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:
The ability of water molecules to form hydrogen bonds with other water molecules and water's ability to dissolve substances that have charges or partial charges are <u>due to water's partial charges.</u>
Explanation:
The partial negative charge on oxygen and partial positive charge on hydrogen enables them to make hydrogen bond and also makes it to dissolve the the other substances having partial charges.
Answer:
Explanation:
All the displacement will be converted into vector, considering east as x axis and north as y axis.
5.3 km north
D = 5.3 j
8.3 km at 50 degree north of east
D₁= 8.3 cos 50 i + 8.3 sin 50 j.
= 5.33 i + 6.36 j
Let D₂ be the displacement which when added to D₁ gives the required displacement D
D₁ + D₂ = D
5.33 i + 6.36 j + D₂ = 5.3 j
D₂ = 5.3 j - 5.33i - 6.36j
= - 5.33i - 1.06 j
magnitude of D₂
D₂²= 5.33² + 1.06²
D₂ = 5.43 km
Angle θ
Tanθ = 1.06 / 5.33
= 0.1988
θ =11.25 ° south of due west.
Answer:
Explanation:
<u>LC Circuit</u>
It's a special circuit made of three basic elements: The AC source, a capacitor, and an inductor. The charge, current, and voltage are oscillating when there is an interaction between the electric and magnetic fields of the elements. The following variables will be used for the formulas:
= charge of the capacitor in any time 
= initial charge of the capacitor
=angular frequency of the circuit
= current through the circuit in any time
The charge in an LC circuit is given by
The current is the derivative of the charge
We are given
It means that
![q(t_1) = q_0 \, cos (\omega t_1 )=q_1\ .......[eq 1]](https://tex.z-dn.net/?f=q%28t_1%29%20%3D%20q_0%20%5C%2C%20cos%20%28%5Comega%20t_1%20%29%3Dq_1%5C%20.......%5Beq%201%5D)
![i(t_1) = - \omega q_0 \, sin(\omega t_1)=i_1.........[eq 2]](https://tex.z-dn.net/?f=i%28t_1%29%20%3D%20-%20%5Comega%20q_0%20%5C%2C%20sin%28%5Comega%20t_1%29%3Di_1.........%5Beq%202%5D)
From eq 1:
From eq 2:
Squaring and adding the last two equations, and knowing that
Operating
Solving for
Now we know the value of 
, we repeat the procedure of eq 1 and eq 2, but now at the second time 
, and solve for 
Solving for
Now we replace the given values. We'll assume that the placeholder is a pi for the frequency, i.e.
Finally
→ 
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