-- Put the rod into the freezer for a while. As it cools,
it contracts (gets smaller) slightly.
-- Put the cylinder into hot hot water for a while. As it heats,
it expands (gets bigger) slightly.
-- Bring the rod and the cylinder togther quickly, before the
rod has a chance to warm up or the cylinder has a chance
to cool off.
-- I bet it'll fit now.
-- But be careful . . . get the rod exactly where you want it as fast
as you can. Once both pieces come back to the same temperature,
and the rod expands a little and the cylinder contracts a little, the fit
will be so tight that you'll probably never get them apart again, or even
move the rod.
Answer:
The length of rod A will be <u>greater than </u>the length of rod B
Explanation:
We, know that the formula for final length in linear thermal expansion of a rod is:
L' = L(1 + ∝ΔT)
where,
L' = Final Length
L = Initial Length
∝ = Co-efficient of linear expansion
ΔT = Change in temperature
Since, the rods here have same original length and the temperature difference is same as well. Therefore, the final length will only depend upon the coefficient of linear expansion.
For Rod A:
∝₁ = 12 x 10⁻⁶ °C⁻¹
For Rod B:
∝₂ = β₂/3
where,
β₂ = Coefficient of volumetric expansion for rod B = 24 x 10⁻⁶ °C⁻¹
Therefore,
∝₂ = 24 x 10⁻⁶ °C⁻¹/3
∝₂ = 8 x 10⁻⁶ °C⁻¹
Since,
∝₁ > ∝₂
Therefore,
L₁ > L₂
So, the length of rod A will be <u>greater than </u>the length of rod B
Answer:
Work= -7.68×10⁻¹⁴J
Explanation:
Given data
q₁=q₂=1.6×10⁻¹⁹C
r₁=2.00×10⁻¹⁰m
r₂=3.00×10⁻¹⁵m
To find
Work
Solution
The work done on the charge is equal to difference in potential energy
W=ΔU
If only I was smart then I could help you :/ no but like for real im madddd dumb sorry :(