Physics

1. Dhererka muraayadda qaloocan waa 10 cm, masaafada shayguna waa 20 cm. Go'aami (a) masaafada sawirka (b) weyneynta sawirka

La yaqaan:

Dhererka diiradda (f) = -10 cm

Calaamadda laga jaray waxay muujinaysaa in barta diiradda ee muraayadda qaloocan ay tahay mid dalwad ah

Masaafada shayga ( d _o ) = 20 cm

Xalka:

Samaynta sawirka iyadoo la adeegsanayo muraayad qaloocan:

Masaafada sawirka (d _i ):

1/ d _i = 1/f – 1/ d _o = -1/10 – 1/20 = -2/20 – 1/20 = -3/20

d _i = -20/3 = -6.7 cm

Calaamadda laga jaray waxay muujinaysaa in sawirku yahay mid casri ah.

Weyneynta sawirka:

m = – d _i / d _o = -(-6.7)/20 = 6.7/20 = 0.3

m = 0,3 jeer ka yar shayga.

Calaamadda dheeraadka ah waxay muujinaysaa in sawirku kor u kacayo.

2. Shay dhererkiisu yahay 10-cm ayaa la dhigayaa hortiisa muraayad qaloocan oo dhererkeedu yahay 20 cm. Go'aami dhererka sawirka haddii masaafada shaygu tahay (a) 10 cm (b) 30 cm (c) 40 cm (d) 50 cm

La yaqaan:

Dhererka diiradda ee muraayadda qaloocan (f) = -20 cm

Calaamadda laga jaray waxay muujinaysaa in barta diiradda la saarayo ay tahay mid casri ah

Gacanka qalooca ( r ) = 2 f = 2(20) = 40 cm

Dhererka shayga (h) = 10 cm

Xalka:

a) dhererka diiradda (f) = -20 cm iyo masaafada shayga (d _o ) = 10 cm

Masaafada sawirka (d _i ) :

1/ d _i = 1/f – 1/ d _o = -1/20 – 1/10 = -1/20 – 2/20 = -3/20

d _i = -20/3 = -6.7

Calaamadda laga jaray waxay muujinaysaa in sawirku yahay mid casri ah ama sawirku uu ka dambeeyo muraayadda.

Weyneynta sawirka ( m ):

m = – d _i / d _o = -(-6.7)/10 = 6.7/10 = 0.67

Calaamadda dheeraadka ah waxay muujinaysaa in sawirku uu toosan yahay.

Sawirku waa 0.67 ka yar yahay shayga.

Dhererka sawirka (h _i ):

m = h _i / h _o

h _i = h _o m = (10 cm)(0.67) = 6.7 cm

b) dhererka diiradda (f) = -20 cm iyo masaafada shayga ( d _o ) = 30 cm

Masaafada sawirka ( d _i ):

1/ d _i = 1/f – 1/ d _o = -1/20 – 1/30 = -3/60 – 2/60 = -5/60

d _i = -60/5 = -12

Calaamadda laga jaray waxay muujinaysaa in sawirku yahay mid casri ah ama sawirku uu ka dambeeyo muraayadda.

Weyneynta sawirka ( m ):

m = – d _i / d _o = -(-12)/30 = 12/30 = 0.4

Calaamadda dheeraadka ah waxay muujinaysaa in sawirku uu toosan yahay.

Sawirka ayaa 0,4 jeer ka yar shayga.

Dhererka sawirka (h _i ):

m = h _i / h _o

h _i = h _o m = (10 cm)(0.4) = 4 cm

c) Dhererka diiradda (f) = -20 cm iyo masaafada shayga ( d _o ) = 40 cm

Masaafada sawirka ( d _i ):

1/ d _i = 1/f – 1/ d _o = -1/20 – 1/40 = -2/40 – 1/40 = -3/40

d _i = -40/3 = -13.3

Calaamadda laga jaray waxay muujinaysaa in sawirku yahay mid casri ah ama sawirku uu ka dambeeyo muraayadda qaloocan.

Weyneynta sawirka ( m ):

m = – d _i / d _o = -(-13.3)/40 = 13.3/40 = 0.3

Calaamadda dheeraadka ah waxay muujinaysaa in sawirku uu toosan yahay.

Sawirku waa 0.3 ka yar yahay shayga.

Dhererka sawirka ( h _{i )} :

m = h _i / h _o

h _i = h _o m = (10 cm)(0.3) = 3 cm

d) Dhererka diiradda (f) = -20 cm iyo masaafada shayga ( d _o ) = 50 cm

Masaafada sawirka ( d _i ):

1/ d _i = 1/f – 1/ d _o = -1/20 – 1/50 = -5/100 – 2/100 = -7/100

d _i = -100/7 = -14.3

Calaamadda laga jaray waxay muujinaysaa in sawirku yahay mid casri ah ama sawirku uu ka dambeeyo muraayadda qaloocan.

Weyneynta sawirka ( m ):

m = – d _i / d _o = -(-14.3)/50 = 14.3/50 = 0.3

Calaamadda dheeraadka ah waxay muujinaysaa in sawirku uu toosan yahay.

Sawirku waa 0.3 ka yar yahay shayga.

Dhererka sawirka (h _i ) :

m = h _i / h _o

h _i = h _o m = (10 cm)(0.3) = 3 cm

3. Shayga saaran waa 20 cm hortiisa muraayadda qaloocan . Haddii dhererka sawirku yahay 1/5 jibaar dhererka shayga, go'aami ( a) dhererka sawirka b) dhererka diiradda c) sifooyinka sawirka

La yaqaan:

Masaafada shayga ( d _o ) = 20 cm

Dhererka sawirka (h _i ) = 1/5 saac = 0.2 saac

Dhererka shayga (h) = h

Xalka:

a) masaafada sawirka ( d _i )

Qaacidda weyneynta sawirka :

m = h _i / h _o = 0.2h / h = 0.2

Calaamadda dheeraadka ah waxay muujinaysaa in sawirku uu toosan yahay.

Sawirku waa 0.2 ka yar yahay shayga.

Masaafada sawirka ( d _i ):

– d _i = md _o

d _i = – m d _o = -(0.2)(20 cm) = -4 cm

Calaamadda laga jaray waxay muujinaysaa in sawirku yahay mid casri ah ama sawirku uu ka dambeeyo muraayadda qaloocan.

b) Dhererka diiradda ( f)

Dhererka diiradda (f):

1/f = 1/ d _o + 1/ d _i = 1/20 – 1/4 = 1/20 – 5/20 = -4/20

f = -20/4 = -5 cm

Calaamadda laga jaray waxay muujinaysaa in barta diiradda la saarayo ay tahay mid casri ah.

c) Astaamaha sawirka:

- Toosan

- Yar

- dalwad

4. Iftiinka oo ku dhaca muraayad qaloocan oo barbar socota dhidibka ayaa la arki doonaa….

A. dhanka barta diiradda ee muraayadda

B. sida laga soo bilaabo barta dhexe ee muraayadda

C. iyada oo loo marayo bartamaha qalooca muraayadda

D. oo ku toosan muraayadda diyaaradda

Solution

Dhibaatadu waxay ku jirtaa sawirka hoose.

Jawaabta saxda ah waa B.

5. Baaskiil wade wuxuu arkaa sawirka mooto gadaasheeda 1/6 jibaar cabbirkeedii asalka ahaa marka masaafada u dhaxaysa mootada iyo mootada ay tahay 30 mitir. Go'aami gacanka qalooca muraayadda dambe…

A. 7.14 mitir

B. 8.57 mitir

C. 12.00 mitir

D. 24.00 mitir

La yaqaan:

Weyneynta sawirka (M) = 1/6 jeer

Masaafada walxaha (d) = 30 mitir

La Rabay: Gacanka qalooca ee muraayadda aragtida dambe (R)

Xalka:

Xisaabi masaafada sawirka (d')

Maadaama la yaqaanay weyneynta sawirka (M) iyo masaafada (yada) shayga, masaafada sawirka waxaa lagu ogaan karaa iyadoo la isticmaalayo qaacidada weyneynta sawirka:

Calaamad taban waxay ka dhigan tahay in sawirku yahay mid casri ah. Sawirku wuxuu 5 mitir ka dambeeyaa muraayadda qaloocan.

Xisaabi dhererka diiradda (f)

Maadaama masaafada shayga (d) iyo masaafada sawirka (d'), markaa dhererka diiradda waxaa lagu xisaabin karaa iyadoo la adeegsanayo qaacidada muraayadda:

Gacanka qalooca (R)

Gacanka qalooca muraayadda qalooca leh waa labanlaab dhererka diiradda muraayadda qalooca leh.
R = 2 f = 2 (6 mitir) = 12 mitir
Gacanka qalooca muraayadda qalooca leh waa 12 mitir.
Jawaabta saxda ah waa C.

6. Muraayadda qaloocan waxaa loo doortay muraayadda gadaal ee mootada sababtoo ah sifooyinka sawirka uu muraayaddu soo saarto waa…

A. dhab ah, toosan, la yareeyay

B. dhab ah, toosan, la ballaariyay

C. dadban, toosan, la yareeyay

D. dadban, toosan, la ballaariyay

Xalka:

Iyada oo lagu saleynayo labada sawir ee kor ku xusan, waxaa lagu soo gabagabeyn karaa in sifooyinka sawirku ay yihiin kuwo casri ah, toosan, la yareeyay.

Jawaabta saxda ah waa C.

7. Shay wuxuu 12 cm ka hor yahay muraayad qaloocan oo leh gacan 6 cm ah. Astaamaha sawirka waa…

A. dhab ah, rogan masaafo dhan 12 cm

B. dhab ah, toosan masaafad dhan 4 cm

C. qof ahaan, toosan masaafad dhan 2.4 cm

D. dalwad, rogrogmi karta masaafad dhan 6 cm

La yaqaan:

Masaafada shayga (d) = 12 cm

Dhexroorka muraayadda qaloocan (r) = 6 cm.

Dhererka muraayadda qaloocan ( f) = 6 cm / 2 = -3 cm

Dhererka muraayadda qaloocan waxaa loo calaamadeeyay taban sababtoo ah waa mid casri ah. Macmal ah sababtoo ah iftiin ma dhaafo.

La Raadinayo: Sifooyinka sawirka

Xalka:

Masaafada sawirka (d'):

1/d' = 1/f – 1/d = -1/3 – 1/12 = -4/12 – 1/12 = -5/12

d' = -12/5 = -2.4 cm

Masaafada sawirka ee la saxiixay ee taban waxay ka dhigan tahay in sawirku yahay mid dalwad ah.

Weyneynta sawirka (m):

m = -d' / d = -(-2.4) / 12 = 2.4 / 12 = 0.2 jeer

Weyneynta sawirka ee calaamadaysan ee togan waxay la macno tahay in sawirku toosan yahay weyneynta sawirkuna ay tahay 0.2 macnaheedu waa cabbirka sawirku wuu ka yar yahay cabbirka shayga (la dhimay).

Jawaabta saxda ah waa C.

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Akhri wax dheeraad ah

1. Shay ayaa la dhigayaa 10 cm muraayad qaloocan . Dhererka diiradda waa 5 cm. Go'aami (a) Masaafada sawirka (b) weyneynta sawirka

La yaqaan:

Dhererka diiradda (f) = 5 cm

Masaafada shayga ( d _o ) = 10 cm

Xalka :

Samaynta sawirka iyadoo la adeegsanayo muraayad qaloocan:

Masaafada sawirka:

1/ d _i = 1/f – 1/ d _o = 1/5 – 1/10 = 2/10 – 1/10 = 1/10

d _i = 10/1 = 10 cm

Masaafada sawirka waa 10 cm.

Weyneynta:

m = – d _i / d _o = -10/10 = -1

1 macnaheedu waa in sawirku uu la mid yahay shayga.

Calaamadda laga jaray waxay muujinaysaa in sawirku rogmanayo. Haddii calaamaddu togan tahay, sawirku waa toosan yahay.

2. Shay dhererkiisu yahay 5-cm ayaa la dhigayaa hortiisa muraayad qaloocan oo leh gacan qalooc ah oo ah 20 cm. Go'aami dhererka sawirka haddii masaafada shaygu tahay 5 cm, 15 cm, 20 cm, iyo 30 cm.

La yaqaan:

Gacanka qalooca ( r ) = 20 cm

Dhererka diiradda (f) = R/2 = 20/2 = 10 cm

Dhererka shayga (h _o ) = 5 cm

Xalka:

a) dhererka diiradda (f) = 10 cm iyo masaafada shayga ( d _o ) = 5 cm

Samaynta sawirka iyadoo la adeegsanayo muraayad qaloocan:

Masaafada sawirka ( d _i ):

1/ d _i = 1/f – 1/ d _o = 1/10 – 1/5 = 1/10 – 2/10 = -1/10

d _i = -10/1 = -10 cm

Calaamadda laga jaray waxay muujinaysaa in sawirku yahay mid casri ah ama sawirku uu ka dambeeyo muraayadda.

Weyneynta sawirka ( m ):

m = – d _i / d _o = -(-10)/5 = 10/5 = 2

Calaamadda dheeraadka ah waxay muujinaysaa in sawirku uu toosan yahay.

Dhererka sawirka ( h _i ):

m = h _i / h _o

h _i = h _o m = (5 cm)(2) = 10 cm

Dhererka sawirka waa 10 cm.

b) Dhererka diiradda (f) = 10 cm iyo masaafada shayga ( d _o ) = 15 cm

Samaynta sawirka iyadoo la adeegsanayo muraayad qaloocan:

Masaafada sawirka ( d _i ):

1/ d _i = 1/f – 1/ d _o = 1/10 – 1/15 = 3/30 – 2/30 = 1/30

d _i = 30/1 = 30 cm

Calaamadda dheeraadka ah waxay muujinaysaa in sawirku yahay mid dhab ah ama sawirku uu yahay 30 cm oo ku yaal muraayadda horteeda, isla dhinaca shayga.

Weyneynta sawirka ( m ):

m = – d _i / d _o = -30/15 = -2

Calaamadda laga jaray waxay muujinaysaa in sawirku rogmanayo.

Sawirku wuxuu 2 jeer ka weyn yahay shayga.

Dhererka sawirka (h _i ):

m = h _i / h _o

h _i = h _o m = (5 cm)(2) = 10 cm

Dhererka sawirka waa 10 cm.

c) Dhererka diiradda (f) = 10 cm iyo masaafada shayga ( d _o ) = 20 cm

Samaynta sawirka iyadoo la adeegsanayo muraayad qaloocan:

Masaafada sawirka (d _i ):

1/ d _i = 1/f – 1/ d _o = 1/10 – 1/20 = 2/20 – 1/20 = 1/20

d _i = 20/1 = 20 cm

Calaamadda togan waxay muujinaysaa in sawirku yahay mid dhab ah ama sawirku uu 20 cm ka horreeyo muraayadda, isla dhinaca shayga.

Weyneynta sawirka ( m ):

m = – d _i / d _o = -20/20 = -1

Calaamadda taban waxay ka dhigan tahay in sawirku rogmanayo.

Dhererka sawirka ( sare ):

m = h _i / h _o

h _i = h m = (5 cm)(1) = 5 cm

d) Dhererka diiradda (f) = 10 cm iyo masaafada shayga ( d _o ) = 30 cm

Masaafada sawirka ( d _i ):

1/ d _i = 1/f – 1/ d _o = 1/10 – 1/30 = 3/30 – 1/30 = 2/30

d _i = 30/2 = 15 cm

Calaamadda lagu daray waxay muujinaysaa in sawirku yahay mid dhab ah ama sawirku uu 15 cm ka horreeyo muraayadda, isla dhinaca shayga.

Weyneynta sawirka ( m) :

m = – d _i / d _o = -15/30 = -0.5

Calaamadda laga jaray waxay muujinaysaa in sawirku rogmanayo.

Sawirku waa 0.5 ka yar yahay shayga.

Dhererka sawirka (h _i ):

m = h _i / h _o

h _i = h _o m = (5 cm)(0.5) = 2.5 cm

3. Sawir ay samaysay muraayad qaloocan ayaa 4 jeer ka weyn shayga. Haddii gacanka qaloocan uu yahay 20 cm, go'aami masaafada shayga ee muraayadda horteeda ah!

La yaqaan:

Weyneynta sawirka ( m ) = 4

Gacanka qalooca ( r ) = 20 cm

Dhererka diiradda (f) = r /2 = 20/2 = 10 cm

La Doonayo : Masaafada shayga ( d _o )

Xalka:

m = – d _i / d _o

4 = – d _{i /} d _o

– d _i = 4 d _o

d _i = – 4 d _o

1/f = 1/ d _o + 1/ d _i

1/10 = 1/ d _o + 1/4 d _o

4/40 = 4/4 d _o + 1/4 d _o

4/40 = 5/4 d _o

(4)(4) = (5)(40)

16 d _o = 200

d _o = 12.5 cm

Masaafada shayga = 12.5 cm.

4. Shay dhererkiisu yahay 1-cm ayaa la dhigayaa 10 cm muraayad qaloocan oo leh dhererka diiradda, f = 15 cm. Go'aami:

A. Masaafada sawirka?

B. Dhererka sawirka?

C. Astaamaha sawirka oo ay samaysay muraayadda qaloocan?

La yaqaan:

Dhererka shayga (h) = 1 cm

Masaafada shayga (d _o ) = 10 cm

Dhererka diiradda ee muraayadda qaloocan (f) = 15 cm

Xalka:

A. Masaafada sawirka (d _i )

1/f = 1/d _o + 1/d _i

1/d _i = 1/f – 1/d _o = 1/15 – 1/10 = 2/30 – 3/30 = -1/30

d _i = -30/1 = -30 cm

Calaamadda taban waxay muujinaysaa in sawirku yahay mid casri ah ama sawirku uu ka dambeeyo muraayadda.

B. Dhererka sawirka (h _i )

Weyneynta sawirka (M):

M = -d _i /d _o = h _i /h _o

M = -(-30)/10 = 30/10 = 3 jeer

Dhererka sawirka (h _i ):

M = h _i / h _o

3 = h _i / 1 cm

h _i = 3 (1 cm)

h _i = 3 cm

Dhererka sawirku waa 3 cm. Calaamadda lagu daray waxay muujinaysaa in sawirku kor u socdo.

C. Sifooyinka sawirka:

Dalwad, kor, ka weyn shayga

5. Weyneynta sawirka, sida ku cad sawirka hoose.

La yaqaan:

Masaafada shayga (d _o ) = 60 cm

Dhererka diiradda (f) = 20 cm

La Doonayo: Weyneynta Sawirka (M)

Xalka:

Masaafada sawirka:

1/f = 1/d _o + 1/d _i

1/d _i = 1/f – 1/d _o = 1/20 cm – 1/60 cm = 3/60 cm – 1/60 cm = 2/60 cm

d _i = 60/2 cm = 30 cm

Weyneynta sawirka (M):

M = d _i / d _o = 30 cm / 60 cm = 1/2 jeer

6. Haddii shayga la dhigo 6 cm muraayad qaloocan, masaafada sawirku waa 12 cm sida ku cad sawirka hoose. Waa maxay masaafada sawirka haddii shayga laga raro meeshii asalka ahayd 1 cm muraayadda.

La yaqaan:

Masaafada shayga (d _o ) = 6 cm

Masaafada sawirka (d _i ) = 12 cm

La Rabay: haddii masaafada shayga (d _o ) = 7 cm markaas masaafada sawirku waa ...

Xalka:

1/f = 1/do + 1/di = 1/6 + 1/12 = 2/12 + 1/12 = 3/12

f = 12/3 = 4 cm

Dhererka diiradda waa mid togan, taasoo la micno ah in barta diiradda la saarayo ay tahay mid dhab ah ama fallaaraha ay ka gudbaan barta.

Masaafada sawirka:

1/d _i = 1/f – 1/d _o = 1/4 – 1/7 = 7/28 – 4/28 = 3/28

d _i = 28/3 = 9.3 cm

7. Dhakhtarka ilkaha ayaa fiirinaya oo hubinaya ilkaha bukaanka isagoo adeegsanaya muraayad leh gacan 8 cm ah. Si uu dhakhtarku si cad u arko godka, waa maxay masaafada u dhaxaysa ilkaha bukaanka iyo muraayadda?

A. wax ka yar 4 cm oo ku yaal muraayad qaloocan oo horteeda ah

B. wax ka yar 4 cm oo ku yaal muraayad qaloocan oo horteeda ah

C. in ka badan 4 cm oo ku yaal muraayadda qaloocan ee hore

D. in ka badan 4 cm oo ku yaal muraayadda qaloocan ee hore

La yaqaan:

Dhexroorka muraayadda (r) = 8 cm

Dhererka muraayadda (f) = r / 2 = 8 / 2 = 4 cm

La Doonayo: Masaafada u dhaxaysa ilkaha bukaanka iyo muraayadda

Xalka:

Muraayadda la isticmaalay ma muraayad qaloocan mise muraayad qaloocan? Si godka ilkaha uu si cad u arko dhakhtarku, muraayadda la isticmaalay waa inay awood u yeelataa inay ballaadhiso sawirka ilkaha sawirkuna waa inuu ahaadaa mid toosan. Muraayadda qaloocan had iyo jeer waxay soo saartaa sawirro qaloocan cabbirka sawirkuna wuu ka yar yahay cabbirka shayga. Taas beddelkeeda muraayad qaloocan waxay soo saari kartaa sawir toosan haddii masaafada shayga (d) ay ka yar tahay dhererka qaloocan (f). Haddii masaafada shaygu ay ka weyn tahay dhererka qaloocan (f) markaa muraayadda qaloocan waxay soo saartaa sawir qaloocan.

Dhererka muraayadda qaloocan (f) waa 4 cm, sidaa darteed ilkaha bukaanka waa inay ka yaraadaan 4 cm hortiisa muraayad qaloocan.

Jawaabta saxda ah waa A.

8. Muraayad qaloocan waxay leedahay gacan qalooc ah oo ah 24 cm. Haddii shayga la dhigo muraayadda horteeda 20 cm markaas go'aami sifooyinka sawirka.

A. Dhab ah, toosan oo la ballaariyay

B. Dhab ah, rogrogmay oo la ballaariyay

C. Daalac, toosan oo la ballaariyay

D. Dalwad, rogrogmi kara oo ka yar

La yaqaan:

Gacanka qalooca ( r) = 24 cm

Dhererka diiradda (f) = R/2 = 24/2 = +12 cm

Dhererka muraayadda qaloocan waa mid togan ama dhab ah sababtoo ah iftiinku wuxuu dhex maraa barta diiradda ee muraayadda.

Masaafada walxaha (d) = 20 cm

La Raadinayo: Sifooyinka sawirka

Xalka:

Sawirku ma mid casri ah baa mise waa mid dhab ah? Xisaabi masaafada sawirka:

1/d + 1/d' = 1/f

1/d' = 1/f – 1/d

1/d' = 1/12 – 1/20

1/d' = 5/60 – 3/60

1/d' = 2/60

d' = 60/2

d' = 30 cm

Masaafada sawirka ee togan ee la calaamadeeyay waxay ka dhigan tahay in sawirku yahay mid dhab ah sababtoo ah waxaa gudbiya iftiin.

Sawirka oo la weyneeyay? Toosan mise rogan? Marka hore xisaabi weyneynta sawirka (M):

M = -d' / d = -30/20 = -1.5

M > 1 macnaheedu waa sawirka waa la weyneeyay, M wuxuu leeyahay calaamad taban macnaheedu waa sawir rogrogay. Markaa sifooyinka sawirku waa kuwo dhab ah, rogrogay, weynaaday.

Jawaabta saxda ah waa B.

9. Muraayad wareegsan ayaa soo saarta sawir cabirkiisu yahay 5 jeer ka weyn shayga shaashadda ku yaal, 5 mitirna u jira shayga. Muraayaddu waa…..

A. qaloocan oo leh dhererka diiradda saaraya 25/24 m

B. qaloocsan oo leh dhererka diiradda oo ah 25/24 m

C. qaloocan oo leh dhererka diiradda saaraya 24/25 m

D. qaloocsan oo leh dhererka diiradda oo ah 24/25 m

La yaqaan:

Weyneynta sawirka (M) = 5 jeer

Masaafada u dhaxaysa shayga iyo sawirka = 5 mitir

Xalka:

Cabbirka sawirka uu soo saaro muraayadda qaloocan ayaa had iyo jeer ka yar cabbirka shayga, sidaas darteed, muraayaddu waa muraayad qaloocan.

Masaafada walxaha (d) = x

Masaafada sawirka (d') = x + 5

Weyneynta sawirka (M) = 5 jeer

Qaacidada weyneynta sawirka:

Qaacidada dhererka diiradda (f):

Jawaabta saxda ah waa A.

[wpdm_package id='858′]

1. Dhibaatooyinka iyo xalalka muraayadda Concave
2. Dhibaatooyinka iyo xalalka muraayadda isku-dhafan
3. Dhibaatooyinka muraayadaha kala duwan iyo xalalka
4. Isku-dubaridka dhibaatooyinka muraayadaha iyo xalalka
5. Qalabka indhaha dhibaatooyinka iyo xalalka indhaha aadanaha
6. Qalabka indhaha dhibaatooyinka iyo xalalka muraayadaha indhaha ee la xiriira
7. Muraayadaha indhaha ee qalabka indhaha
8. Dhibaatooyinka iyo xalalka qalabka indhaha ee weyneynaya muraayadaha
9. Microscope-ka qalabka indhaha - dhibaatooyinka iyo xalalka
10. Dhibaatooyinka iyo xalalka teleskoobyada qalabka indhaha

Akhri wax dheeraad ah

1. Biyo 1-kg ah oo heerkulkoodu yahay 100 oC oo ^lagu qasay biyo 1-kg ah oo heerkulkoodu yahay 10 ^o C oo ku jira nidaam go'doon ah. Kulaylka gaarka ah ee biyuhu waa 4200 J/kg ^o C. Go'aami heerkulka ugu dambeeya ee isku darka!

La yaqaan:

Cufka biyaha kulul (m1 ₎ = 1 kg

Heerkulka biyaha kulul (T ₁ ) = 100 ^o C

Cufka biyaha qabow (m2 ₎ = 1 kg

Heerkulka biyaha qabow (T2 ₎ = 10 ^oC

La Rabay: Heerkulka kama dambaysta ah (T)

Xalka:

Kulayl lumay = Kulayl la helay (nidaam go'doon ah)

m c Δ T = m c Δ T

m Δ T = m Δ T

m ₁ (T ₁ – T) = m ₂ (T – T ₂ )

(1)(100 – T) = (1)(T – 10)

100 – T = T – 10

100 + 10 = T + T

110 = 2T

T = 110 / 2

T = 55

Heerkulka kama dambaysta ah waa 55 ^o C.

2. Baloog rasaas ah oo 3 kg ah oo miisaankiisu yahay 80^o Kuleylka gaarka ah ee rasaasta waa 1400 J.kg.^-1C^-1 kulaylka gaarka ah ee biyuhuna waa 4200 J.kg^-1C^-1Heerkulka ugu dambeeya ee dheelitirka kulaylka waa 20^oC. Go'aami heerkulka bilowga ah ee biyaha!

La yaqaan:

Cufnaanta ( _m1 ) = 3 kg

Kulaylka gaarka ah ee rasaasta (c1 ₎ = 1400 J.kg ^-1 C ^-1

Heerkulka rasaasta (T ₁ ) = 80 ^o C

Cufka biyaha (m2 ₎ = 10 kg

Kulaylka gaarka ah ee biyaha (c ₂ ) = 4200 J.kg ^-1 C ^-1

Heerkulka dheelitirka kulaylka (T) = 20 ^o C

La Rabay: Heerkulka bilowga ah ee biyaha ( _T2 )

Xalka:

Kulayl lumay = Kulayl la helay

Q lead = Q biyo

m ₁ c ₁ ΔT = m ₂ c ₂ ΔT

(3)(1400)(80-20) = (10)(4200)(20-T)

(4200)(60) = (42,000)(20-T)

252,000 = 840,000 – 42,000 T

42,000 T = 840,000 – 252,000

42,000 T = 588,000

T = 588,000 / 42,000

T = 14

Heerkulka biyaha ee bilowga ah waa 14 ^o C.

3. Baloog naxaas ah oo ku yaal 100^oC lagu shubay 128 garaam oo biyo ah 30 ^oC. Kulaylka gaarka ah ee biyuhu waa 1 cal.g^-1oC^-1 kulaylka gaarka ah ee naxaasta waa 0.1 cal.g^-1oC^-1. Haddii heerkulka dheelitirka kulaylka uu yahay 36 ^oC, go'aami cufka naxaasta!

La yaqaan:

Heerkulka naxaasta (T ₁ ) = 100 ^o C

Kulaylka gaarka ah ee naxaasta (c1 ₎ = 0.1 c al.g ^-1o C ^-1

Cufka biyaha (m2 ₎ = 128 garaam

Heerkulka biyaha (T ₂ ) = 30 ^o C

Kulaylka gaarka ah ee biyaha (c2 ₎ = 1 c al.g ^-1o C ^-1

Heerkulka dheelitirka kulaylka (T) = 36 ^o C

La Raadinayo: Cufka naxaasta ( _m1 )

Xalka:

Kulayl lumay = Kulayl la helay

Q naxaas = Q biyo

m ₁ c ₁ ΔT = m ₂ c ₂ ΔT

(m ₁ )(0.1)(100-36) = (128)(1)(36-30)

(m ₁ )(0.1)(64) = (128)(1)(6)

(m ₁ )(6.4) = 768

m ₁ = 768 / 6.4

m ₁ = 120

Miisaanka naxaasta waa 120 garaam.

4. Baloog baraf ah oo M-kg ah 0^oC lagu shubay 340 garaam oo biyo ah 20^oC weel yar. Haddii kulaylka isku-darka biyo = 80 kaloori g^-1, kulaylka gaarka ah ee biyuhu waa 1 kaloori g^{-1 o}C^-1. Dhammaan barafku wuu dhalaalaa iyo heerkulka dheelitirka kulaylka is 5^oC, go'aami cufka barafka!

La yaqaan:

Cufka biyaha (m) = 340 garaam

Heerkulka barafka (T _baraf ) = 0 ^o C

Heerkulka biyaha (T _biyo ) = 20 ^o C

Heerkulka dheelitirka kulaylka (T) = 5 ^o C

Kulaylka isku-darka biyaha ( L ) = 80 c al g ^-1

Kulaylka gaarka ah ee biyaha (c _biyo ) = 1 kalori g ^{-1 o} C ^-1

La Raadinayo: Cufnaanta barafka (M)

Xalka:

Kulayl lumay = Kulayl la helay

Biyo Q = baraf Q

m c ( ΔT) = m _es L _es + m c ( Δ T)

(340)(1)(20-5) = M (80) + M (1)(5-0)

(340)(15) = 80M + 5M

5100 = 85M

M = 5100/85

M = 60 garaam

[wpdm_package id='714′]

1. Beddelidda miisaanka heerkulka
2. Balaadhinta toosan
3. Ballaarinta aagga
4. Ballaarinta mugga
5. Heat
6. U dhigma farsamo ee kulaylka
7. Awood gaar ah oo kul iyo kulayl ah
8. Kulayl qarsoon, kulayl isku-dhafan, kulayl uumi-baxa
9. Ilaalinta tamarta ee wareejinta kulaylka

Akhri wax dheeraad ah

4 Kulayl qarsoon Kulayl isku-dhafan Kulayl uumi-baxa - Dhibaatooyinka iyo Xalalka

1. Xisaabi inta kulaylka lagu daray 1 garaam oo dahab ah si aad u beddesho marxaladda laga bilaabo adag ilaa dareere. Kulaylka isku-darka dahabka waa 64.5 x 10 ³ J/kg.

La yaqaan:

Cufnaanta (m) = 1 garaam = 1 x 10 ^-3 kg

Kulaylka isku-dhafka (L _F ) = 64.5 x 10 ³ J/kg

La Rabay: Kulayl (Q)

Xalka:

Q = m L _F

Q = (1 x 10 ^-3 kg)(64.5 x 10 ³ J/kg)

Q = 64.5 Joule

2. Xisaabi inta kulaylka lagu sii daayo 1 garaam oo meerkuri ah si aad marxaladda uga beddesho dareere una beddesho mid adag. Kulaylka isku-dhafka meerkurigu waa 11.8 x 10 ³ J/kg.

La yaqaan:

Cufnaanta (m) = 1 garaam = 1 x 10 ^-3 kg

Kulaylka isku-dhafka (L _F ) = 11.8 x 10 ³ J/kg

La Rabay: Kulayl (Q)

Xalka:

Q = m L _F

Q = (1 x 10 ^-3 kg)(11.8 x 10 ³ J/kg)

Q = 11.8 Joule

3. Go'aami xaddiga kulaylka uu nuugo 1 kg oo biyo ah si aad marxaladda uga beddesho dareere una beddesho uumi (uumi). Kulaylka uumi-baxa biyaha = 2256 x 10 ³ J/kg

La yaqaan:

Cufnaanta (m) = 1 kg

Kulaylka uumiga (L _V ) = 2256 x 10 ³ J/kg

La Rabay: Kulayl (Q)

Xalka:

Q = m L _V

Q = (1 kg)( 2256 x 10 ³ J/kg)

Q = 2256 x 10 ³ Joule

4. Go'aami xaddiga kulaylka uu sii daayo naytaroojiin si uu marxaladda uga beddelo uumiga una beddelo dareere. Kulaylka uumiga ee naytaroojiin = 200 x 10 ³ J/kg

La yaqaan:

Cufnaanta (m) = 1 garaam = 1 x 10 ^-3 kg

Kulaylka uumiga (L _V ) = 200 x 10 ³ J/kg

La yaqaan: Kulayl (Q)

Xalka:

Q = m L _V

Q = (1 x 10 ^-3 kg)( 200 x 10 ³ J/kg)

Q = 200 Joule

1. Beddelidda miisaanka heerkulka
2. Balaadhinta toosan
3. Ballaarinta aagga
4. Ballaarinta mugga
5. Heat
6. U dhigma farsamo ee kulaylka
7. Awood gaar ah oo kul iyo kulayl ah
8. Kulayl qarsoon, kulayl isku-dhafan, kulayl uumi-baxa
9. Ilaalinta tamarta ee wareejinta kulaylka

Akhri wax dheeraad ah

1. Jirka oo miisaankiisu yahay 2 kg wuxuu nuugaa kulaylka 100 kaloori marka heerkulkiisu ka kordho 20 ^o C ilaa 70 ^o C. Waa maxay kulaylka gaarka ah ee jirka?

La yaqaan:

Cufnaanta (m) = 2 kg = 2000 g

Kulayl (Q) = 100 c al

Isbeddelka heerkulka ( ΔT ) = 70 ^o C – 20 ^o C = 50 ^o C

La Doonayo: Kulaylka gaarka ah (c)

Xalka:

c = Q / m Δ T

c = 100 kalori / (2000 garaam) (50 ^o C)

c = 100 kalori / 100,000 gr ^o C

c = 10 ² kalori / 10 ⁵ g ^o C

c = (10 ² kaloori)( 10 ^-5 gr ^{-1 o} C ^-1 )

c = 10 ^-3 kalori gr ^{-1 o} C ^-1

c = 10 ^-3 kalori/ g ^o C

Kulaylka gaarka ah ee jirka waa 10 ^-3 kalori/gr ^o C

2. Kulaylka gaarka ah ee biyuhu waa 4180 J/kg C ^o . Immisa awood kuleyl ah oo 2 kg ah oo biyo ah …

La yaqaan:

Kulaylka gaarka ah (c) = 4180 J/kg C ^o

Cufnaanta (m) = 2 kg

La Rabo: Awoodda kulaylka (C)

Xalka:

C = mc

C = (2 kg)(4180 J/kg C ^o )

C = (2)(4180 J/C ^o )

C = 8360 J/C ^o

3. Kulaylka gaarka ah ee aluminiumku waa 900 J/kg C ^o . Immisa awood kuleyl ah oo 2 garaam ah oo aluminium ah…..

La yaqaan:

Kulaylka gaarka ah ee aluminiumka (c) = 900 J/kg C ^o = 9 x 10 ² J/kg C ^o

Cufnaanta (m) = 2 garaam = 2/1000 kg = 2/10 ³ kg = 2 x 10 ^-3 kg

La Rabo: Awoodda kulaylka (C)

Xalka:

C = mc

C = (2 x 10 ^-3 kg)( 9 x 10 ² J/kg C ^o )

C = 18 x 10 ^-3 x 10 ² J/C ^o

C = 18 x 10 ^-1 J/C ^o

C = 1.8 J/C ^o

[wpdm_package id='710′]

1. Beddelidda miisaanka heerkulka
2. Balaadhinta toosan
3. Ballaarinta aagga
4. Ballaarinta mugga
5. Heat
6. U dhigma farsamo ee kulaylka
7. Awood gaar ah oo kul iyo kulayl ah
8. Kulayl qarsoon, kulayl isku-dhafan, kulayl uumi-baxa
9. Ilaalinta tamarta ee wareejinta kulaylka

Akhri wax dheeraad ah

1. 2 k kalori (kiilo c kaloori e ) = ….. c kaloori e?

Solution

1 kcal = 1000 kcal

2 k kalori = 2 (1000 c kaloori e ) = 2000 c kaloori e

2. 4 C kalori e = ….. c kalorie e?

Solution

1 Calori e ( a Capital C ) = 1 k cal = 1000 kalori e

4 C kalori e = 4 (1000 c kalori e ) = 4000 c kalori e

3. 10 c calories e = ….. Joule?

Solution

1 kaloori e = 4.186 Joules

10 c kaloori e = 10 (4.186 Joules) = 41.86 Joules

4. 5 kcal = ….. Joule?

Solution

1 k cal = 1000 c alorie = 4186 Joules

5 k kal = 5 (4186 Joules) = 20930 Joules

5. 2000 Joules = ….. k cal?

Solution

4186 Joules = 1 kcal

8372 Joules = 8372 / 4186 = 2 kcal

2000 Joules = 2000 / 4186 = 0.4777 kcal

[wpdm_package id='706′]

1. Beddelidda miisaanka heerkulka
2. Balaadhinta toosan
3. Ballaarinta aagga
4. Ballaarinta mugga
5. Heat
6. U dhigma farsamo ee kulaylka
7. Awood gaar ah oo kul iyo kulayl ah
8. Kulayl qarsoon, kulaylka isku-darka, kulaylka uumiga
9. Ilaalinta tamarta ee wareejinta kulaylka

Akhri wax dheeraad ah

9 Kulayl Gaar ah oo Cufnaanta Kulaylka Isbeddelka heerkulka - Dhibaatooyinka iyo Xalalka

1. Raasad 2 kg ah ayaa laga kululeeyaa 50^oC ilaa 100^oC. The kuleyl gaar ah rasaastu waa 130 J.kg^{-1 o}C^-1. Waa imisa? kulaylka ma waxaa nuugtaa rasaasta?

La yaqaan:

Cufnaanta (m) = 2 kg

Kulaylka gaarka ah (c) = 1 3 0 J.kg ^-1 C ^-1

Isbeddelka heerkulka ( Δ T) = 100 ^o C – 50 ^o C = 50 ^o C

La Rabay: Kulayl (Q)

Xalka:

Q = mc Δ T

Q = kuleyl , m = cuf, c = kulaylka gaarka ah , Δ T = isbeddelka heerkulka

Kulaylka uu nuugo rasaasta:

Q = (2 kg)(1 3 0 J.kg ^-1 C ^-1 )(50 ^o C)

Q = (100)(1 3 0)

Q = 1 3,000 Joules

Q = 1.3 x ¹⁰⁴ Joules

2. Kulaylka gaarka ah ee naxaasta waa 390 J/k g ^o C, isbeddelka heerkulka waa 40 ^o C. Haddii naxaasta ay nuugto 40 Joules oo kuleyl ah, waa maxay cufka naxaasta!

La yaqaan:

Kulaylka gaarka ah ee naxaasta (c) = 390 J/k g ^o C

Isbeddelka heerkulka ( ΔT ) = 40 ^o C

Kulayl ( Q) = 40 J

La Raadinayo: Cufnaanta (m) ee naxaasta

Xalka:

Q = mc Δ T

40 J = (m)( 390 J/k g ^o C)(40 ^o C)

40 = (m)( 390 /kg )(40)

40 = (m)( 390 /kg )(4)

40 = (m)( 1560 /kg )

m = 40 / 1560

m = 0.026 kg

m = 26 garaam

3. Heerkulka bilowga ah ee 20 garaam oo biyo ah waa 30 ^o C. Kulaylka gaarka ah ee biyaha waa 1 kalori g ^{-1 o} C ^-1 . Haddii biyuhu ay nuugaan 300 kaloori oo kuleyl ah, go'aami heerkulka ugu dambeeya!

La yaqaan:

Cufnaanta (m) = 20 g

Heerkulka bilowga ah (T ₁ ) = 30 ^o C

Kulaylka gaarka ah ee biyaha (c) = 1 kaloori gr ^{-1 o} C ^-1

Kulayl (Q) = 300 kalori

La Doonayo: Heerkulka ugu dambeeya ee biyaha

Xalka:

Q = mc ΔT

300 kalori = (20 garaam)( 1 kalori gr ^{-1 o} C ^-1 )( T ₂ -30 )

300 = (20)(1)( T ₂ -30)

300 = 20 ( T ₂ -30 )

300 = 20 T ₂ – 600

300 + 600 = 20T ₂

900 = 20T ₂

T ₂ = 900 / 20

T ₂ = 45

Isbeddelka heerkulka waa 45 ^o C – 30 ^o C = 15 ^o C.

4. Tisbeddelka heerkulka biyaha badda is 1^oC marka biyuhu nuugaan 3900 Joules oo kuleyl ah. Kulaylka gaarka ah ee biyaha badda waa 3.9 10³ J/kg°C, Waa maxay cufnaanta biyaha badda.

La yaqaan:

Isbeddelka heerkulka ( ΔT ) = 1 ^oC

Kulayl (Q) = 3900 Joule

Kulaylka gaarka ah ee biyaha badda (c) = 3.9 x 10 ³ J/ k g°C = 3900 J/Kg°C

La Raadinayo: Cufnaan (m)

Xalka:

Q = mc Δ T

Q = kuleyl , m = cuf, c = kuleyl gaar ah , Δ T = isbeddelka heerkulka

m = Q / c Δ T = 3900 / (3900)(1) = 3900 / 3900 = 1 kg

5. Naxaas 2-kg ah ayaa nuugta 39,000 J oo kulayl ah 30°C. Haddii kulaylka gaarka ah ee naxaasta uu yahay 390 J/kg °C, waa maxay heerkulka ugu dambeeya ee naxaasta…

La yaqaan:

Cufnaanta (m) = 2 kg

Heerkulka bilowga ah (T ₁ ) = 30 ^o C

Kulayl (Q) = 39,000 Joule

Kulayl gaar ah (c) oo naxaas ah = 390 J/kg ^o C

La Rabay : Heerkulka kama dambaysta ah ( _T2 )

Xalka:

Q = mc ΔT

Q = kuleyl , m = cuf, c = kuleyl gaar ah , ΔT = isbeddelka heerkulka

Q = mc ΔT = mc (T ₂ – T ₁ )

39,000 = (2)( 390)(T ₂ – 30)

100 = (2)(1)(T ₂ – 30)

100 = (2)(T ₂ – 30)

50 = T ₂ – 30

T ₂ = 50 + 30

T ₂ = 80 ^o C

6. Biyo 5-kg ah ayaa lagu kululeeyaa 15°C ilaa 40°C. Waa maxay kulaylka ay nuugaan biyuhu. Kulaylka gaarka ah ee biyuhu waa 4.2 × 10 ³ J/Kg° C.

La yaqaan:

Cufnaanta (m) = 5 kg

Heerkulka bilowga ah (T ₁ ) = 15°C

Heerkulka kama dambaysta ah (T2 ₎ = 40°C

Kulayl gaar ah oo biyo ah (c) = 4.2 × 10 ³ J/kg° C

La Rabay: Kulayl (Q)

Xalka:

Q = mc ΔT

Q = (5 kg)( 4.2 × 10 ³ J/kg°C)(40°C – 15°C)

Q = (5)( 4.2 × 10 ³ J)(25)

Q = 525 x 10 ³ J

Q = 525,000 Joule

7. Biyo 2-kg ah ayaa lagu kululeeyaa 24°C ilaa 90°C. Waa maxay kulaylka ay nuugaan biyuhu. Kulaylka gaarka ah ee biyuhu waa 4.2 × 10 ³ J/Kg° C.

La yaqaan:

Cufnaanta (m) = 2 kg

Heerkulka bilowga ah (T ₁ ) = 24°C

Heerkulka kama dambaysta ah (T2 ₎ = 90°C

Kulayl gaar ah oo biyo ah (c) = 4,200 Joules/kg°C

La Doonayo :: Kulayl (Q)

Xalka:

Q = m c ΔT

Q = (2 kg)( 4,200 Joules/kg°C)(90°C – 24°C)

Q = (2 kg)( 4,200 Joules/kg°C)(66°C)

Q = (132)( 4,200 Joules)

Q = 554,400 Joule

8. Biyo 5 -garaam ah ayaa lagu kululeeyaa 10°C ilaa 40°C. Waa maxay kulaylka ay nuugaan biyuhu. Kulaylka gaarka ah ee biyuhu waa 1 × 10 ³ kalori /g r ° C.

La yaqaan:

Cufnaanta (m) = 5 garaam

Heerkulka bilowga ah (T ₁ ) = 10 ^o C

Heerkulka kama dambaysta ah (T2 ₎ = 40 ^o C

Kulayl gaar ah oo biyo ah (c) = 1 c al/ gr°C

La Doonayo : Kulayl

Xalka:

Q = mc ΔT

Q = (5 garaam)( 1 kalori/gr°C)(40 ^o C – 10 ^o C)

Q = (5)( 1 c al)(30)

Q = 150 kaloori

9. Biyo 0.2-kg ah ayaa nuugaya 42,000 Joules oo kuleyl ah 25 ^oC . Kulaylka gaarka ah ee biyuhu waa 4200 J/kg ^o C, waa maxay heerkulka ugu dambeeya ee biyaha.

La yaqaan:

Cufka biyaha (m) = 0.2 kg

Kulayl (Q) = 42,000 Joule

Kulayl gaar ah oo biyo ah (c) = 4200 J/kg ^o C

Heerkulka bilowga ah (T ₁ ) = 25 ^o C

La Rabay: Heerkulka kama dambaysta ah ( _T2 )

Xalka:

Q = mc ΔT = mc (T ₂ – T ₁ )

Q= kulaylka, m = cuf, c = kuleyl gaar ah, ΔT = isbeddelka heerkulka, T₁ = heerkulka bilowga ah, T₂ = heerkulka kama dambaysta ah

Q = mc (T ₂ – T ₁ )

42,000 = (0.2)(4200)(T ₂ – 25)

42,000 = 840 (T ₂ – 25)

42,000 = 840 T ₂ – 21,000

42,000 + 21,000 = 840 T ₂

63,000 = 840 T ₂

T ₂ = 63,000 / 840

T ₂ = 75 ^o C

1. Beddelidda miisaanka heerkulka
2. Balaadhinta toosan
3. Ballaarinta aagga
4. Ballaarinta mugga
5. Heat
6. U dhigma farsamo ee kulaylka
7. Awood gaar ah oo kul iyo kulayl ah
8. Kulayl qarsoon, kulayl isku-dhafan, kulayl uumi-baxa
9. Ilaalinta tamarta ee wareejinta kulaylka

Akhri wax dheeraad ah

Ballaarinta toosan waxaa la kulmaa oo keliya walxaha adag; ballaarinta mugga waxaa la kulma dhammaan walxaha, labadaba adag, dareere, iyo gaas. Isla'egta ballaarinta mugga waxay la mid tahay isla'egta ballaarinta toosan.

Sharaxaad: V _o = Mugga bilowga ah, V = Mugga kama dambaysta ah, ΔV = V – V _o = Isbeddelka mugga, T _o = Heerkulka bilowga ah, T = Heerkulka kama dambaysta ah, ΔT = T _o = Isbeddelka heerkulka, β = isku-darka ballaarinta mugga. Cutubyada β = (C ^o ) ^-1

Isle'egta ballaarinta mugga ee kor ku xusan waxay khuseysaa oo keliya marka isbeddellada mugga walxaha (adag, dareere, iyo gaas) ay ka yar yihiin mugga asalka ah ee shayga. Haddii isbeddelka mugga shay uu ka weyn yahay mugga bilowga ah ee shayga, isle'egta ballaarinta mugga ma bixiso natiijooyinka saxda ah. Badanaa, isbeddellada mugga ee ay la kulmaan walxaha adag ma aha kuwo aad u weyn. Dhanka kale, isku-dhafka ballaarinta mugga ee dareeraha iyo gaaska waa weyn yahay. Isku-dhafka ballaarinta mugga ee walxaha gaaska sidoo kale waa mid fudud in la beddelo haddii heerkulku isbeddelo. Sidaa darteed qaacidada kore waxaa loo isticmaalaa oo keliya ballaarinta walxaha adag.

1. Heerkulka 30 ^o C mugga wareegga aluminiumku waa 30 cm ^3. Isugeynta ballaarinta toosan waa 24 x 10 ^{-6 o} C ^-1 . Haddii mugga kama dambaysta ahi yahay 30.5 cm ³ , waa maxay heerkulka kama dambaysta ah ee wareegga aluminiumku?

La yaqaan:

Isugeynta ballaarinta toosan ( α ) = 24 x 10 ^{-6 o} C ^-1

Isugeynta ballaarinta mugga ( β ) = 3 α = 3 x 24 x 10 ^{-6 o} C ^-1 = 72 x 10 ^{-6 o} C ^-1

Heerkulka bilowga ah (T ₁ ) = 30 ^o C

Mugga bilowga ah (V1 ₎ = 30 cm ³

Mugga ugu dambeeya (V2 ₎ = 30.5 cm ³

Isbeddelka mugga ( Δ V) = 30.5 cm ³ – 30 cm ³ = 0.5 cm ³

La Rabay: Heerkulka kama dambaysta ah ( _T2 )

Xalka:

Δ V = β (V1 ₎ ( ΔT )

Δ V = β (V ₁ )(T ₂ – T ₁ )

0.5 cm³ = (72 x 10^{-6 o}C^-1)(30 cm³)(T₂ - 30^oC)

0.5 = (2160 x 10 ^-6 ) (T ₂ – 30 )

0.5 = (2.160 x 10 ^-3 ) (T 2 – 30 )

0.5 = (2.160 x 10 ^-3 ) (T ₂ – 30 )

0.5 / (2.160 x 10 ^-3 ) = T ₂ – 30

0.23 x 10 ³ = T ₂ – 30

0.23 x 1000 = T ₂ – 30

230 = T ₂ – 30

230 + 30 = T ₂

T ₂ = 260 ^o C

2. Isugeynta ballaarinta toosan ee wareegga birta waa 9 x 10 ^{-6 o} C ^-1 . Dhexroorka gudaha ee wareegga birta ee 20 ^o C waa 2.2 cm. Haddii dhexroorka kama dambaysta ahi yahay 2.8 cm, waa maxay heerkulka kama dambaysta ah!

La yaqaan:

Isugeynta ballaarinta toosan ( α ) = 9 x 10 ^{-6 o} C ^-1

Isugeynta ballaarinta mugga ( β) = 3 α = 3 x 9 x 10 ^{-6 o} C ^-1 = 27 x 10 ^{-6 o} C ^-1

Heerkulka bilowga ah (T ₁ ) = 20 ^o C

Dhexroorka bilowga ah (D ₁ ) = 2.2 cm

Dhexroorka ugu dambeeya (D ₂ ) = 2.8 cm

Gacanka bilowga ah (r ₁ ) = D ₁ / 2 = 2.2 cm ³ / 2 = 1.1 cm ³

Gacanka ugu dambeeya (r ₂ ) = D ₂ / 2 = 2.8 cm ³ / 2 = 1.4 cm ³

Mugga bilowga ah (V ₁ ) = 4/3 π r ₁ ³ = (4/3)(3.14)(1.1 cm) ³ = (4/3)(3.14)(1.331 cm ³ ) = 5.57 cm ³

Mugga ugu dambeeya (V2 ₎ = 4/3 π r ₂ ³ = (4/3)(3.14)(1.4 cm) ³ = (4/3)(3.14)(2.744 cm ³ ) = 11.48 cm ³

Isbeddelka mugga ( Δ V) = 11.48 cm ³ – 5.57 cm ³ = 5.91 cm ³

La Rabay: Heerkulka _{ugu dambeeya} ( T2 )

Xalka:

Δ V = β (V1 ₎ ( ΔT )

5.91 cm³ = (27 x 10^{-6 o}C^-1)(5.57 cm³)(T₂ - 20^oC)

5.91 = (150.39 x 10 ^-6 ) (T ₂ – 20)

5.91 / 150.39 x 10 ^-6 = T ₂ – 20

0.039 x 10 ⁶ = T ₂ – 20

39 x 10 ³ = T ₂ – 20

39,000 = T ₂ – 20

39,000 + 20 = T ₂

T ₂ = 39,020 ^o C

3. 2000-cm³ weel aluminium, oo lagu buuxiyay biyo 0^oC. Kadibna lagu kululeeyo 90^oC. Haddii isku-dhafka ballaarinta toosan ee aluminiumku yahay 24 x 10^-6 (^oC)^-1 iyo isku-darka ballaarinta mugga biyaha waa 6.3 x 10^-4 (^oC)^-1, go'aami mugga biyaha daatay.

La yaqaan:

Mugga hore ee weelka aluminiumka iyo biyaha (Vo ₎ = 2000 cm ³ = 2 x 10 ³ cm ³

Heerkulka bilowga ah ee weelka aluminiumka iyo biyaha (T ₁ ) = 0 ^o C

Heerkulka kama dambaysta ah ee weelka aluminiumka iyo biyaha (T2 ₎ = 90 ^o C

Isugeynta ballaarinta toosan ee aluminiumka (α) = 24 x 10 ^-6 ( ^o C ) ^-1

Isugeynta ballaarinta mugga ee aluminiumka (γ) = 3α = 3 (24 x 10 ^-6 ( ^o C ) ^-1 ) = 72 x 10 ^{-6 o} C ^-1

Isugeynta ballaarinta mugga biyaha (γ) = 6.3 x 10 ^-4 ( ^o C ) ^-1

La Doonayo: Mugga biyaha daatay

Xalka:

Isle'egta ballaarinta mugga:

V = Vo _o + γ Vo _o ΔT

V – Vo _o = γ Vo _o ΔT

ΔV = γ V _o ΔT

V = mugga kama dambaysta ah, V _o = mugga bilowga ah, ΔV = isbeddelka mugga, γ = isku-dhafka ballaarinta mugga, ΔT = isbeddelka heerkulka

Xisaabi isbeddelka mugga weelka aluminiumka:

ΔV = γ V _o ΔT = (72 x 10 ^-6 )(2 x 10 ³ )(90) = 12960 x 10 ^-3 = 12.960 cm ³

Xisaabi isbeddelka mugga biyaha:

ΔV = γ V _o ΔT = (6. 3 x 10 ^-4 )(2 x 10 ³ )(90) = 1134 x 10 ^-1 = 113.4 cm ³

Isbeddelka mugga biyuhu wuu ka weyn yahay weelka aluminiumka sidaa darteed biyo ayaa daatay.

Xisaabi mugga biyaha daatay:

113.4 cm ³ – 12.960 cm ³ = 100.44 cm ³

[wpdm_package id='702′]

1. Beddelidda miisaanka heerkulka
2. Balaadhinta toosan
3. Ballaarinta aagga
4. Ballaarinta mugga
5. Heat
6. U dhigma farsamo ee kulaylka
7. Awood gaar ah oo kul iyo kulayl ah
8. Kulayl qarsoon, kulaylka isku-darka, kulaylka uumiga
9. Ilaalinta tamarta ee wareejinta kulaylka

Akhri wax dheeraad ah

1. Heerkulka 20 ^o C, dhererka xaashi bir ah waa 50 cm ballaciisuna waa 30 cm. Haddii isku-dhafka ballaarinta toosan ee birta uu yahay 10 ^{-5 o} C ^-1 , go'aami isbeddelka aagga iyo aagga ugu dambeeya ee 60 ^o C.

La yaqaan:

Heerkulka bilowga ah (T ₁ ) = 20 ^o C

Heerkulka kama dambaysta ah (T2 ₎ = 60 ^o C

Isbeddelka heerkulka ( ΔT ) = 60 ^o C – 20 ^o C = 40 ^o C

Bedka bilowga ah (A ₁ ) = dhererka x ballaca = 50 cm x 30 cm = 1500 cm ²

Isugeynta ballaarinta toosan ee birta ( α) = 10 ^{-5 o} C ^-1

Isugeynta ballaarinta aagga ee birta ( β) = 2α = 2 x 10 ^{-5 o} C ^-1

La Doonayo: Isbeddelka aagga ( ΔA )

Xalka:

Isbeddelka aagga ( ΔA ):

ΔA = β A ₁ ΔT

ΔA = ( 2 x 10 ^{-5 o} C ^-1 )(1500 cm ² )(40 ^o C)

ΔA = (80 x 10 ^-5 ) (1500 cm ² )

ΔA = 1 20,000 x 10 ^-5 cm ²

ΔA = 1. 2 x 10 ⁵ x 10 ^-5 cm ²

ΔA = 1.2 cm ²

Aagga ugu dambeeya ( A ₂ ):

A ₂ = A ₁ + ΔA

A ₂ = 1500 cm ² + 1. 2 cm ²

A ₂ = 1501. 2 cm ²

2. Heerkulka 30 ^° C, bedka xaashida aluminiumka waa 40 cm ² , isku-dhafka ballaarinta toosanna waa 24 x 10 ^-6 / ^o C. Go'aami heerkulka kama dambaysta ah haddii aagga kama dambaysta ahi yahay 40.2 cm ².

La yaqaan:

Heerkulka bilowga ah (T ₁ ) = 30 ^o C

Isugeynta ballaarinta toosan ( α ) = 24 x 10 ^{-6 o} C ^-1

Isugeynta ballaarinta aagga ( β) = 2a = 2 x 24 x 10 ^{-6 o} C ^-1 = 48 x 10 ^{-6 o} C ^-1

Bedka bilowga ah (A ₁ ) = 40 cm ²

Bedka ugu dambeeya (A ₂ ) = 40.2 cm ²

Isbeddelka aagga ( ΔA) = 40.2 cm ² – 40 cm ² = 0.2 cm ²

La Rabay: Go'aami heerkulka kama dambaysta ah ( _T2 )

Xalka:

Qaacidada isbeddelka aagga ( ΔA) :

ΔA = β A ₁ Δ T

Heerkulka kama dambaysta ah (T2 ₎ :

ΔA = β A ₁ (T ₂ – T ₁ )

0.2 cm² = (48 x) 10^{-6 o}C^-1(40 cm)²)(T₂ - 30^oC)

0.2 = (1920 x 10 ^-6 ) (T ₂ – 30 )

0.2 = (1.920 x 10 ^-3 ) (T ₂ – 30)

0.2 = (2 x 10 ^-3 ) (T ₂ – 30)

0.2 / (2 x 10 ^-3 ) = T ₂ – 30

0.1 x 10 ³ = T ₂ – 30

1 x 10 ² = T ₂ – 30

100 = T ₂ – 30

100 + 30 = T ₂

T ₂ = 130

Heerkulka kama dambaysta ah = 130 ^o C

3. Gacanka giraanta ee 20 ^o C waa 20 cm. Haddii gacanka ugu dambeeya ee 100 ^o C uu yahay 20.5 cm, go'aami isku-dhafka ballaarinta aagga iyo isku-dhafka ballaarinta toosan…

La yaqaan:

Heerkulka bilowga ah (T ₁ ) = 30 ^o C

Heerkulka kama dambaysta ah (T2 ₎ = 100 ^o C

Isbeddelka heerkulka ( ΔT ) = 100 ^o C – 30 ^o C = 70 ^o C

Gacanka bilowga ah (r ₁ ) = 20 cm

Gacanka ugu dambeeya (r ₂ ) = 20.5 cm

La Doonayo: Isugeynta ballaarinta aagga ( β )

Xalka:

Bedka bilowga ah (A ₁ ) = π r ₁ ² = (3.14)(20 cm) ² = (3.14)(400 cm ² ) = 1256 cm ²

Bedka ugu dambeeya (A ₂ ) = π r ₂ ² = (3.14)(20.5 cm) ² = (3.14)(420.25 cm ² ) = 1319.585 cm ²

Isbeddelka aagga ( ΔA) = 1319.585 cm ² – 1256 cm ² = 63.585 cm ²

Qaacidada isbeddelka aagga ( ΔA) :

ΔA = β A ₁ ΔT

Isugeynta ballaarinta aagga:

ΔA = β A ₁ ΔT

63.585 cm ² = b (1 256 cm ² )(70 ^o C)

63.585 = b ( 87,920 ^o C)

β = 63.585 / 87,920 ^o C

β = 0.00072 / ^o C

β = 7.2 x 1 0 ^-4 / ^o C

β = 7.2 x 1 0 ^{-4 o} C ^-1

Isugeynta ballaarinta toosan ( α ):

β = 2 α

α = β/2

α = (7.2 x 1 0 ^-4 ) / 2

α = 3.6 x 1 0 ^{-4 o} C ^-1

[wpdm_package id='698′]

1. Beddelidda miisaanka heerkulka
2. Balaadhinta toosan
3. Ballaarinta aagga
4. Ballaarinta mugga
5. Heat
6. U dhigma farsamo ee kulaylka
7. Awood gaar ah oo kul iyo kulayl ah
8. Kulayl qarsoon, kulayl isku-dhafan, kulayl uumi-baxa
9. Ilaalinta tamarta ee wareejinta kulaylka

Akhri wax dheeraad ah