Mgbasa ahịrị - nsogbu na ngwọta

1. Ígwè dị 40 cm n'ogologo na 20 ^o C. Ọnụọgụ nke mgbasawanye linear maka ígwè bụ 12 x 10 ^-6 (C ^o ) ^-1 . Mmụba n'ogologo na ogologo ikpeazụ mgbe ọ dị na 70 ^o C ga-abụ…

A maara:

Mgbanwe na okpomọkụ ( ΔT ) = 70 ^o C – 20 ^o C = 50 ^o C

Ogologo mbụ (L ₁ ) = 40 cm

Ọnụọgụ nke mgbasawanye ahịrị maka ígwè (α ) = 12 x 10 ^-6 (C ^o ) ^-1

Achọrọ: Mgbanwe n'ogologo ( Δ L) na ogologo ikpeazụ (L ₂ )

Ngwọta:

a) Mgbanwe n'ogologo ( ΔL)

Δ L = α L ₁ ΔT

Δ L = ( 12 × 10 ^{-6 o} C ^-1 ) (40 cm) (50 ^o C)

Δ L = (10 ^-6 ) (24 x 10 ³ ) cm

Δ L = 24 x 10 ^-3 cm

Δ L = 24/10 ³ cm

Δ L = 24/1000 cm

Δ L = 0.024 cm

b) Ogologo ikpeazụ ( _L2 )

L ₂ = L ₁ + ΔL

L ₂ = 40 cm + 0.024 cm

L ₂ = 40.024 cm

[irp]

2. Mkpanaka ígwè a na-ekpo ọkụ site na 30 ^o C ruo 80 ^o C. Ogologo ígwè ikpeazụ bụ 115 cm, ọnụọgụgụ nke mgbasawanye ahịrị bụkwa 3 × 10 ^{-3 o} C ^-1 . Gịnị bụ ogologo mbụ na mgbanwe n'ogologo ígwè ahụ?

Ngwọta:

Mgbanwe na okpomọkụ ( ΔT) = 80 ^o C – 30 ^o C = 50 ^o C

Ogologo ikpeazụ (L ₂ ) = 115 cm

Ọnụọgụ nke mgbasawanye ahịrị ( α) = 3×10 ^{-3 o} C ^-1

Achọrọ: ogologo mbụ (L ₁ ) na mgbanwe n'ogologo ( Δ L)

Ngwọta:

a) Ogologo mbụ (L ₁ )

Usoro mgbanwe n'ogologo maka mgbasawanye ahịrị:

Δ L = α L ₁ ΔT

Usoro nke ogologo ikpeazụ :

L ₂ = L ₁ + ΔL

L ₂ = L ₁ + α L ₁ ΔT

L ₂ = L ₁ (1 + α ΔT)

115 cm = L ₁ (1 + (3.10 ^{-3 o} C ^-1 )(50 ^o C)

115 cm = L ₁ (1 + 150.10 ^-3 )

115 cm = L ₁ (1 + 0.15)

115 cm = L ₁ (1.15)

L 1 = 115 cm / 1.15 L 1 = 100 cm b) mgbanwe n'ogologo ( ΔL ) Δ L = L 2 – L 1 ΔL = 115 cm – 100 cm Δ L = 15 cm

[irp]

3. Na 25 ^o C, ogologo iko ahụ bụ 50 cm. Mgbe a kpochara ya ọkụ, ogologo iko ahụ bụ 50.9 cm. Ọnụọgụ nke mgbasawanye ahịrị bụ α = 9 x 10 ^-6 C ^-1 . Chọpụta okpomọkụ ikpeazụ nke iko ahụ…

A maara:

Ogologo mbụ (L ₁ ) = 50 cm

Ogologo ikpeazụ (L ₂ ) = 50.09 cm

Mgbanwe n'ogologo ( Δ L) = 50.2 cm – 50 cm = 0.09 cm

Ọnụọgụ nke mgbasawanye ahịrị (α) = 9 x 10 ^{-6 o} C ^-1

Okpomọkụ mbụ (T1 ₎ = 25 ^oC

Achọrọ: Okpomọkụ ikpeazụ ( _T2 )

Ngwọta:

ΔL = α L ₁ ΔT

Δ L = α L ₁ (T ₂ – T ₁ )

0.09 cm = ( 9 x 10 ^{-6 o} C) (50 cm )(T ₂ – 25 ^o C)

0.09 = (45 x 10 ^-5 )(T ₂ – 25)

0.09 / (45 x 10 ^-5 ) = T ₂ – 25

0.002 x 10 ⁵ = T ₂ – 25

2 x 10 ² = T ₂ – 25

200 = T ₂ – 25

T ₂ = 200 + 25

T ₂ = 225 ^o C

Okpomọkụ ikpeazụ bụ 225 ^o C.

[irp]

4. Ogologo mbụ nke ígwè ahụ bụ mita 1, ogologo ikpeazụ ya bụ mita 1.02. Mgbanwe okpomọkụ ahụ bụ Kelvin 50. Chọpụta ọnụọgụgụ nke mgbasawanye ahịrị!

A maara:

Ogologo mbụ (L ₁ ) = mita 1

Ogologo ikpeazụ (L ₂ ) = mita 1.02

Mgbanwe n'ogologo ( Δ L) = L ₂ – L ₁ = mita 1.02 – mita 1 = mita 0.02

Mgbanwe na okpomọkụ (Δ T ) = 50 Kelvin = 50 ^o C

Achọrọ: Ọnụọgụ nke mgbasawanye ahịrị

Ngwọta:

Δ L = α L ₁ ΔT

0.02 m = α (1 m ) (50 ^o C)

0.02 = α ( 50 ^o C)

α = 0.02 / 50 ^o C

α = 0.0004 ^o C ^-1

α = 4 x 10 ^{-4 o} C ^-1

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1. Ịgbanwe nha okpomọkụ
2. Mgbasawanye ahịrị
3. Mgbasawanye mpaghara
4. Mgbasawanye olu
5. okpomọkụ
6. Ihe nhata igwe nke okpomọkụ
7. Ike okpomọkụ na okpomọkụ kpọmkwem
8. Okpomọkụ zoro ezo, okpomọkụ nke njikọta, okpomọkụ nke vaporization
9. Nchekwa ike maka nnyefe okpomọkụ