A vacuum tube can be used to__. A. change alternating current into direct current B. increase the strength of a signal.. C. turn
2 answers:
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Answer:
option D
Explanation:
given,
A conductor is carrying current = 2.0 A is 0.5 mm thick
Hall voltage = 4.5 x 10-6 V
uniform magnetic field = 1.2 T
density of the charge = n =?
hall voltage =
n = 6.67 × 10²⁷ charges/m
hence the correct answer is option D
Answer:
The beat frequency when each string is vibrating at its fundamental frequency is 12.6 Hz
Explanation:
Given;
velocity of wave on the string with lower tension, v₁ = 35.2 m/s
the fundamental frequency of the string, F₁ = 258 Hz
<u>velocity of wave on the string with greater tension;</u>
where;
v₁ is the velocity of wave on the string with lower tension
T₁ is tension on the string
μ is mass per unit length
Where;
T₁ lower tension
T₂ greater tension
v₁ velocity of wave in string with lower tension
v₂ velocity of wave in string with greater tension
From the given question;
T₂ = 1.1 T₁
<u>Fundamental frequency of wave on the string with greater tension;</u>
<u /><u />
Beat frequency = F₂ - F₁
= 270.6 - 258
= 12.6 Hz
Therefore, the beat frequency when each string is vibrating at its fundamental frequency is 12.6 Hz