Answer:
0.073 N-m
Explanation:
i = 12 A, l = 0.8 m, B = 0.12 T
The circumference of the loop is 0.8 m.
Let r be the radius of the loop.
2 x 3.14 x r = 0.8
r = 0.127 m
Maximum Torque = i x A x B
Maximum Torque = 12 x 3.14 x 0.127 x 0.127 x 0.12 = 0.073 N-m
The highest frequency sound to which the machine can be adjusted is :
<u>Given data :</u>
Pressure = 10 Pa
Speed of sound = 344 m/s
Displacement altitude = 10⁻⁶ m
<h3>Determine the highest frequency sound ( f ) </h3>
applying the formula below
Pmax =
--- ( 1 )
Therefore :
f = ( Pmax * V ) / 
= ( 10 * 344 ) / 2
* 1.31 * 10⁵ * 10⁻⁶
= 4179.33 Hz
Hence we can conclude that The highest frequency sound to which the machine can be adjusted is : 4179.33 Hz .
Learn more about Frequency : brainly.com/question/25650657
<u><em>Attached below is the missing part of the question </em></u>
<em>A loud factory machine produces sound having a displacement amplitude in air of 1.00 μm, but the frequency of this sound can be adjusted. In order to prevent ear damage to the workers, the maximum pressure amplitude of the sound waves is limited to 10.0 Pa. Under the conditions of this factory, the bulk modulus of air is 1.31×105 Pa. The speed of sound in air is 344 m/s. What is the highest-frequency sound to which this machine can be adjusted without exceeding the prescribed limit?</em>
Normal force, weight, Kinetic friction, and air resistance are a few I think of the top of my head.
I hope this helps
Answer:
higher frequency waves will have more energy
Explanation:
D. Energy that is transformed is neither destroyed nor created