Answer:
T = 188.5 s, correct is C
Explanation:
This problem must be worked on using conservation of angular momentum. We define the system as formed by the fan and the paper, as the system is isolated, the moment is conserved
initial instant. Before the crash
L₀ = r m v₀ + I₀ w₀
the angular speed of the fan is zero w₀ = 0
final instant. After the crash
L_f = I₀ w + m r v
L₀ = L_f
m r v₀ = I₀ w + m r v
angular and linear velocity are related
v = r w
w = v / r
m r v₀ = I₀ v / r + m r v
m r v₀ = (I₀ / r + mr) v
v = 
let's calculate
v = 
v = 
v = 0.02 m / s
To calculate the time of a complete revolution we can use the kinematics relations of uniform motion
v = x / T
T = x / v
the distance of a circle with radius r = 0.6 m
x = 2π r
we substitute
T = 2π r / v
let's calculate
T = 2π 0.6/0.02
T = 188.5 s
reduce
t = 188.5 s ( 1 min/60 s) = 3.13 min
correct is C
The capacitance of the capacitor is 
Explanation:
The capacitance of a parallel-plate capacitor is given by the equation
where
k is the dielectric constant of the medium
is the vacuum permittivity
A is the area of the plates
d is the separation between the plates
For the capacitor in this problem, we have:
k = 2.1 is the dielectric constant
is the separation between the plates
(I assumed that 5.4 m is a typo, since it is not a realistic size for the side of the plate)
Therefore, the capacitance of the capacitor is
Learn more about capacitors:
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Explanation:
Using Ohm's Law and a bit of substitution, we can use voltage divided by current to solve for resistance. Doing that, we'll get 6 Ohm.
A decagram is 1000 times bigger than a centigram
Answer:
the coefficient of volume expansion of the glass is 
Explanation:
Given that:
Initial volume of the glass flask = 1000 cm³ = 10⁻³ m³
temperature of the glass flask and mercury= 1.00° C
After heat is applied ; the final temperature = 52.00° C
Temperature change ΔT = 52.00° C - 1.00° C = 51.00° C
Volume of the mercury overflow = 8.50 cm^3 = 8.50 × 10⁻⁶ m³
the coefficient of volume expansion of mercury is 1.80 × 10⁻⁴ / K
The increase in the volume of the mercury = 10⁻³ m³ × 51.00 × 1.80 × 10⁻⁴
The increase in the volume of the mercury = 
Increase in volume of the glass = 10⁻³ × 51.00 × 
Now; the mercury overflow = Increase in volume of the mercury - increase in the volume of the flask
the mercury overflow = 
Thus; the coefficient of volume expansion of the glass is
