Matsalolin da aka Magance a fannin Fizik na Asali

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

Tizinin sarari kyauta = ε_o

Pyankin da ya makara = A ba

The distance tsakanin faranti = d

SE busca: Ƙarfin capacitor (C)

Magani:

2. Capacitor wanda ke da mafi girman ƙarfin bisa ga zane-zanen da ke ƙasa is...

Magani:

Tsarin capacitor mai layi ɗaya:

Ƙarfin capacitors:

Capacitor wanda shine mafi girman ƙarfiny shine capacitor C_3.

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:

Aboki 1:

Yankin saman = 2 A

Nisa tsakanin faranti =d₁

Aboki 2:

Yankin saman = A ba

Nisa tsakanin faranti = 2 kwana₁

SE busca: Kwatanta ƙarfin capacitors 1 da 2

Magani:

5.

Ƙayyade kwatancen ƙarfin Masu haɗa na'urori masu layi ɗaya I da II.

Magani:

Kwatanta ƙarfin Masu haɗa faranti masu layi ɗaya I da II:

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

Idan A₁ = ½ A₂ da d₂ = 3 kwana₁ sa'an nan ƙayyade rabon ƙarfin da ke tsakanin layi ɗaya-na'urar capacitor na faranti tsakanin hoton 2 da hoton 1.

An sani:

Mai haɗa farantin layi ɗaya I:

A₁ = 1

d₁ = 1

Layin capacitor II:

A₂ = 2

d₂ = 3

SE busca: rabon ƙarfin capacitor mai layi ɗaya tsakanin hoton 2 da hoton 1

Magani:

Mai haɗa farantin layi ɗaya I:

Layin capacitor II:

Trabon ƙarfin da ke tsakanin layi ɗaya-na'urar capacitor na farantin II da I:

Karin bayani

1. Ayyade da ƙarfin lantarki a wani wuri da ke da nisan 1 cm daga caji 5.0 μC. Madaidaitan Coulomb (k) = 9 x 10⁹ Nm²C^-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 μC = 5.0 x 10^-6 C

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

Ana so: Ƙarfin wutar lantarki (V)

Magani:

Ƙarfin wutar lantarki:

Ƙarfin wutar lantarki shine 4.5 x 10⁶ karfin wuta

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 Q₁ = -5.0 μC = -5.0 x 10^-6 C

The distance na aya A daga Q₁ = 10 cm = 0.1 m = 10^-1 m

cajin Q₂ = 6.0 μC = 6.0 x 10^-6 C

Nisa daga wurin a Q₂ = 10 cm = 0.1 m = 10^-1 m

Coulomb na yau da kullun (k) = 9 x 10⁹ Nm²C^-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 = V₂ - V₁

V = (54 – 45) x 10⁴

V = 9 x 10⁴

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 Q₁ = -5.0 μC = -5.0 x 10^-6 C

Nisa daga aya ta A Q₁ = 40 cm = 0.4 m = 4 x 10^-1 m

cajin Q₂ = 6.0 μC = 6.0 x 10^-6 C

Nisa daga aya ta A Q₂ = 50 cm = 0.5 m = 5 x 10^-1 m

Coulomb na yau da kullun (k) = 9 x 10⁹ Nm²C^-2

Ana so: Ƙarfin wutar lantarki a lokacin A

Magani:

Ƙarfin wutar lantarki 1 :

Ƙarfin wutar lantarki 2 :

Ƙarfin wutar lantarki a wuri na A :

V = V₁ + V₂

V = (-11.25 + 10.8) x 10⁴

V = -0.45 x 10⁴

V = -4.5 x 10³

Ƙarfin wutar lantarki a wurin A shine - 4.5 x 10³ karfin wuta

Karin bayani

1. Ana hanzarta na'urar lantarki daga hutawa ta hanyar bambancin V na 12. Menene canjin da ke cikin makamashin wutar lantarki na electron?

An sani:

The cajin akan na'urar lantarki (e) = -1.60 x 10^-19 Coulomb

Ƙarfin wutar lantarki = irin ƙarfin lantarki (V) = Volt 12

SE busca: Canjin makamashin lantarki na lantarki (ΔPE)

Magani:

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

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

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

An sani:

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

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

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

Ana so: Canjin makamashin 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)(V 10)

ΔPE = 16 x 10^-19 Joule

ΔPE = 1.6 x 10^-18 Joule

3. An raba caji maki biyu da nisan santimita 10. Caji a kan maki A =+9 μC kuma caji akan batu B = -4 μC. k = 9 x 10⁹ Nm²C^-2, 1 μC = 10^-6 C. Menene canjin makamashin 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 (q₁) = -4 μC = -4 x 10^-6 C

ku = 9 x 10⁹ Nm²C^-2

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

Ana so: Canjin makamashin lantarki (ΔEP)

Magani:

Karin bayani

1. Kayan aiki filin lantarki E = 8000 N/C yana ratsawa ta cikin murabba'in fili mai faɗi A = 10 m². Ayyade da kwararar wutar lantarki.

An sani:

Girman filin lantarki (E) = 8000 N/C

Yanki (A) = 10 m²

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

SE busca: Fitowar wutar lantarki (Φ)

Magani:

Tsarin kwararar lantarki:

Φ = EA cos q

Φ = kwararar wutar lantarki (Nm²/C), E = filin lantarki (N/C), A = yanki (m²), q = kusurwa tsakanin layin filin lantarki tare da layin al'ada.

Gudun wutar lantarki:

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

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

An sani:

Filin wutar lantarki (E) = 5000 N/C

Yanki (A) = 2 m²

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

Ana so: Fitowar wutar lantarki (Φ)

Magani:

Gudun wutar lantarki:

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

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 lantarki (Q) = 10 μC = 10 x 10^-6 C

Ana so: Fitowar wutar lantarki (Φ)

Magani:

Filin lantarki:

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

Faɗin saman:

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

Gudun wutar lantarki:

Layukan filin lantarki suna daidaita zuwa yankin, don haka kusurwar da ke tsakanin alkiblar filin lantarki da layin da aka zana a tsaye zuwa yankin, ita ce 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 ku 10⁵

Φ = 1.13 x ku 10⁶ Nm²/C

Karin bayani

1. Lissafa girma da alkiblar filin lantarki a wani wuri A da ke a 5 cm daga cajin maki Q = + 10 μC. ku = 9 x 10⁹ Nm²C^-2, 1 μC = 10^-6 C)

An sani:

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

The distance tsakanin aya ta A da aya ta caji Q (r)_A) = 5 cm = 0.05 m = 5 x 10^-2 m

ku = 9 x 10⁹ Nm²C^-2

SE busca: Girma da alkiblar filin lantarki a lokaci 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.

2. Lissafa girma da alkiblar filin lantarki a wurin P da ke da nisan santimita 10 daga cajin maki Q = -20 μC. ku = 9 x 10⁹ Nm²C^-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 wutar lantarki (r)_P) = 10 cm = 0.1 m = 1 x 10^-1 m

ku = 9 x 10⁹ Nm²C^-2

SE busca: Girma da alkiblar filin lantarki a wurin 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.

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 (q)_A) = -2 μC = -2 x 10^-6 C

Caji B (q)_B) = +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 wurin P.

Magani:

Cajin A yana da korau don haka alkiblar wuraren filin lantarki tana zuwa ga Q_A (Zuwa hannun hagu).

Caji B yana da kyau don haka alkiblar wuraren filin lantarki ba ta da nisa da Q_B (Zuwa hannun hagu).

Jimlar filin lantarki a aya ta A:

E = E_A +E_B

E = (4.5 x 10⁵(9 x 10.)⁵)

E = 13.5 x 10⁵ N / C

Alkiblar filin lantarki tana nuni zuwa ga Q_A (Zuwa hannun hagu).

4. Girman filin lantarki sifili ne a…

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

An sani:

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

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

ku = 9 x 10⁹ Nm²C^-2

Nisa tsakanin caji A da kuma kuɗin 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 ba shi da nisa 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.

5Dangane da hoton da ke ƙasa, wGa batun P don haka filin lantarki a wurin P sifili ne? (k = 9 x 10⁹ Nm²C^-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

ku = 9 x 10⁹ Nm²C^-2

Nisa tsakanin caji 1 da caji 2 = 3 cm

Nisa tsakanin Q₁ da kuma maki P (r)_1P) = a

Nisa tsakanin Q₂ da kuma maki P (r)_2P) = 3 + a

Ana so: wurin da aka nuna P don haka filin lantarki a wurin P sifili ne

Magani:

PMaki na P yana gefen hagu na Q₁.

Filin wutar lantarki da Q ya samar₁ a wurin P:

Cajin gwajin yana da kyau kuma Q yana da kyau₁ yana da kyau don haka alkiblar filin lantarki tana hagu.

Filin wutar lantarki da Q ya samar₂ a wurin P:

Cajin gwajin yana da kyau kuma Q yana da kyau₂ yana da mummunan yanayi 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 Q₂ da kuma maki P (r)_2P) = 3 + a = 3 – 1.8 = 1.2 cm.

Maki na P yana kan 1.2 cm dama na Q₁.

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 (q)_A) = +8 μC = +8 x 10^-6 C

Caji B (q)_B) = -5 μC = -5 x 10^-6 C

ku = 9 x 10⁹ Nm²C^-2

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

Nema : Girman ƙarfin wutar lantarki

Magani:

Tsarin Dokar Coulomb :

Girman ƙarfin wutar lantarki:

2. Ƙwayoyin cuta guda biyu da aka caje kamar yadda aka nuna a 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

ku = 9 x 10⁹ Nm²C^-2

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

Ana so: Tgirman ƙarfin lantarki

Magani:

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 (q)_A) = -5 μC = -5 x 10^-6 C

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

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

ku = 9 x 10⁹ Nm²C^-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 F_BA Ana amfani da barbashi B ta hanyar barbashi A da ƙarfin F_BC an yi amfani da barbashi na B ta hanyar barbashi na C.

Ƙarfin F_BA an yi amfani da barbashi B ta hanyar barbashi A:

 

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

Ƙarfin F_BC an yi amfani da 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).

4. +Q₁ = 10 μC, +Q₂ = 50 μC da Q₃ an raba su kamar yadda aka nuna a cikin hoton da ke ƙasa. Menene cajin lantarki akan barbashi na 3 idan ƙarfin lantarki na lantarki akan barbashi na 2 sifili ne.

An sani:

Caji 1 (q₁) = +10 μC = +10 x 10^-6 C

Caji 2 (q₂) = +50 μC = +50 x 10^-6 C

Nisa tsakanin caji 1 da 2 (r₁₂) = 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 (F)₂) = 0

Nema : caji 3 (q₃)

Magani:

Ƙarfin da ke kan barbashi na 2 shine vector Jimlar ƙarfin F₂₁ an yi amfani da shi akan barbashi 2 ta hanyar barbashi 1 da ƙarfin F₂₃ an yi amfani da shi a kan barbashi 2 ta hanyar barbashi 3.

Ƙarfin F₂₁ an yi amfani da 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 tare da lokacin inertia of 2 kg m² yana juyawa a 1 rad/s. Mene ne ƙarfin kusurwa na abin?

An sani:

Lokacin inertia (I) = 2 kg m²

Gudun kusurwa (ω) = 1 rad/s

Ana so: Karfin kusurwa (L)

Magani:

Tsarin motsin kusurwa:

L = Ni ω

L = ba ƙarfin kusurwa (kg m²/s), I = lokacin inertia (kg m²), ω = saurin kusurwa (ra/s)

Motsin kusurwa:

L = Ni ω = (2)(1) = 2 kg m²/s

2. Na 2-kg kura ta silinda mai radius na 0.1 m yana juyawa a daidai gudun kusurwa na 2 rad/s. Menene saurin kusurwa na pulley?

An sani:

Mass na kura (M) = 2kg

Radius na kura (r) = 0.1 m

Gudun kusurwa (ω) = 2 rtalla/s

Ana so: Karfin kusurwa

Magani:

Tsarin lokacin inertia don silinda mai ƙarfi:

I = 1/2 m²

Ni = lokacin inertia (kg m²), m = taro (kg), r = radius (M)

Lokacin inertia:

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

Gudun kusurwa:

L = Ni ω = (0.01)(2) = 0.02 kg m²/s

3. Layin da ya dace da juna na kilogiram 2 tare da radius na 0.2 m yana juyawa a 4 rad/s. Menene ma'aunin kusurwa na ƙwallon.

An sani:

Mass na ƙwallo (M) = 2 kg

Radius na ball (r) = 0.2 m

Gudun kusurwa (ω) = 4 rad/s

Ana so: Karfin kusurwa

Magani:

Tsarin lokacin inertia don siffa mai kama da juna:

I = (2/5) Mr.²

I = lokacint na rashin kuzari (kg m²), 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 m²

Motsin kusurwa na ƙwallo:

L = Ni ω = (0.032)(4) = 0.128 kg m²/s

4. A 1 kg ƙwayar cuta tana juyawa a daidai gudun kusurwa na 2 rad/s. Menene saurin kusurwa idan radius na da'irar shine 10 cm.

An sani:

Mass na abu (M) = 1kg

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

Gudun kusurwa (ω) = 2 rad/s

Ana so: Karfin kusurwa

Magani:

Tsarin lokacin inertia ga barbashi:

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

Motsin kusurwa:

L = I ω = (0.01)(2) = 0.02 kg m²/s

Karin bayani

1. Wani abu yana da lokacin inertia na 1 kg m² yana juyawa a daidai gudun kusurwa na 2 rad/s. Mene ne makamashin motsi na juyawa na abin?

An sani:

Lokacin inertia (I) = 1 kg m²

The tururuwa mai daidaituwa (ω) = 2 rad/s

SE busca: Makamashin motsi na juyawa (KE)

Magani:

Tsarin makamashin motsi na juyawa:

KE = 1/2 I ω²

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

2. A 20-kg kura ta silinda tare da radius na 0.2 m yana juyawa a daidai gudun kusurwa na 4 rad/s. Menene kuzarin motsi na juyawa na pulley?

An sani:

Mass na silinda pulley (M) = 20kg

Radius na silinda (r) = 0.2 m

Gudun kusurwa (ω) = 4 rad/s

Ana so: Menene makamashin motsi na juyawa

Mafita;

Tsarin lokacin inertia na silinda:

I = 1/2 m²

Ni = lokacin inertia (kg m²), m = nauyi (kg), r = radius (mita)

Lokacin inertia na silinda pulley:

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

Ƙarfin motsi na juyawa na kura:

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

3. A-10 kg ƙwallon da ke da radius na 0.1 m yana juyawa akai-akai na 10 rad/s. Menene kuzarin motsi na ƙwallon.

An sani:

Mass na ƙwallo (M) = 10 kg

Radius na ball (r) = 0.1 m

Aikacewar gudu (ω) = 10 rtalla/s

Ana so: Makamashin motsi na juyawa

Magani:

Tsarin lokacin inertia:

I = (2/5) Mr.²

Ni = lokacin inertia (kg m²), m = nauyi (kg), r = radius (M)

Lokacin rashin kuzari na ƙwallon:

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

Ƙarfin motsi na juyawa na ƙwallon:

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

4. A 0.5-ƙwayoyin kilogiram suna juyawa a daidai gudun kusurwa na 2 rad/s. Menene kuzarin motsi na juyawa na barbashi idan radius na da'irar ya kasance 10 cm.

An sani:

Mass na barbashi (M) = 0.5kg

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

Gudun kusurwa (ω) = 2 rad/s

Ana so: Makamashin motsi na juyawa

Magani:

Lokacin inertia ga ƙwayoyin cuta:

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

Makamashin motsi na juyawa:

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

Karin bayani

1. Ana amfani da ƙarfin F a kan igiyar da aka naɗe a kan injin silinda. karfin juyi is 2 N m da kuma lokacin inertia is 1 kg m², menene hanzarin kusurwa na silinda.

An sani:

Karfin juyi (τ) = 2 N m

Lokacin inertia (I) = 1 kg m²

SE busca: Saurin kusurwa na silinda

Magani:

Στ = Ni α

Στ = karfin juyi mai yawa, I = lokacin rashin kuzari, α = hanzarin kusurwa

Haɓaka silinda a kusurwa:

α = Στ / I = 2/1 = 2 rad/s²

2. Ana amfani da ƙarfin F a kan igiyar da aka naɗe a kan injin silinda. Girman ƙarfin shine 10 N, radius na silinda shine 0.2 m kuma lokacin inertia shine 1 kg m2, Wshin wannan shine saurin kusurwa na silinda?

An sani:

Ƙarfi (F) = 10 N

Radius na silinda (R) = 0.2 m

Lokacin inertia (I) = 1 kg m²

SE busca: Saurin kusurwa na silinda.

Magani:

τ = FR

τ = karfin juyi, F = karfi, R = radius na silinda

Karfin juyi:

τ = FR = (1)0 N)(0.2 m) = 2 N m

Στ = Ni α

Στ = karfin juyi mai yawa, I = lokacin rashin kuzari, α = hanzarin kusurwa

Haɓaka silinda a kusurwa:

α = Στ / I = 2 / 1 = 2 rad/s²

3. Ana amfani da ƙarfin F a kan igiyar da aka naɗe a kan injin silinda. Girman ƙarfin shine 10 N, radius na silinda shine 0.2 m kuma nauyin silinda shine 20 kg m^2,. Whula ita ce hanzarin kusurwa na silinda.

An sani:

Ƙarfi (F) = 10 N

Radius na silinda (R) = 0.2 m

Nauyin silinda (M) = 20 kg

Ana so: Haɓaka kusurwa na silinda

Magani:

τ = FR = (1)0 N)(0.2 m) = 2 N m

Lokacin inertia:

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

Haɓaka silinda a kusurwa:

α = Στ / I = 2 / 0.4 = 5 rad/s²

4. Toshe mai nauyin kilogiram 1 da aka rataye daga igiya da aka naɗe a kan silinda pulley. Lokacin inertia na pulley shine kilogiram 1 m² kuma radius na pulley shine 0.2 m. Menene saurin kusurwa na pulley. Hanzarta saboda nauyi shine 10 m/s².

An sani:

Lokacin inertia na kura (I) = 1 kg m²

Mass na toshe (m) = 1 kg

Saurin gudu saboda nauyi (g) = 10 m/s²

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

Radius na kura (R) = 0.2 m

Ana so: Hanzarta kusurwa

Magani:

Karfin juyi:

τ = FR = w R = (1)0 N)(0.2 m) = 2 N m

Lokacin inertia:

I = 1 kg m²

Haɓaka kusurwa:

α = Στ / I = 2 / 1 = 2 rad/s²

5. Bulo mai nauyin kilogiram 1 da aka rataye daga igiya da aka naɗe a kan pulley ɗin silinda. Nauyin pulley ɗin shine kilogiram 20. kuma radius na pulley ɗin shine 0,2 m. Menene hanzarin kusurwa na pulley ɗin da kuma faɗuwa kyauta hanzarta toshewar. Saurin da ake samu sakamakon nauyi shine 10 m/s².

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/s²

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

Ana so: hanzarta kusurwar kurfin da kuma hanzarta 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 MR² = 1⁄2 (20)(0.2)² = (10)(0.04) = 0.4 kg m²

Saurin kusurwa na kusurwoyi:

α = Στ / I = 2 / 0.4 = 5 rad / s²

Haɓaka faɗuwar free fall na tonon:

a = R α = (0.2)(5) = 1 m/s²

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

Mass ƙwallon (m) = gram 100 = 100/1000 = 0.1 kg

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

SE busca: Lokacin rashin kuzari na ƙwallo (I)

Magani:

Ni = Mr.² = (0.1 kg)(0.3 m)²

I = (0.1 kg)(0.09 m²)

I = 0.009 kg m²

2. Kwallo mai nauyin gram 100, m₁, da kuma ƙwallon gram 200, m₂, 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 ball 1 da kuma axis na juyawa (r)₁) = 30 cm = 30/100 = 0.3 m

Nauyin ƙwallon (m)₂) = gram 200 = 200/1000 = 0.2 kg

The distance na ball 2 da kuma axis na juyawa (r₂) = 30 cm = 30/100 = 0.3 m

Ana so: lokacin rashin ƙarfin ƙwallo

Amsa:

Ni = m₁ r₁² +m₂ r₂²

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

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

Ni = 0.009 kg m² + 0.018 kg m²

Ni = 0.027 kg m²

3. Kwallo mai nauyin gram 200, m₁ da kuma ƙwallon gram 100, m_2, An haɗa shi da sanda mai tsawon santimita 60. Yi watsi da nauyin sandar. Axis na juyawa yana wurin ƙwallon m₂Menene lokacin rashin ƙarfin ƙwallo. Yi watsi da nauyin sandar.

An sani:

Nauyin ƙwallon 1 (m)₁) = 2gram 00 = 200/1000 = 0.2 kg

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

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

Nisa tsakanin ball 2 da kuma axis na juyawa (r)₂= 0m

Ana so: Lokacin rashin kuzari na ƙwallo

Magani:

Ni = m₁ r₁² +m₂ r₂²

I = (0.2 kg)(0,6 m)² + (0.2 kg)(0)²

I = (0.2 kg)(0.36 m²) + 0

I = 0.072 kg m²

4. 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 = m₁ = m₂ = m₃ = m₄ = 1gram 00 = 100/1000 = 0.1 kg

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

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

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

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

An sani: Lokacin inertia

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 m²) + (0.1 kg)(0.09 m²) + (0.1 kg)(0.09 m²) + (0.1 kg)(0.09 m²)

I = 0.036 kg m²

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

SE busca: Lokacin inertia

Magani:

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

I = (1/12) ML²

I = (1/12) (2 kg)(2 m)²

I = (1/12) (2 kg)(4 m)²)

I = (1/12)(8 kg m²)

I = 8/12 kg m²

I = 2/3 kg m²

6. 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

SE busca: Lokacin inertia

Magani:

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

I = (1/3) ML²

I = (1/3) (2 kg)(2 m)²

I = (1/3) (2 kg)(4 m)²)

I = (1/3)(8 kg m²)

I = 8/3 kg m²

7. 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

SE busca: Lokacin inertia

SE busca: Lokacin inertia

Magani:

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

I = (1/2) MR²

I = (1/2) (10 kg)(0.1 m)²

I = (1/2) (10 kg)(0.01 m)²)

I = (1/2)(0.1 kg m²)

I = 0.05 kg m²

8. 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

SE busca: wani lokaci na rashin kuzari

Magani:

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

I = (2/5) MR²

I = (2/5)(20 kg)(0.1 m)²

I = (2/5)(20 kg)(0.01 m²)

I = (2/5)(0.2 kg m²)

I = 0.4/5 kg m²

I = 0.08 kg m²

9. 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 inertia

Magani:

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

I = (1/12) M (a² + b²)

I = (1/12)(2)(0.5)² + 0.2²)

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

I = (1/6)(0.29)

I = 0.29/6 kg m²

Karin bayani