The equilibrium temperature of aluminium and water is 33.2°C
We know that specific heat of aluminium is 0.9 J/gm-K, and that of water is 1 J/gm-K
Now we can calculate the equilibrium temperature
(mc∆T)_aluminium=(mc∆T)_water
15.7*0.9*(53.2-T)=32.5*1*(T-24.5)
T=33.2°C
Answer:
19.53 cm
Explanation:
The computation of the height is as follows:
Here we applied the conservation of the energy formula
As we know that
P.E of the block = P.E of the spring
m g h = ( 1 ÷ 2) k x^2
where
m = 0.15
g = 9.81
k = 420
x = 0.037
So now put the values to the above formula
(0.15) (9.81) (h) = 1 ÷2 × 420 × (0.037)^2
1.4715 (h) = 0.28749
h = 0.19537 m
= 19.53 cm
the answer should be:
When the buoyant force is equal to the force of gravity
Answer:
a) 4.9*10^-6
b) 5.71*10^-15
Explanation:
Given
current, I = 3.8*10^-10A
Diameter, D = 2.5mm
n = 8.49*10^28
The equation for current density and speed drift is
J = I/A = (ne) Vd
A = πD²/4
A = π*0.0025²/4
A = π*6.25*10^-6/4
A = 4.9*10^-6
Now,
J = I/A
J = 3.8*10^-10/4.9*10^-6
J = 7.76*10^-5
Electron drift speed is
J = (ne) Vd
Vd = J/(ne)
Vd = 7.76*10^-5/(8.49*10^28)*(1.60*10^-19)
Vd = 7.76*10^-5/1.3584*10^10
Vd = 5.71*10^-15
Therefore, the current density and speed drift are 4.9*10^-6
And 5.71*10^-15 respectively
