Answer:
A) a = 73.304 rad/s²
B) Δθ = 3665.2 rad
Explanation:
A) From Newton's first equation of motion, we can say that;
a = (ω - ω_o)/t. We are given that the centrifuge spins at a maximum rate of 7000rpm.
Let's convert to rad/s = 7000 × 2π/60 = 733.04 rad/s
Thus change in angular velocity = (ω - ω_o) = 733.04 - 0 = 733.04 rad/s
We are given; t = 10 s
Thus;
a = 733.04/10
a = 73.304 rad/s²
B) From Newton's third equation of motion, we can say that;
ω² = ω_o² + 2aΔθ
Where Δθ is angular displacement
Making Δθ the subject;
Δθ = (ω² - ω_o²)/2a
At this point, ω = 0 rad/s while ω_o = 733.04 rad/s
Thus;
Δθ = (0² - 733.04²)/(2 × 73.304)
Δθ = -537347.6416/146.608
Δθ = - 3665.2 rad
We will take the absolute value.
Thus, Δθ = 3665.2 rad
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
To what question please put the question ill be glad to help
The correct answer to this is (A. Units Only).
It shows that there is a velocity of 35, but the units are missing.
Answer:
Rate of heat is 45.1 kJ/min
Explanation:
Heat required to evaporate the water is given by
Q = mL
here we know that
now we have
now the power is defined as rate of energy
