Answer:
Explanation:
a )
According to graph the object moves with constant velocity of 10 m /s during first 8 s and then its velocity changes to - 5 m /s from 8 th second upto 12 th second .
Initial displacement = 8 m
displacement during 8 s = velocity x time
= 8 x 10 = 80 m
total displacement = 88 m
b )
displacement during period from 8 th to 12 th s
= - 5 x 4 = - 20 m
total displacement after 12 s
= 88 - 20 = 68 m
c ) average speed = total distance covered during 12 s / total time
= (80 + 20) / 12
= 8.33 m /s
average velocity = total displacement during 12 s / total time
= 80 - 20 / 12
= 60 / 12
= 5 m /s .
Below are the choices:
<span> A) The box will slow down.
B) The box's velocity will be 1 m/s.
C) The box's velocity will not change.
D) The box will experience acceleration
</span>
The answer is D) The box will experience acceleration
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Answer:
the answer is equal to yes
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>
Answer:
b
Explanation: because it wont effect the animals home