No. I do not agree with Stefan. Quite the contrary. I disagree
with his description of "<span>angle of incidence" as the angle between
the surface of the mirror and the incoming ray.
The correct description of "angle of incidence" is </span><span>the angle between
the NORMAL TO the surface of the mirror and the incoming ray.
Thus, the true angle of incidence is the complement of the angle that
Stefan calculates or measures.</span>
Answer:
a. 600 ml, 12
Explanation:
The movement described in the question exhibits that of a polygon. Exhibiting a constant distance and angle with only varying direction until the starting point is reached.
The sum of exterior angles of a polygon = 360 degrees.
Exterior angle of a polygon = (360 ÷ number of sides)
Therefore,
Number of sides = 360 ÷ exterior angle
Exterior angle = 30 degrees
Hence,
Number of sides = 360 ÷ 30 = 12 sides
Since distance traveled of 50 miles is the same for each displacement ;
Total displacement = distance traveled * number of sides
Total displacement = 50 * 12 = 600 miles.
Answer:
The second distance of the sound from the source is 431.78 m..
Explanation:
Given;
first distance of the sound from the source, r₁ = 1.48 m
first sound intensity level, I₁ = 120 dB
second sound intensity level, I₂ = 70.7 dB
second distance of the sound from the source, r₂ = ?
The intensity of sound in W/m² is given as;
![dB = 10 Log[\frac{I}{I_o} ]\\\\For \ 120 dB\\\\120 = 10Log[\frac{I}{1*10^{-12}}]\\\\12 = Log[\frac{I}{1*10^{-12}}]\\\\10^{12} = \frac{I}{1*10^{-12}}\\\\I = 10^{12} \ \times \ 10^{-12}\\\\I = 1 \ W/m^2](https://tex.z-dn.net/?f=dB%20%3D%2010%20Log%5B%5Cfrac%7BI%7D%7BI_o%7D%20%5D%5C%5C%5C%5CFor%20%5C%20120%20dB%5C%5C%5C%5C120%20%3D%2010Log%5B%5Cfrac%7BI%7D%7B1%2A10%5E%7B-12%7D%7D%5D%5C%5C%5C%5C12%20%3D%20%20Log%5B%5Cfrac%7BI%7D%7B1%2A10%5E%7B-12%7D%7D%5D%5C%5C%5C%5C10%5E%7B12%7D%20%3D%20%5Cfrac%7BI%7D%7B1%2A10%5E%7B-12%7D%7D%5C%5C%5C%5CI%20%3D%2010%5E%7B12%7D%20%5C%20%5Ctimes%20%5C%2010%5E%7B-12%7D%5C%5C%5C%5CI%20%3D%201%20%5C%20W%2Fm%5E2)
![For \ 70.7 dB\\\\70.7 = 10Log[\frac{I}{1*10^{-12}}]\\\\7.07 = Log[\frac{I}{1*10^{-12}}]\\\\10^{7.07} = \frac{I}{1*10^{-12}}\\\\I = 10^{7.07} \ \times \ 10^{-12}\\\\I = 1 \times \ 10^{-4.93} \ W/m^2](https://tex.z-dn.net/?f=For%20%5C%2070.7%20dB%5C%5C%5C%5C70.7%20%3D%2010Log%5B%5Cfrac%7BI%7D%7B1%2A10%5E%7B-12%7D%7D%5D%5C%5C%5C%5C7.07%20%3D%20%20Log%5B%5Cfrac%7BI%7D%7B1%2A10%5E%7B-12%7D%7D%5D%5C%5C%5C%5C10%5E%7B7.07%7D%20%3D%20%5Cfrac%7BI%7D%7B1%2A10%5E%7B-12%7D%7D%5C%5C%5C%5CI%20%3D%2010%5E%7B7.07%7D%20%5C%20%5Ctimes%20%5C%2010%5E%7B-12%7D%5C%5C%5C%5CI%20%3D%201%20%5Ctimes%20%5C%2010%5E%7B-4.93%7D%20%5C%20W%2Fm%5E2)
The second distance, r₂, can be determined from sound intensity formula given as;
Therefore, the second distance of the sound from the source is 431.78 m.
Using v=u+at, Where v is final velocity(m/s), u is initial velocity(m/s), a is acceleration(m/s^2) and t is time(s).
v = 0 + 3.2*6
v=19.2 m/s.
Yes/..................................................
