He tauira o ngā pātai whakawhanui roa

5 Ngā tauira pātai mō te whakawhanui i te roa

1. He pae rino i te pāmahana o te 20^oE 40 cm te roa o C. Ko te tauwehenga whānui raina o te maitai he 10^{-5 o}C^-1. Ko te pikinga o te roa o te maitai me te roa whakamutunga o te maitai i te pāmahana 70^oKo C te……
Kōrero
E mōhiotia ana:
Te pikinga o te pāmahana (ΔT) = 70^oC - 20^oC=50^oC
Te roa tīmatanga (L)₁) = 40 henemita
Tauwehenga o te whakawhanuitanga rārangi o te maitai (a) = 10^{-5 o}C^-1
I pātaihia: Te pikinga o te roa (ΔL) me te roa whakamutunga (L₂)
Whakautu:
a) Te pikinga o te roa (ΔL)
ΔL = αL₁ ΔT
ΔL = (10^{-5 o}C^-1)(40 henimita)(50^oC)
ΔL = (10^-5)(2 x 10³) henimita
ΔL = 2 x 10^-2 cm
ΔL = 2 / 10² cm
ΔL = 2 / 100 cm
ΔL = 0,02 henimita

b) Te roa whakamutunga (L)₂)
L₂ =L₁ +ΔL
L₂ = 40 henimita + 0,02 henimita
L₂ = 40,02cm

2. Ka pāngia tētahi konganuku e te huringa pāmahana o te 30^oKa huri a C hei 80^oC. Mena ko te roanga whakamutunga o te konganuku he 115 cm, ā, ko te tauwehenga whānui rārangi o te konganuku he 3.10^{-3 o}C^-1, kātahi ko te roa tīmatanga o te konganuku me te pikinga o te roa o te konganuku ko…..
Kōrero
E mōhiotia ana:
Te pikinga o te pāmahana (ΔT) = 80^oC - 30^oC=50^oC
Te roa whakamutunga (L)₂) = 115 henemita
Ko te tauwehenga whānui rārangi o te konganuku (a) = 3.10^{-3 o}C^-1
I pātaihia: Te roa tīmatanga (L)₁) me te pikinga roa (ΔL)
Whakautu:
a) Te roa tīmatanga (L)₁)
Te tātai mō te whakanui i te roa:
ΔL = αL₁ ΔT
Tātai roa whakamutunga:
L₂ =L₁ +ΔL
L₂ =L₁ + α L₁ ΔT
L₂ =L₁ (1 + α ΔT)
115 henimita = R₁ (1 + (3.10^{-3 o}C^-1)(50^oC)
115 henimita = R₁ (1 + 150.10^-3)
115 henimita = R₁ (1 + 0,15)
115 henimita = R₁ (1,15)
L₁ = 115 henimita / 1,15
L₁ = 100cm

b) Te pikinga o te roa (ΔL)
ΔL = L₂ - L₁
ΔL = 115 henimita – 100 henimita
ΔL = 15 henimita

3. Ina 25 te pāmahana^oKo te roa o te karāhe he 50 cm. I muri i te whakamahana, ka eke te roa o te karāhe ki te 50,9 cm. Ko te tauwehenga whānui rārangi o te karāhe ko α = 9 x 10^-6 C^-1Ko te pāmahana whakamutunga o te karāhe ko….
Kōrero
E mōhiotia ana:
Te roa tīmatanga (L)₁) = 50 henemita
Te roa whakamutunga (L)₂) = 50,09 henemita
Te pikinga o te roa (ΔL) = 50,2 cm – 50 cm = 0,09 cm
Ko te tauwehenga whānui rārangi o te karāhe (α) = 9 x 10^{-6 o te ata}C^-1
Te pāmahana tīmatanga (T₁) = 25^oC
I pātaihia: Te pāmahana whakamutunga o te karāhe (T₂)
Whakautu:
ΔL = αL₁ ΔT
ΔL = αL₁ (T₂ - T₁)
0,09 henimita = (9 x 10^{-6 o}C)(50 henimita)(T₂ - 25 ^oC)
0,09 = (45 x 10^-5)(T₂ - 25)
0,09 / (45 x 10^-5) = T₂ - 25
0,002 x 10⁵ =T₂ - 25
2 x 10² =T₂ - 25
200 = T₂ - 25
T₂ = 200 + 25
T₂ = 225^oC
Ko te pāmahana whakamutunga o te karāhe he 225^oC.

4. Ka piki ake te roa o tētahi konganuku he 1 mita te roa i te tīmatanga ki te 1,02 m i muri i te huringa pāmahana o te 50 Kelvin. Tātaihia te tauwehenga whānui rārangi o te konganuku!
Kōrero
E mōhiotia ana:
Te roa tīmatanga (L)₁) = 1 mita
Te roa whakamutunga (L)₂) = 1,02 mita
Te huringa o te roa (ΔL) = L₂ - L₁ = 1,02 mita – 1 mita = 0,02 mita
Huringa pāmahana (ΔT) = 50 Kelvin = 50^oC
I pātaihia: Tauwehenga o te whakawhānui raina o te konganuku
Whakautu:
ΔL = αL₁ ΔT
0,02 mita = α (1 mita)(50^oC)
0,02 = α (50^oC)
α = 0,02 / 50^oC
α = 0,0004 ^oC^-1
α = 4 x 10^{-4 o te ata}C^-1

5. Ko te roa o ia ara rerewē he 8 mita, kua tāutahia ki te pāmahana o te 30 ^oC. Ko te tauwehenga whānui o te rēriwe he 12 × 10^-6 /C^o. Mēnā kei te pāmahana o te 60 ^oC ka pā ngā reera e rua tetahi ki tetahi, kātahi ka roa te āputa i waenganui i ngā reera e rua i te pāmahana o te 30 ^oKo C…
A. 5,76 mm
B. 3,24 mm
C. 1,20 mm
D. 0,8 mm
E. 0,6 mm
Kōrero:

E mōhiotia ana :
Te roa tīmatanga (L)_o) = 8 mita
Te pāmahana tīmatanga (T) = 30 ^oC
Te pāmahana whakamutunga (T) = 60 ^oC
Tauwehenga o te whakawhānui raina = 12 x 10^-6  /C^o
I pātaihia :
Te roa o te āputa i waenganui i ngā rēri e rua
Whakautu :
Tātaihia te pikinga o te roa o ia tokotoko reriwe:
Ka piki ake te roa o ia pae reera mā te 2,88 mm, nō reira ko te roa o te āputa i waenganui i ngā pae reera e rua = 2 x 2,88 mm = 5,76 mm
Ko te whakautu tika ko A.