Tse tsejoang:

Boima ba pulley (M) = 20 kg

Radius ea pulley (R) = 0,2 m

Boima ba boloko (m) = 1 kg

Ho potlaka ka lebaka la matla a khoheli (g) = 10 m/s ²

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

Ho batloa: ho potlaka ha khoele ea pulley le ho potlaka ha boloko ho sa lefelloeng.

Tharollo:

Torque:

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

Motsotso oa inertia ea pulley ea silindara:

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

Ho potlaka ha angular ha pulley:

α = Στ / I = 2 / 0.4 = 5 rad/s ²

Ho potlaka ha ho oa ha boloko ka bolokolohi:

a = R α = (0.2)(5) = 1 m/s ²

Bala haholoanyane

Motsotso oa inertia ea karoloana

1. Bolo ea ligrama tse 100 e hokahaneng pheletsong e 'ngoe ea thapo e bolelele ba 30 cm. Nako ea ho se sebetse hantle ha bolo e mabapi le axis ea potoloho AB ke efe? Hlokomoloha boima ba thapo.

Tse tsejoang:

Mokhahlelo oa potoloho ho AB

Bolo e boima (m) = 100 grams = 100/1000 = 0.1 kg

Sebaka se pakeng tsa bolo le mothapo oa potoloho (r) = 30 cm = 0.3 m

Ho batloa: Motsotso oa ho se tsitse ha bolo (I)

Tharollo:

Ke = monghadi ² = (0.1 kg)(0.3 m) ²

I = (0.1 kg)(0.09 ^m2 )

I = 0.009 kg m ²

2. Bolo ea ligrama tse 100, m ₁ , le bolo ea ligrama tse 200, m ₂ , li hokahantsoe ke thupa e bolelele ba 60 cm. Boima ba thupa ha bo hlokomolohuoe. Mothapo oa potoloho o bohareng ba thupa. Nako ea ho se sebetse hantle ha bolo ke efe mabapi le mothapo oa potoloho?

Tse tsejoang:

Boima ba bolo 1 (m ₁ ) = 100 grams = 100/1000 = 0.1 kg

Sebaka sa bolo 1 le axis ea potoloho (r ₁ ) = 30 cm = 30/100 = 0.3 m

Boima ba bolo (m ₂ ) = 200 grams = 200/1000 = 0.2 kg

Sebaka sa bolo 2 le axis ea potoloho (r ₂ ) = 30 cm = 30/100 = 0.3 m

Ho batloa: motsotso oa ho hloka botsitso ha libolo

Karabo:

Ke = 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 m2 ⁾ + (0.2 kg)( 0.09 ^m2 )

I = 0.009 kg m ² + 0.018 kg m ²

I = 0.027 kg m ²

3. Bolo ea ligrama tse 200, m ₁ le bolo ea ligrama tse 100, m _2, li hokahantsoe ke thupa e bolelele ba 60 cm. Hlokomoloha boima ba thupa. Mothapo oa potoloho o ho bolo ea m _2. Motsotso oa ho se sebetse hantle ha libolo ke ofe? Hlokomoloha boima ba thupa.

Tse tsejoang:

Boima ba bolo 1 (m ₁ ) = 2 00 grams = 200/1000 = 0.2 kg

Sebaka se pakeng tsa bolo ea 1 le axis ea potoloho (r ₁ ) = 60 cm = 60/100 = 0.6 m

Boima ba bolo 2 (m ₂ ) = 100 grams = 100/1000 = 0.1 kg

Sebaka se pakeng tsa bolo ea 2 le axis ea potoloho (r ₂ ) = 0 m

Ho batloa: Motsotso oa ho hloka botsitso ha libolo

Tharollo:

Ke = 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 ^m2 ) + 0

I = 0.072 kg m ²

4. Boima ba bolo ka 'ngoe ke ligrama tse 100, tse hokahaneng ka thapo. Bolelele ba thapo ke 60 cm 'me bophara ba thapo ke 30 cm. Nako ea ho se sebetse hantle ha bolo ke efe mabapi le axis ea potoloho? Hlokomoloha boima ba thapo.

Tse tsejoang:

Boima ba bolo = m ₁ = m ₂ = m ₃ = m ₄ = 1 00 grams = 100/1000 = 0.1 kg

Sebaka se pakeng tsa bolo le mothapo oa potoloho (r ₁ ) = 30 cm = 30/100 = 0.3 m

Sebaka se pakeng tsa bolo ea 2 le axis ea potoloho (r ₂ ) = 30 cm = 30/100 = 0.3 m

Sebaka se pakeng tsa bolo ea 3 le axis ea potoloho (r ₃ ) = 30 cm = 30/100 = 0.3 m

Sebaka se pakeng tsa bolo ea 4 le axis ea potoloho (r ₄ ) = 30 cm = 30/100 = 0.3 m

Tse tsejoang: Motsotso oa ho se inertia

Tharollo:

Ke = 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 + (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 m ²

Motsotso oa ho hloka botsitso ha ntho e thata

5. Nako ea ho se sebetse hantle ha thupa e telele ea 2 kg e lekanang e bolelele ba limithara tse 2 ke efe. Mothapo oa ho potoloha o bohareng ba thupa.

Tse tsejoang:

Boima ba thupa (M) = 2 kg

Bolelele ba thupa (L) = 2 m

Ho batloa: Motsotso oa ho se inertia

Tharollo:

Foromo ea motsotso oa inertia ha motsoako oa potoloho o le bohareng ba molamu o molelele o ts'oanang:

Ke = (1/12) ML ²

Ke = (1/12) (2 kg)( 2 m) ²

I = (1/12) (2 kg)(4 ^m2 )

I = (1/12)(8 kg ^m2 )

I = 8/12 kg m ²

I = 2/3 kg m ²

6. Nako ea ho se sebetse hantle ha thupa e bolelele ba 2 kg e lekanang e bolelele ba limithara tse 2 ke efe? Mothapo oa potoloho o fumaneha pheletsong e 'ngoe ea thupa.

Tse tsejoang:

Boima ba thupa (M) = 2 kg

Bolelele ba thupa e thata (L) = 2 m

Ho batloa: Motsotso oa ho se inertia

Tharollo:

Foromo ea motsotso oa inertia ha motsoako oa potoloho o le pheletsong e 'ngoe ea molamu:

Ke = (1/3) ML ²

Ke = (1/3) (2 kg)( 2 m) ²

I = (1/3) (2 kg)(4 ^m2 )

I = (1/3)(8 kg ^m2 )

I = 8/3 kg m ²

7. Silindara e tiileng ea boima ba lik'hilograma tse 10 e nang le radius ea 0.1 m. Mothapo oa potoloho o bohareng ba silindara e tiileng, o bontšitsoeng setšoantšong se ka tlase. Nako ea ho se sebetse ha silindara ke efe?

Tse tsejoang:

Boima ba silindara e tiileng (M) = 10 kg

Radius ea silindara (L) = 0.1 m

Ho batloa: Motsotso oa ho se inertia

Ho batloa: Motsotso oa ho se inertia

Tharollo:

Foromo ea motsotso oa inertia ha motsoako oa potoloho o le bohareng ba silindara:

Ke = (1/2) MR ²

I = (1/2) (10 kg)(0.1 m) ²

I = (1/2) (10 kg)(0.01 ^m2 )

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

I = 0.05 kg m ²

8. Selikalikoe se lekanang sa 20 kg se bolelele ba 0.1 m. Mothapo oa potoloho o bohareng ba selikalikoe o bontšitsoe setšoantšong se ka tlase.

Tse tsejoang:

Boima ba selika-likoe (M) = 20 kg

Radius ea selika-likoe (L) = 0.1 m

Ho batloa: motsotso oa ho se inertia

Tharollo:

Foromo ea motsotso oa inertia ha motsoako oa potoloho o le bohareng ba lebala:

Ke = (2/5) MR ²

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 m ²

I = 0.08 kg m ²

9. Poleiti e tšesaane e khutlonnetsepa ea 2-kg e bolelele ba 0.5 m le bophara ba 0.2 m. Mothapo oa potoloho o bohareng ba poleiti e khutlonnetsepa e bontšitsoeng setšoantšong se ka tlase. Nako ea ho se sebetse hantle ha khutlonnetsepa ke efe?

Tse tsejoang:

Boima ba poleiti e khutlonnetsepa (M) = 2 kg

Bolelele ba poleiti (a) = 0.5 m

Bophara ba poleiti (b) = 0.2 m

Ho Batloa: Nako ea Boima

Tharollo:

Foromo ea motsotso oa inertia ha motsoako oa potoloho o le bohareng ba poleiti:

Ke = (1/12 ) M (a ² + b ² )

Ke = (1/12)(2)(0.5 ² + 0.2 ² )

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

Ke = (1/6)(0.29)

I = 0.29/6 kg m ²

Bala haholoanyane