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The highest frequency sound to which the machine can be adjusted is :
- 4179.33 Hz
<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:
I = 1.06886 N s
Explanation:
The expression for momentum is
I = F t = Δp
therefore the momentum is a vector quantity, for which we define a reference system parallel to the floor
Let's find the components of the initial velocity
sin 28.2 = v_y / v
cos 28.2= vₓ / v
v_y = v sin 282
vₓ = v cos 28.2
v_y = 42.8 sin 28.2 = 20.225 m / s
vₓ = 42.8 cos 28.2 = 37.72 m / s
since the ball is heading to the ground, the vertical velocity is negative and the horizontal velocity is positive, it can also be calculated by making
θ = -28.2
v_y = -20.55 m / s
v_x = 37.72 m / s
X axis
Iₓ = Δpₓ =
since the ball moves in the x-axis without changing the velocity, the change in moment must be zero
Δpₓ = m - m v₀ₓ = 0
v_{fx} = v₀ₓ
therefore
Iₓ = 0
Y axis
I_y = Δp_y = p_{fy} -p_{oy}
when the ball reaches the floor its vertical speed is downwards and when it leaves the floor its speed has the same modulus but the direction is upwards
v_{fy} = - v_{oy}
Δp_y = 2 m v_{oy}
Δp_y = 2 0.0260 (20.55)
= 1.0686 N s
the total impulse is
I = Iₓ i ^ + I_y j ^
I = 1.06886 j^ N s
Answer:
a) λ = 1.12 m
b) f = 5.41 Hz
c) v = 154.54 m/s
d) A = 0.22m
e)
Explanation:
You have the following equation for a wave traveling on a cord:
(1)
The general expression for a wave is given by:
(2)
By comparing the equation (1) and (2) you have:
A: amplitude of the wave = 0.22m
k: wave number = 5.6 m^-1
w: angular velocity = 34 rad/s
a) The wavelength is given by substitution in the following expression:
b) The frequency is:
c) The velocity of the wave is:
d) The amplitude is 0.22m
e) To calculate the maximum and minimum speed of the particles you obtain the derivative of the equation of the wave, in time:
cos function has a minimum value -1 and maximum +1. Then, you obtain for maximum and minimum velocity:
1) In the initial situation, the total mechanical energy of the system is given only by the kinetic energy of the ball that is moving with speed v:
where
is the mass of the ball.
In the final situation, where the system (ball+pendulum) rises a vertical distance of h=0.145 m, the system is stationary (v=0) so the total mechanical energy of the system is the gravitational potential energy:
where
is the mass of the pendulum.
For the law of conservation of energy,
, so we can find the initial speed v of the ball:
2) The kinetic energy lost in the collision is the initial kinetic energy of the ball:
where
In the final situation, where the system (ball+pendulum) rises a vertical distance of h=0.145 m, the system is stationary (v=0) so the total mechanical energy of the system is the gravitational potential energy:
where
For the law of conservation of energy,
2) The kinetic energy lost in the collision is the initial kinetic energy of the ball: