Correct question is;
A thermal tap used in a certain apparatus consists of a silica rod which fits tightly inside an aluminium tube whose internal diameter is 8mm at 0°C.When the temperature is raised ,the fits is no longer exact. Calculate what change in temperature is necessary to produce a channel whose cross-sectional is equal to that of the tube of 1mm. (linear expansivity of silica = 8 × 10^(-6) /K and linear expansivity of aluminium = 26 × 10^(-6) /K).
Answer:
ΔT = 268.67K
Explanation:
We are given;
d1 = 8mm
d2 = 1mm
At standard temperature and pressure conditions, the temperature is 273K.
Thus; Initial temperature; T1 = 273K,
Using the combined gas law, we have;
P1×V1/T1 = P2×V2/T2
The pressure is constant and so P1 = P2. They will cancel out in the combined gas law to give:
V1/T1 = V2/T2
Now, volume of the tube is given by the formula;V = Area × height = Ah
Thus;
V1 = (πd1²/4)h
V2 = (π(d2)²/4)h
Thus;
(πd1²/4)h/T1 = (π(d2)²/4)h/T2
π, h and 4 will cancel out to give;
d1²/T1 = (d2)²/T2
T2 = ((d2)² × T1)/d1²
T2 = (1² × T1)/8²
T2 = 273/64
T2 = 4.23K
Therefore, Change in temperature is; ΔT = T2 - T1
ΔT = 273 - 4.23
ΔT = 268.67K
Thus, the temperature decreased to 268.67K
Answer:
1- b: 2- a : 3- c : 4- d
Explanation:
it starts 2 move away from strting point, then no motion, then moves toward the start, the slows up.
Gravitational potential energy, relative to some level =
(mass of the object)
times
(height above the reference level)
times
(acceleration due to gravity) .
Answer:
Explanation:
Since work done is in the form of potential energy, we will use the formula of potential energy here.
We know that,
<h3>P.E. = mgh </h3>
Where,
m = mass = 20 kg
g = acceleration due to gravity = 10 m/s²
h = vertical height = 20 m
So,
<h3>Work done = mgh</h3>
Work done = (20)(10)(20)
Work done = 4000 joules
Work done = 4 kJ
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Answer:
I think the 1st statement is right.
Explanation:
Wind patterns doesn't stay the same.
Waves don't follow the same patterns.
Waves move further up the shore.
I didn't hear about "waves adding" before..so i guess 1st statement is right.
