How does a cockcroft walton generator work?
1 answer:
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Answer:
_s = 37.77 m / s
Explanation:
This is an exercise of the Doppler effect that the change in the frequency of the sound due to the relative speed of the source and the observer, in this case the observer is still and the source is the one that moves closer to the observer, for which relation that describes the process is
f ’= f₀
where d ’= 530 Make
when the ambulance passes away from the observer the relationship is
f ’’ = f₀
where d ’’ = 424 beam
let's write the two expressions
f ’ (v-v_s) = fo v
f ’’ (v + v_s) = fo v
let's solve the system, subtract the two equations
v (f ’- f’ ’) - v_s (f’ + f ’’) = 0
v_s = v
the speed of sound is v = 340 m / s
let's calculate
v_s = 340
v_s = 340 )
v_s = 37.77 m / s
Answer:
1) μ = 1.33 10⁻³ kg / m , F = - 14,256 , 2) v= 103.53 m/s, 3) f = 138.04 Hz , 4) 1, 25, 50, 76, 101 , 5) A = 0.00869 m , 6) # _position = (# _account-1) (1.5m / 100 accounts)
Explanation:
1) Linear density is the mass per unit length
μ = m / L
μ = 2 1 10⁻³ / 1,5
μ = 1.33 10⁻³ kg / m
this is the density when the chain is stretched, which is when the pulse occurs
we can find the tension with
F = - k (x₁-x₀)
where k is the spring constant
F = - 28.8 (1.5 -1.005)
F = - 14.256 N
the negative sign indicates that the force is restorative
2) the pulse speed is
v = √ T /μ
v = √ 14,256 / 1,33 10⁻³
v = 103.53 m / s
3) If standing waves are formed with fixed points at the ends and 4 antinodes, the wavelength is
2 λ = L
λ = L / 2
wave speed is related to frequency and wavelength
v = λ f
f = v / λ
f = v 2 / L
f = 103.53 2 / 1.5
f = 138.04 Hz
4) The marbles are numbered, the marbles that remain motionless are
the first (1) and the last (101)
Let's look for the distance to each node, for this we must observe that in each wavelength there is a node at the beginning, one in the center and one at the end, therefore the nodes are in
#_node = m λ / 2 = m L / 4
#_node position (m)
1 1.5 / 4 = 0.375
2 2 1.5 / 4 = 0.75
3 3 1.5 / 4 = 1,125
Since there are 101 marbles in the initial length, this number does not change with increasing length, so there is 101 marble in 1.5 m. Let's find with a direct proportion rule the number of marbles at these points with nodes
#_canica = 0.375 m (101 marble / 1.5 m) 0.375 67.33
# _canica = 25
#_canica = 0.75 67.33
#_canica = 50
# _canica = 1,125 67.33
#canica = 75.7 = 76
in short the number of the fixed marbles is
1, 25, 50, 76, 101 canic
5) The movement of the account is oscillatory at this point, which is why it is described by
y = A cos wt
= -A w sin wt
the speed is maximum for when the breast is worth ±1
v_{y} = Aw
A = v_{y} / w
angular velocity related to frequency
w = 2π f
A = v_{y} / 2πf
A = 7.54 / (2π 138.04)
A = 0.00869 m
6) for the position of each account we can use a direct proportion rule
in total there are 100 accounts distributed in the 1.50 m distance, the #_account is in the # _position. Note that it starts to be numbered 1, so this number must be subtracted from the index of the amount
# _position = (# _account-1) (1.5m / 100 accounts)
#_canic position(m)
1 0
2 0.015
3 0.045
4 0.06
7) the wave has a constant velocity, but every wave is oscillated perpendicular to this velocity, with an oscillatory movement described by the expression
y = Acos wt
the maximum speed is
= -Aw sin wt
speed is maximum when the sine is ±1
v_{y} = A w
to calculate the amplitude of the count we use that for a standing wave
y = 2Asin kx
y / A = 2 sin (2π /λ x)
the wavelength is
λ = 0.75 m
the position is
x (30) = 29 1.5 / 100 = 0.435 m
y (30) A = 2 sin (2pi 0.435 / 0.75)
y (30) / A = 0.96 m