Answer:
92.33Hz
Explanation:
A tuba is considered as having one open end and the other end is closed. Then:
f_n = n×v / 4L
so here
Where n = 4
f_4 = 116.5 Hz
116.5 Hz = nv/4L
116.5 Hz = 4v/4L
4×v / 4L = v / L
116.5 Hz = v/L............1
Let's assume the speed of sound is 343 m/s. Then substituting 343 m/s for v in equation 1
116.5 Hz = v/L
116.5 Hz = 343 m/s/L
Making L the subject
116.5 Hz × L = 343 m/s
L = 343m/s / 116.5Hz
= 2.944 m
Add 0.721 m to the length, and
f_4 = 343m/s / (2.944+0.721)m
f_4 = 343m/s ÷ 3.715
= 92.328 Hz
Approximately = 92.33Hz
Hence the new frequency for the 4th harmonic is 92.33Hz
Answer:
680.6 N
Explanation:
The net force here is the resultant force. Taking that the two given forces act perpendicular to each other then the resultant is the hypotenuse. Therefore, where d is the magnitude of downward force and s is the magnitude of the sideways force. Substituting 631 N for downward force and 255 N for sideways force then
Therefore, the magnitude of net force is equivalent to 680. 6 N
The amplitude of a wave corresponds to its maximum oscillation of the wave itself.
In our problem, the equation of the wave is
We can see that the maximum value of y(x,t) is reached when the cosine is equal to 1. When this condition occurs,
and therefore this value corresponds to the amplitude of the wave.
Resistance of our body is given as
voltage applied across the body is
now by ohm's law current pass through our body is given by
So current from our body will be 4 * 10^-3 A