David wants to experiment with the device, so he connects an ammeter into the circuit and measures 11.5 AA when the device is co
1 answer:
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Answer:
(a) the fundamental frequency of this string is 65 Hz
(b) the harmonics of the given frequencies are third and fourth respectively.
(c) the length of the string is 2.74 m
Explanation:
Given;
mass density of the string, μ = 3 x 10⁻³ kg/m
tension of the string, T = 380 N
resonating frequencies, 195 Hz and 260 N
For the given resonant frequencies;
(c) From any of the equations, solve for Length of the string (L);
(a) the fundamental frequency is calculated as;
(b) harmonics of the given frequencies;
the first harmonic (n = 1) = f₀ = 65 Hz
the second harmonic (n = 2) = 2f₀ = 130 Hz
the third harmonic (n = 3) = 3f₀ = 195 Hz
the fourth harmonic (n = 4) = 4f₀ = 260 Hz
Thus, the harmonics of the given frequencies are third and fourth respectively.
Answer:
y=-1.66 m
Explanation:
We know that
Range R
Here given that
u= 275 m/s
R=75 m
Now by putting the values
θ=0.13°
Now horizontal component of velocity u will be u cosθ.
Horizontal component = 275 cos0.13 = 274.99 m/s
So the time required to cover 180 m in horizontal direction t
180 = 274.99 x t
t=0.65 sec
Now vertical component of velocity u will be u sinθ.
Horizontal component = 275 sin0.13 = 0.62 m/s
So now vertical displacement y will be
y=-1.66 m