Answer:
Circular tube
Explanation:
Now for better understanding lets take an example
Lets take
Diameter of solid bar= 
cm
Outer diameter of tube =6 cm
Inner diameter of tube=2 cm
So from we can say that both tubes have equal cross sectional area.
We know that buckling load is given as 
If area moment of inertia(I) is high then buckling load will be high.
We know that area moment of inertia(I)
For circular tube 
For circular bar 
Now by putting the values
For circular tube 
For circular bar 
So we can say that for same cross sectional area the area moment of inertia(I) is high for tube as compare to bar.So buckling load will be higher in tube as compare to bar.
Answer:
The energy, that is dissipated in the resistor during this time interval is 153.6 mJ
Explanation:
Given;
number of turns, N = 179
radius of the circular coil, r = 3.95 cm = 0.0395 m
resistance, R = 10.1 Ω
time, t = 0.163 s
magnetic field strength, B = 0.573 T
Induced emf is given as;
where;
ΔФ is change in magnetic flux
ΔФ = BA = B x πr²
ΔФ = 0.573 x π(0.0395)² = 0.002809 T.m²
According to ohm's law;
V = IR
I = V / R
I = 3.0848 / 10.1
I = 0.3054 A
Energy = I²Rt
Energy = (0.3054)² x 10.1 x 0.163
Energy = 0.1536 J
Energy = 153.6 mJ
Therefore, the energy, that is dissipated in the resistor during this time interval is 153.6 mJ
Answer:
a) C= 1/120
b) P(X>=5) = 0.333
Explanation:
The attached file contains the explanation for the answers
Answer:
250.7mw
Explanation:
Volume of the reservoir = lwh
Length of reservoir = 10km
Width of reservoir = 1km
Height = 100m
Volume = 10x10³x10³x100
= 10⁹m³
Next we find the volume flow rate
= 0.1/100x10⁹x1/3600
= 277.78m³/s
To get the electrical power output developed by the turbine with 92 percent efficiency
= 0.92x1000x9.81x277.78x100
= 250.7MW
