Answer:
w = √[g /L (½ r²/L2 + 2/3 ) ]
When the mass of the cylinder changes if its external dimensions do not change the angular velocity DOES NOT CHANGE
Explanation:
We can simulate this system as a physical pendulum, which is a pendulum with a distributed mass, in this case the angular velocity is
w² = mg d / I
In this case, the distance d to the pivot point of half the length (L) of the cylinder, which we consider long and narrow
d = L / 2
The moment of inertia of a cylinder with respect to an axis at the end we can use the parallel axes theorem, it is approximately equal to that of a long bar plus the moment of inertia of the center of mass of the cylinder, this is tabulated
I = ¼ m r2 + ⅓ m L2
I = m (¼ r2 + ⅓ L2)
now let's use the concept of density to calculate the mass of the system
ρ = m / V
m = ρ V
the volume of a cylinder is
V = π r² L
m = ρ π r² L
let's substitute
w² = m g (L / 2) / m (¼ r² + ⅓ L²)
w² = g L / (½ r² + 2/3 L²)
L >> r
w = √[g /L (½ r²/L2 + 2/3 ) ]
When the mass of the cylinder changes if its external dimensions do not change the angular velocity DOES NOT CHANGE
Answer:
9.25 x 10^-4 Nm
Explanation:
number of turns, N = 8
major axis = 40 cm
semi major axis, a = 20 cm = 0.2 m
minor axis = 30 cm
semi minor axis, b = 15 cm = 0.15 m
current, i = 6.2 A
Magnetic field, B = 1.98 x 10^-4 T
Angle between the normal and the magnetic field is 90°.
Torque is given by
τ = N i A B SinФ
Where, A be the area of the coil.
Area of ellipse, A = π ab = 3.14 x 0.20 x 0.15 = 0.0942 m²
τ = 8 x 6.20 x 0.0942 x 1.98 x 10^-4 x Sin 90°
τ = 9.25 x 10^-4 Nm
thus, the torque is 9.25 x 10^-4 Nm.
<h2>
Average speed of transit train is 60 mph</h2>
Explanation:
Average speed of passenger train = 45 mph
Time taken from station A to station B for passenger train = 10:00 - 6:00 = 4 hours
Distance between station A to station B = 45 x 4 = 180 miles.
Time taken from station A to station B for transit train = 4 - 1 = 3 hours
Distance between station A to station B = Average speed of transit train x Time taken from station A to station B for transit train
180 = Average speed of transit train x 3
Average speed of transit train = 60 mph
Average speed of transit train is 60 mph
Normally a storm surge.
Experience: I lived through Andrew and Wilma
Answer:
a) the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s
b) the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s
Explanation:
Given the data in the question;
as the equation of standing wave on a string is fixed at both ends
y = 2AsinKx cosωt
but k = 2π/λ and ω = 2πf
λ = 4 × 0.150 = 0.6 m
and f = v/λ = 260 / 0.6 = 433.33 Hz
ω = 2πf = 2π × 433.33 = 2722.69
given that A = 2.20 mm = 2.2×10⁻³
so 
= A × ω
= 2.2×10⁻³ × 2722.69 m/s
= 5.9899 m/s
therefore, the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s
b)
A' = 2AsinKx
= 2.20sin( 2π/0.6 ( 0.075) rad )
= 2.20 sin( 0.7853 rad ) mm
= 2.20 × 0.706825 mm
A' = 1.555 mm = 1.555×10⁻³
so
= A' × ω
= 1.555×10⁻³ × 2722.69
= 4.2338 m/s
Therefore, the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s
