Answer:
1. a) 0.5 h = 1,800 s
h) 20 cm = 0.2 m
b) 2.0 h = 7,200 s
i) 5.0 kg = 5,000 g
c) 3.5 h = 12,600 s
j) 1.5 kg = 1,500 g
d) 1/4 h = 900 s
k) 450.0 g = 0.45 kg
e) 3.0 m = 300 cm
l) 20.0 g = 0.02 kg
f) 2.5 m = 250 cm
m) 500.0 g = 0.5 kg
g) 0.5 m = 500 mm
n) 1000.0 g = 1 kg
Answer:
The induced emf in the short coil during this time is 1.728 x 10⁻⁴ V
Explanation:
The magnetic field at the center of the solenoid is given by;
B = μ(N/L)I
Where;
μ is permeability of free space
N is the number of turn
L is the length of the solenoid
I is the current in the solenoid
The rate of change of the field is given by;
The induced emf in the shorter coil is calculated as;
where;
N is the number of turns in the shorter coil
A is the area of the shorter coil
Area of the shorter coil = πr²
The radius of the coil = 2.5cm / 2 = 1.25 cm = 0.0125 m
Area of the shorter coil = πr² = π(0.0125)² = 0.000491 m²
E = 14 x 0.000491 x 0.02514
E = 1.728 x 10⁻⁴ V
Therefore, the induced emf in the short coil during this time is 1.728 x 10⁻⁴ V
A 3rd harmonic of a tube open at both ends will have displacement antinodes at both ends.
In a tube of length L with two open ends, the longest standing wave has displacement antinodes (pressure nodes) at both ends. The fundamental or first harmonic is what it is known as. The second harmonic is the longest standing wave in a tube of length L with two open ends.
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