Answer:
The distance is 1.026 m.
Explanation:
mass of rod, M = 1.23 kg
Length, L = 1.25 m
mass, m = 10 kg
Time period, T = 2 s
Let the distance is d.
The formula of the time period is given by
KNOWN: Long, 30mm-diameter cylinder with embedded electrical heater; power required
to maintain a specified surface temperature for water and air flows.
FIND: Convection coefficients for the water and air flow convection processes, hw and ha,
respectively.
ASSUMPTIONS: Flow is cross-wise over cylinder which is very long in the direction
normal to flow.
The convection heat rate from the cylinder per unit length of the cylinder has
the form
q' = h*(pi*D)*(Ts-Tinf)
and solving for the heat transfer convection coefficient, find
Water
hw = q'/((pi*D)*(Ts-Tinf))
hw = (38*10^3 W/m) / ((pi*(0.030m))*(80-25)C)=
7330.77314 W/m^2K
Air
ha = (400W/m) / ((pi*(0.030m))*(80-25)C)=<span>
77.166033 </span> W/m^2K
COMMENTS: Note that the air velocity is 10 times that of the water flow, yet
hw ≈ 95 × ha.
These values for the convection coefficient are typical for forced convection heat transfer with
liquids and gases
Watter is a better convective heat transfer media than air
Answer: a)1.27 *10^6 A/m^2; b) 57.9*10^6 Ω*m; c) 94.16*10^-6 m/s;
d) 4.29*10^-3 m^2/V*s
Explanation: In order to explain this problem we have to take into account the following expressions:
The current density is equal:
J=I/A where A is the area of the wire.
J=4A/(π*1*10^-6m^2)=1.27 *10^6 A/m^2
The resistence of the wire is given by:
R=L/(σ*A) where σ and L are the conductivity and the length of the wire, respectively.
Then we have:
σ= L/(R*A)=1/(5.5*10^-3*π*1*10^-6)=57.9 *10^6 Ω*m
The drift velocity of free electrons is given by:
Vd=i/(n*A*e) where n is the numer of electrons per volume. e is the electron charge.
then we have:
Vd=J/(n*e)= 1.27 *10^6/(8.43*10^28*1.6*10^-19)=94.16 *10^-6 m/s
Finally, the mobility of free electrons is equal to:
μ=Vd/E ; E=J/σ
μ=E*σ/J= 94.16*10^-6*57.9 *10^6/1.27 *10^6=4.29*10^-3 m^2/V*s
Answer:
I = 16 kg*m²
Explanation:
Newton's second law for rotation
τ = I * α Formula (1)
where:
τ : It is the moment applied to the body. (Nxm)
I : it is the moment of inertia of the body with respect to the axis of rotation (kg*m²)
α : It is angular acceleration. (rad/s²)
Kinematics of the wheel
Equation of circular motion uniformly accelerated :
ωf = ω₀+ α*t Formula (2)
Where:
α : Angular acceleration (rad/s²)
ω₀ : Initial angular speed ( rad/s)
ωf : Final angular speed ( rad
t : time interval (rad)
Data
ω₀ = 0
ωf = 1.2 rad/s
t = 2 s
Angular acceleration of the wheel
We replace data in the formula (2):
ωf = ω₀+ α*t
1.2= 0+ α*(2)
α*(2) = 1.2
α = 1.2 / 2
α = 0.6 rad/s²
Magnitude of the net torque (τ )
τ = F *R
Where:
F = tangential force (N)
R = radio (m)
τ = 80 N *0.12 m
τ = 9.6 N *m
Rotational inertia of the wheel
We replace data in the formula (1):
τ = I * α
9.6 = I *(0.6
)
I = 9.6 / (0.6
)
I = 16 kg*m²
Answer:
Check the explanation
Explanation:
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