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:

Bokhoni ba capacitor =C

Ttumello ea sebaka se lokolohileng = ε_o

Psebaka sa morao =A

The distanta pakeng tsa lipoleiti = d

Oa Batla: Bokgoni ba capacitor (C)

Tharollo:

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

Tharollo:

Foromo ea capacitor ea poleiti e tšoanang:

Bokhoni ba li-capacitor:

Capacitor e nang le bokgoni bo boholo ka ho fetisisay ke capacitor C_3.

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:

Moqapi 1:

Sebaka se kahare = 2A

Sebaka se pakeng tsa lipoleiti =d₁

Moqapi 2:

Sebaka se kahare =A

Sebaka se pakeng tsa lipoleiti = 2 d₁

Oa Batla: Papiso ea bokhoni ba li-capacitor 1 le 2

Tharollo:

5.

Fumana papiso ea bokhoni ba Li-capacitor tsa poleiti e tšoanang I le II.

Tharollo:

Papiso ea bokhoni ba Li-capacitor tsa poleiti e tšoanang I le II:

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

Haeba A₁ = ½ A₂ d₂ = 3 d₁ ebe ikemiselitse karolelano ea bokhoni ba parallel-capacitor ea poleiti pakeng tsa setšoantšo sa 2 le setšoantšo sa 1.

Tse tsejoang:

Capacitor ea poleiti e bapileng I:

A₁ = 1

d₁ = 1

Capacitor ea poleiti e bapileng II:

A₂ = 2

d₂ = 3

Oa Batla: karolelano ea bokhoni ba capacitor ea poleiti e tšoanang pakeng tsa setšoantšo sa 2 le setšoantšo sa 1

Tharollo:

Capacitor ea poleiti e bapileng I:

Capacitor ea poleiti e bapileng II:

TKarolelano ea bokhoni ba ho bapa-poleiti ea capacitor II le I:

Bala haholoanyane

1. Etsa qeto ea bokgoni ba motlakase sebakeng se bohōle ba 1 cm ho tloha tefisong ea 5.0 μC. Coulomb e sa fetoheng (k) = 9 x 10⁹ Nm²C^-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

lefisa (q) = 5.0 μC = 5.0 x 10^-6 C

Coulomb kamehla (k) = 9 x 10⁹ Nm²C^-2

Ho batloa: Bokhoni ba motlakase (V)

Tharollo:

Bokhoni ba motlakase:

Bokhoni ba motlakase ke 4.5: x 10⁶ Volt

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:

lefisa Q₁ = -5.0 μC = -5.0 x 10^-6 C

The distanta ea ntlha ea A ho tloha Q₁ = 10 cm = 0.1 m = 10^-1 m

lefisa Q₂ = 6.0 μC = 6.0 x 10^-6 C

Sebaka sa ntlha a ho tloha ho Q₂ = 10 cm = 0.1 m = 10^-1 m

Kamehla ea Coulomb (k) = 9 x 10⁹ Nm²C^-2

Ho batloa: Bokhoni ba motlakase ntlheng ea A

Tharollo:

Bokhoni ba motlakase 1 :

Bokhoni ba motlakase 2 :

Bokhoni ba motlakase ntlheng ea A :

V = V₂ - V₁

V = (54 - 45) x 10⁴

V = 9 x 10⁴

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:

lefisa Q₁ = -5.0 μC = -5.0 x 10^-6 C

Sebaka sa ntlha A ho tloha ho Q₁ = 40 cm = 0.4 m = 4 x 10^-1 m

lefisa Q₂ = 6.0 μC = 6.0 x 10^-6 C

Sebaka sa ntlha A ho tloha ho Q₂ = 50 cm = 0.5 m = 5 x 10^-1 m

Kamehla ea Coulomb (k) = 9 x 10⁹ Nm²C^-2

Ho batloa: Bokhoni ba motlakase ntlheng A

Tharollo:

Bokhoni ba motlakase 1 :

Bokhoni ba motlakase 2 :

Bokhoni ba motlakase ntlheng ea A :

V = V₁ + V₂

V = (-11.25 + 10.8) x 10⁴

V = -0.45 x 10⁴

V = -4.5 x 10³

Bokgoni ba motlakase ntlheng ya A ke -4.5 x 10³ Volt

Bala haholoanyane

1. Electron e potlakiswa ho tloha phomolong ka phapang e ka bang teng ea 12 V. Phetoho ke efe ho matla a motlakase ea elektrone?

Tse tsejoang:

The tefiso hodima elektrone (e) = -1.60 x 10^-19 Coulomb

Bokhoni ba motlakase = Palo ea li-volts (V) = 12 Volt

Oa Batla: 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 mme 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 nang le tjhaja e ntle ho ea poleiting e nang le tjhaja e mpe?

Tse tsejoang:

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

The distanta pakeng tsa lipoleiti (li) = 2 cm = 0,02 m

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

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^-18 Joule

3. Litefiso tse peli tsa lintlha li arotsoe ka sebaka sa 10 cm. Litefiso ntlheng ea A =+9 μC le tefiso ntlheng ea B = -4 μC. k = 9 x 10⁹ Nm²C^-2, 1 μC = 10^-6 C. Phetoho ea matla a motlakase a tjhaja ntlheng ea B ke efe haeba e potlakisetsoa ntlheng ea A?

Tse tsejoang:

Tefiso A (q)₁) = +9 μC = +9 x 10^-6 C

lefisa B (q₁) = -4 μC = -4 x 10^-6 C

k = 9 x 10⁹ Nm²C^-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. Yunifomo tšimo ea motlakase E = 8000 N/C e fetang sebakeng se bataletseng sa sekwere A = 10 m². Etsa qeto ea phallo ea motlakase.

Tse tsejoang:

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

Sebaka (A) = 10 m²

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

Oa Batla: Motlakase oa motlakase ((Φ)

Tharollo:

Mokhoa oa ho phalla ha motlakase:

Φ = EA cos q

Φ = phallo ea motlakase (Nm²/C), E = tšimo ea motlakase (N/C), A = sebaka (m²), q = sekhutlo pakeng tsa mola wa tshimo ya motlakase le mola o tlwaelehileng.

Phallo ea motlakase:

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

2. Tšimo ea motlakase e tšoanang E = 5000 N/C e fetang sebakeng se bataletseng sa sekwere A = 2 m². Fumana phallo ea motlakase.

Tse tsejoang:

Tšimo ea motlakase (E) = 5000 N/C

Sebaka (A) = 2 m²

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

Ho batloa: Motlakase oa motlakase ((Φ)

Tharollo:

Phallo ea motlakase:

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

Tharollo:

Tšimo ea motlakase:

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.

Sebaka sa bokaholimo:

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

Phallo ea motlakase:

Methapo ea motlakase e otlolohile sebakeng seo, e le hore sekhutlo se pakeng tsa tataiso ea tšimo ea motlakase le mola o huloang o otlolohileng sebakeng seo, ke 0^o.

Φ = EA cos q

Φ = (3.60: x 10⁵)(3.14)(cos 0)

Φ = (11.304 x 10⁵)(1)

Φ = 11.304 x 10⁵

Φ = 1.13 x 10⁶ Nm²/C

Bala haholoanyane

1. Bala boholo le tataiso ea tšimo ea motlakase ntlheng A e fumanehang ho 5 cm ho tloha tefisong ea ntlha Q = + 10 μC. k = 9 x 10⁹ Nm²C^-2, 1 μC = 10^-6 C)

Tse tsejoang:

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

The distanta pakeng tsa ntlha A le ntlha ea tjhaja Q (r)_A) = 5 cm = 0.05 m = 5 x 10^-2 m

k = 9 x 10⁹ Nm²C^-2

Oa Batla: Boholo le tataiso ea tšimo ea motlakase ntlheng 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 = -20 μC. k = 9 x 10⁹ Nm²C^-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 10⁹ Nm²C^-2

Oa Batla: Boholo le tataiso ea tšimo ea motlakase ntlheng 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 ho le letšehali).

Tefiso ea B e ntle hoo tataiso ea lintlha tsa tšimo ea motlakase e leng hole le Q_B (Ka ho le letšehali).

Tšimo eohle ea motlakase ntlheng ea A:

E = E_A +E_B

E = (4.5 x 10⁵) + (9 x 10⁵)

E = 13.5 x 10⁵ N / C.

Tsela eo tšimo ea motlakase e tsamaeang ka eona e supa Q_A (Ka ho 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 bo leng ho lefela ntlheng ea P, pakeng tsa liqoso tseo ka bobeli.

Tse tsejoang:

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

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

k = 9 x 10⁹ Nm²C^-2

Sebaka se pakeng tsa litefiso A le tefiso 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.

5Ho latela setšoantšo se ka tlase, wntlha P ke ena hoo tšimo ea motlakase ntlheng ea P e leng lefela? (k = 9 x 10⁹ Nm²C^-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:

Q₁ = +9 μC = +9 x 10^-6 C

Q₂ = -4 μC = -4 x 10^-6 C

k = 9 x 10⁹ Nm²C^-2

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

Sebaka se pakeng tsa Q₁ le ntlha P (r_1P) = a

Sebaka se pakeng tsa Q₂ le ntlha P (r_2P) = 3 + a

Ho batloa: sebaka sa ntlha P e le hore Tšimo ea motlakase ntlheng ea P ke lefela

Tharollo:

PNtlha ea P e ka letsohong le letšehali la Q₁.

Tšimo ea motlakase e hlahisitsoeng ke Q₁ ntlheng ea P:

Tefiso ea teko e ntle 'me Q₁ e ntle hoo tsela eo tšimo ea motlakase e shebaneng le eona e leng ka letsohong le letšehali.

Tšimo ea motlakase e hlahisitsoeng ke Q₂ ntlheng ea P:

Tefiso ea teko e ntle 'me Q₂ e mpe hoo tsela eo tšimo ea motlakase e eang 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 Q₂ le ntlha P (r_2P) = 3 + a = 3 – 1.8 = 1.2 cm.

Ntlha ea P e ka letsohong le letona la Q ka 1.2 cm₁.

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:

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

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

k = 9 x 10⁹ Nm²C^-2

The distanta pakeng tsa tefiso A le B (r_AB) = 10 cm = 0.1 m

o ne a batla : 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._P = +10 μC le Q_q = +20 μC li arotsoe 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 10⁹ Nm²C^-2

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

Ho batloa: Tboholo 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 10⁹ Nm²C^-2

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

Sebaka se pakeng tsa likaroloana 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 e sebelisoang holim'a karoloana ea B ke karoloana ea A le matla a F_BC e sebediswa hodima karoloana ya B ke karoloana ya C.

Matla a F_BA e sebediswang hodima karoloana B ke karoloana 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 e sebediswang hodima 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. +Q₁ = 10 μC, +Q₂ = 50 μC le Q₃ li arotsoe joalo ka ha ho bontšitsoe setšoantšong se ka tlase. Tefiso ea motlakase e sa fetoheng holim'a karoloana ea 3 ke efe haeba matla a motlakase a sa fetoheng holim'a karoloana ea 2 e le lefela.

Tse tsejoang:

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

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

Sebaka se pakeng tsa litefiso 1 le 2 (r₁₂) = 2 cm = 0.02 m = 2 x 10^-2 m

Sebaka se pakeng tsa tjhaja 2 le tjhaja 3 (r₂₃) = 6 cm = 0.06 m = 6 x 10^-2 m

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

o ne a batla : tefiso 3 (q₃)

Tharollo:

Matla a letlooa holim'a karoloana ea 2 ke vector kakaretso ea matla F₂₁ e sebelisoang holim'a karoloana ea 2 ke karoloana ea 1 le matla a F₂₃ e sebediswa hodima karoloana ya 2 ke karoloana ya 3.

Matla a F₂₁ e sebediswa hodima karoloana ya 2 ke karoloana ya 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 oa inertia of 2 lik'hilograma m² e potoloha ho Rad/s e le 1. ke seo botsitso ba angular ea ntho eo?

Tse tsejoang:

Motsotsoana oa inertia (I) = 2 kg m²

Lebelo la angular (ω) = 1 rad/s

Ho batloa: Angular lebelo (L)

Tharollo:

Foromo ea momentum ea angular:

L = KE ω

L= botsitso ba angular (kg m²/s), I = motsotso oa inertia (kg m²), ω = lebelo la angular (rad/s)

Motsamao o potolohang:

L = KE ω = (2)(1) = 2 kg m²/s

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

Tse tsejoang:

Boholo ea pulley (M) = 2 kg

Radius ea pulley (r) = 0.1 m

Lebelo la angular (ω) = 2 rpapatso/lipapatso

Ho batloa: Angular lebelo

Tharollo:

Foromo ea motsotso oa inertia bakeng sa silindara e tiileng:

I = 1/2 Mong²

ke = motsotso oa inertia (kg m²), m = boima (kg), r = radius (M)

Motsotso oa ho se inertia:

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

Lebelo la angular:

L = KE ω = (0.01)(2) = 0.02 kg m²/s

3. Selikalikoe se lekanang sa 2 kg se nang le radius ea 0.2 limithara e potoloha ho Rad/s e le 4. Motsamao o potolohang oa bolo ke ofe?

Tse tsejoang:

Boholo ea bolo (M) = 2 kg

Radius ea bolo (r) = 0.2 m

Lebelo la angular (ω) = 4 rad/s

Ho batloa: Angular lebelo

Tharollo:

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

I = (2/5) Monghali.²

Ke = motsotsot ea inertia (kg m²), m = boima (kg), r = radius (M)

Motsotso oa inertia bakeng sa sekala se ts'oanang:

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

Motsamao o potolohang oa selika-likoe:

L = I ω = (0.032)(4) = 0.128 kg m²/s

4. A 1 kg karoloana e potoloha ka lebelo le sa fetoheng la Rad/s e le 2. Lebelo la angular ke lefe haeba radius ea selikalikoe e le 1?Cm 0.

Tse tsejoang:

Boholo ea ntho (M) = 1 kg

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

Lebelo la angular (ω) = 2 rad/s

Ho batloa: Angular lebelo

Tharollo:

Foromo ea motsotso oa inertia bakeng sa likaroloana:

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

Motsamao o potolohang:

L = I ω = (0.01)(2) = 0.02 kg m²/s

Bala haholoanyane

1. Ntho e na le motsotso oa inertia ea 1 kg m² e potoloha ka lebelo le sa fetoheng la Rad/s e le 2. ke seo matla a kinetic a potolohang ea ntho eo?

Tse tsejoang:

Motsotso oa inertia (I) = 1 kg m²

The lebelo la maqhubu (ω) = 2 rad/s

Oa Batla: Matla a kinetic a potolohang (KE)

Tharollo:

Foromo ea matla a kinetic a potolohang:

KE = 1/2 I ω²

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

2. Boima ba 20-kg pulley ea silindara ka 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:

Boholo ea pulley ea silindara (M) = 20 kg

Radius ea silindara (r) = 0.2 m

Lebelo la angular (ω) = 4 rad/s

Ho batloa: Matla a kinetic a potolohang ke eng

Tharollo;

Foromo ea motsotso oa inertia ea silindara:

I = 1/2 Mong²

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

Motsotso oa inertia ea pulley ea silindara:

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

Matla a kinetic a potolohang a pulley:

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

3. A-10 lik'hilograma bolo e nang le radius ea 0.1 limithara e potoloha ka lebelo le sa fetoheng la Rad/s e le 10. Matla a kinetic a bolo ke afe?

Tse tsejoang:

Boholo ea bolo (M) = 10 kg

Radius ea bolo (r) = 0.1 m

Lebelo la angular (ω) = 10 rpapatso/lipapatso

Ho batloa: Matla a kinetic a potolohang

Tharollo:

Foromo ea motsotso oa inertia:

I = (2/5) Monghali.²

ke = motsotso oa inertia (kg m²), m = boima (kg), r = radius (M)

Motsotso oa ho se tsitse ha bolo:

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

Matla a kinetic a potolohang a bolo:

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

4. A 0.5-lik'hilograma tsa dikarolwana di potoloha ka lebelo le sa fetoheng la Rad/s e le 2. Matla a kinetic a potolohang a karoloana ke afe haeba radius ea selikalikoe e le Cm 10.

Tse tsejoang:

Boholo ea karoloana (M) = 0.5 kg

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:

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

Matla a kinetic a potolohang:

KE = 1/2 I ω² = 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 ho potoloha pulley ya silindara. torque is 2 N m le motsotso oa inertia is 1 lik'hilograma m², ke eng lebelo la angular ea silindara.

Tse tsejoang:

Torque (τ) = 2 N m

Motsotso oa ho se inertia (I) = 1 kg m²

Oa Batla: Ho potlaka ha silindara ka tsela e potolohang

Tharollo:

Στ = KE α

Στ = torque e hloekileng, I = motsotso oa ho se be le nako, α = ho potlaka ha angular

Ho potlakisa silindara ka mokhoa o potolohang:

α = Στ / I = 2 / 1 = 2 rad/s²

2. Matla F a sebediswa thapong e phuthetsweng ka pulley ya silindara. Boholo ba matla ke 10 N, radius ea silindara ke 0.2 m 'me motsotso oa inertia ke 1 lik'hilograma m2, WKe hobane'ng ha lebelo la silindara le potolohang?

Tse tsejoang:

Matla (F) = 10 N

Radius ea silindara (R) = 0.2 m

Motsotso oa ho se inertia (I) = 1 kg m²

Oa Batla: Ho potlaka ha silindara ka tsela e potolohang.

Tharollo:

τ = FR

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

Torque:

τ = FR = (10 N(0.2 m) = 2 N m

Στ = KE α

Στ = torque e hloekileng, I = motsotso oa ho se be le nako, α = ho potlaka ha angular

Ho potlakisa silindara ka mokhoa o potolohang:

α = Στ / I = 2 / 1 = 2 rad/s²

3. Matla F a sebediswa thapong e phuthetsweng ka pulley ya silindara. Boholo ba matla ke 10 N, radius ea silindara ke 0.2 m 'me boima ba silindara ke 20 kg m^2,. Wkatiba ke lebelo le potolohang la silindara.

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 mokhoa o potolohang

Tharollo:

τ = FR = (10 N(0.2 m) = 2 N m

Motsotso oa ho se inertia:

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

Ho potlakisa silindara ka mokhoa o potolohang:

α = Στ / I = 2 / 0.4 = 5 rad/s²

4. Sekotwana sa 1-kg se leketlileng thapong se phuthetsweng ka pulley ya silindara. Nako ya ho se sebetse hantle ha pulley ke 1 kg m² 'me radius ea pulley ke 0.2 m. Ho potlaka ha angular ea pulley ke eng. Ho potlakisa ka baka la matla a khoheli ke 10 m/s².

Tse tsejoang:

Nako ea ho se sebetse ha pulley (I) = 1 kg m²

Boholo ea boloko (m) = 1 kg

Ho potlaka ka lebaka la matla a khoheli (g) = 10 m/s²

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

Radius ea pulley (R) = 0.2 m

Ho batloa: Ho potlaka ha angular

Tharollo:

Torque:

τ = FR = w R = (10 N(0.2 m) = 2 N m

Motsotso oa ho se inertia:

I = 1 kg m²

Ho potlaka ha Angular:

α = Στ / I = 2 / 1 = 2 rad/s²

5. Sekotwana sa 1-kg se leketlileng thapong se phuthetsweng ka pulley ya silindara. Boima ba pulley ke 20 kg 'me radius ea pulley ke 0,2 m. Ho potlaka ha angular ha pulley le ho oela mahala Ho potlaka ha boloko. Ho potlaka ka lebaka la matla a khoheli ke 10 m/s².

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/s²

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

Ho batloa: ho potlaka ha angular ha pulley le ho potlaka ha ho wa ha boloko ka bolokolohi.

Tharollo:

Torque:

τ = FR = w R = (10 N(0.2 m) = 2 N m

Motsotso oa inertia ea pulley ea silindara:

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

Ho potlaka ha angular ha pulley:

α = Στ / I = 2 / 0.4 = 5 rad / s²

Ho potlaka ha ho oa ha boloko ka bolokolohi:

a = R α = (0.2)(5) = 1 m/s²

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

Boholo bolo (m) = ligrama tse 100 = 100/1000 = 0.1 kg

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

Oa Batla: Motsotso oa ho se tsitse ha bolo (I)

Tharollo:

Ke = Monghali.² = (0.1 kg)(0.3 m)²

I = (0.1 kg)(0.09 m²)

I = 0.009 kg m²

2. Bolo ea ligrama tse 100, m₁, le bolo ea ligrama tse 200, m₂, e 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 (m₁) = Ligrama tse 100 = 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 (m₂) = digrama tse 200 = 200/1000 = 0.2 kg

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

Ho batloa: motsotso oa ho se tsitse ha libolo

Karabo:

Ke = m₁ r₁² +m₂ r₂²

Ke = (0.1 kg)(0.3 limithara)² + (0.2 kg)(0.3 limithara)²

Ke = (0.1 kg)(0.09 limithara²) + (0.2 kg)(0.09 limithara²)

ke = 0.009 kg m² + 0.018 kg m²

ke = 0.027 kg m²

3. Bolo ea ligrama tse 200, m₁ le bolo ea ligrama tse 100, m_2, e hokahantsoe ka thupa e bolelele ba 60 cm. Hlokomoloha boima ba thupa. Mothapo oa potoloho o sebakeng sa bolo m₂. Motsotso oa ho se tsitse ha libolo ke ofe. Hlokomoloha boima ba molamu.

Tse tsejoang:

Boima ba bolo 1 (m₁= 2Ligrama tse 00 = 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 (m₂) = digrama tse 100 = 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 se tsitse ha libolo

Tharollo:

Ke = m₁ r₁² +m₂ r₂²

Ke = (0.2 kg)(0,6 limithara)² + (0.2 kg)(0)²

Ke = (0.2 kg)(0.36 limithara²) + 0

I = 0.072 kg m²

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 = m₁ = m₂ = m₃ = m₄ = 1Ligrama tse 00 = 100/1000 = 0.1 kg

Sebaka se pakeng tsa bolo le axis ea 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: Motsotsoana oa inertia

Tharollo:

Ke = m₁ r₁² +m₂ r₂² +m₃ r₃² +m₄ r₄²

Ke = (0.1 kg)(0.3 limithara)² + (0.1 kg)(0.3 limithara)² + (0.1 kg)(0.3 m)² + (0.1 kg)(0.3 m)²

Ke = (0.1 kg)(0.09 limithara²) + (0.1 kg)(0.09 m²) + (0.1 kg)(0.09 m²) + (0.1 kg)(0.09 m²)

I = 0.036 kg m²

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

Oa Batla: Motsotsoana oa inertia

Tharollo:

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

Ke = (1/12) ML²

I = (1/12) (2 kg)(2 limithara)²

I = (1/12) (2 kg)(4 m²)

I = (1/12)(8 kg m²)

I = 8/12 kg m²

I = 2/3 kg m²

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

Oa Batla: Motsotsoana oa inertia

Tharollo:

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

Ke = (1/3) ML²

I = (1/3) (2 kg)(2 limithara)²

I = (1/3) (2 kg)(4 m²)

I = (1/3)(8 kg m²)

I = 8/3 kg m²

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

Oa Batla: Motsotso oa inertia

Oa Batla: Motsotso oa inertia

Tharollo:

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

Ke = (1/2) MR²

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 m²

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

Oa Batla: motsotso oa ho se inertia

Tharollo:

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

Ke = (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. 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: Motsotsoana oa inertia

Tharollo:

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

Ke = (1/12) M (a² +b²)

Ke = (1/12)(2)(0.5² + 0.2²)

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

Ke = (1/6)(0.29)

I = 0.29/6 kg m²

Bala haholoanyane