The concepts necessary to solve this problem are framed in the expression of string vibration frequency as well as the expression of the number of beats per second conditioned at two frequencies.
Mathematically, the frequency of the vibration of a string can be expressed as
Where,
L = Vibrating length string
T = Tension in the string
Linear mass density
At the same time we have the expression for the number of beats described as
Where
= First frequency
= Second frequency
From the previously given data we can directly observe that the frequency is directly proportional to the root of the mechanical Tension:
If we analyze carefully we can realize that when there is an increase in the frequency ratio on the tight string it increases. Therefore, the beats will be constituted under two waves; one from the first string and the second as a residue of the tight wave, as well
Replacing 
for n and 202Hz for 
The frequency of the tightened is 205Hz
Answer:
0.54 A
Explanation:
Parameters given:
Number of turns, N = 15
Area of coil, A = 40 cm² = 0.004 m²
Change in magnetic field, ΔB = 5.1 - 1.5 = 3.6 T
Time interval, Δt = 2 secs
Resistance of the coil, R = 0.2 ohms
To get the magnitude of the current, we have to first find the magnitude of the EMF induced in the coil:
|V| = |(-N * ΔB * A) /Δt)
|V| = | (-15 * 3.6 * 0.004) / 2 |
|V| = 0.108 V
According to Ohm's law:
|V| = |I| * R
|I| = |V| / R
|I| = 0.108 / 0.2
|I| = 0.54 A
The magnitude of the current in the coil of wire is 0.54 A
Answer:
4.96 × 10⁵ Pa
Explanation:
F = mg
This force is evenly distributed on the three leg
radius, r = d/2
= 2.8 / 2
= 1.4 cm = 0.014 m
total cross sectional area of the three legs, A = 3*pi*r^2
Pressure due to weight,
P = Weight/A
P = 4.96 × 10⁵ Pa
The time spent in the air by the ball at the given momentum is 6.43 s.
The given parameters;
- <em>momentum of the ball, P = 0.9 kgm/s</em>
- <em>weight of the ball, W = 0.14 N</em>
The impulse experienced by the ball is calculated as follows;
where;
is impulse
is change in momentum
The time of motion of the ball is calculated as follows;
Thus, the time spent in the air by the ball at the given momentum is 6.43 s.
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