The main difference between the drift current and diffusion current is the corresponding reaction of the semiconductor materials in the electric fields. In electric fields, when semiconductor material's movement go along with the direction of the electric field it is called drift current, but when it tends to go from high concentrated area to lower concentrated area then it is diffusion current.
Answer:
Explanation:
It is given that the sphere is insulated from ground and a large charge is placed on the sphere. The charge on the hollow sphere will always remain on the outer surface of the sphere and there will be no charge on the inner surface of the sphere.
If a person touches the inner surface of the sphere then he will not be harmed as there is no charge on the inner surface of the sphere.
If a person carries the charge of the opposite sign of the same magnitude then the sphere and person get neutralized upon touching the sphere.
If a person does not touches the sphere then the charge on the outer surface will be zero and there will be a positive charge on the inner surface of the sphere
The impulse required to decrease the speed of the boat is equal to the variation of momentum of the boat:
where
m=225 kg is the mass of the boat
is the variation of velocity of the boat
By substituting the numbers into the first equation, we find the impulse:
and the negative sign means the direction of the impulse is against the direction of motion of the boat.
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
