The bob (weight) at the end of a pendulum has a mass of 0.3 kilograms. The bob is pulled to position B and allowed to swing. It
2 answers:
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Answer:
f = 931.1 Hz
Explanation:
Given,
Mass of the wire, m = 0.325 g
Length of the stretch, L = 57.7 cm = 0.577 m
Tension in the wire, T = 650 N
Frequency for the first harmonic = ?
we know,
μ is the mass per unit length
μ = 0.325 x 10⁻³/ 0.577
μ = 0.563 x 10⁻³ Kg/m
now,
v = 1074.49 m/s
The wire is fixed at both ends. Nodes occur at fixed ends.
For First harmonic when there is a node at each end and the longest possible wavelength will have condition
λ=2 L
λ=2 x 0.577 = 1.154 m
we now,
v = f λ
f = 931.1 Hz
The frequency for first harmonic is equal to f = 931.1 Hz
Answer:
(a)
(b)
(c)
Explanation:
Given that,
(a) An electron accelerated from rest through a potential difference of 100 V. The De Broglie wavelength in terms of potential difference is given by :
Where
m and e are the mass of and charge on an electron
On solving,
V = 100 V
(b) V = 1 kV = 1000 V
(c) If
Hence, this is the required solution.