A clock while keeping correct time at 30°C has a pendulum rod made of brass. The number of seconds it gains or loses per second when the temperature falls to 10°C is
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A clock while keeping correct time at 30°C has a pendulum rod made of brass. The number of seconds it gains or loses per second when the temperature falls to 10°C is
NEED IT URGENT!
SPAM = REPORT !!
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The rate at which a clock gains or loses time due to temperature changes is determined by the coefficient of thermal expansion of the material used for the pendulum rod. Brass has a coefficient of thermal expansion, which means it expands or contracts with temperature changes.
To calculate the change in timekeeping due to temperature, you can use the formula:
[tex]{ \red{ \boxed{ \tt \: Δt = L \times α \times ΔT}}}[/tex]
Where:
Let's assume:
Now, plug these values into the formula:
[tex] \sf \: Δt = 1 × (19 \times 10^{-6}) × 20[/tex]
[tex]{ \boxed{ \bold{Δt ≈ 0.00038 \: seconds \: per \: second}}}[/tex]
So, the clock will gain or lose approximately 0.00038 seconds per second when the temperature falls from 30°C to 10°C.
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[tex]{ \orange{ \sf \: hope \: it \: helps \: you}}[/tex]
Answer:
We know, α=dl/(dt×l)
or, dl=α×dt×l -----(1)
Now T=2π
l/g
--- (2)
or, T+dT=2π
(l+dl)/g
or, T(1+dT/T)=2π
l/g
1+dl/l
----- (3)
Using (2)
1+dT/T=(l+dl/l)
1/2
1+dT/T=1+(1/2)dl/l
dT/T=(1/2)(dl/l)
dT=(1/2)(dl/l)T
Using (1)
dT=(1/2)[(αdtl)/l]XT
=(1/2)×18×10
−6
××1
=18×10
−5
Explanation:
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