Answer:
A longitudinal wave is a wave in which particles of the medium move in a direction parallel to the direction that the wave moves.
Explanation:
hope this helped
Refer to the diagram shown below.
The suspended wire is in the shape of a parabola defined by the equation
y = ax²
where a = a positive constant.
The derivative of y with respect to x is y' = 2ax.
The vertex is at (0,0) and the line of symmetry is x = 0.
The suspended length is 41 ft, therefore half the suspended length is 20.5 ft.
The length between x = 0 and x = 20 is given by
Because we do not know the value of a, we shall find it numerically.
Define the function
The plot for f(a) versus a yields an approximate solution (from Matlab) of a = 0.01 (shown in the figure).
Therefore
y = 0.01x²
When x = 20 ft, h = 0.01(400) = 4 ft
Because the vertex of the parabola is 19 ft above ground, the support points for the wire are 19 + h = 23 ft above ground.
Answer: 23.00 ft
Answer:
vB = 0.5418 m/s (→)
aB = - (0.3189/L) m/s²
ωcd = (0.2117/L) rad/s
Explanation:
a) Given:
vA = 0.23 m/s (↑) (constant value)
If
tan θ = vA/vB
For the instant when θ = 23° we have
vB = vA/ tan θ
⇒ vB = 0.23 m/s/tan 23°
⇒ vB = 0.5418 m/s (→)
b) If tan θ = vA/vB ⇒ vA = vB*tan θ
⇒ d(vA)/dt = d(vB*tan θ)/dt
⇒ 0 = tan θ*d(vB)/dt + vB*Sec²θ*dθ/dt
Knowing that
aB = d(vB)/dt
ωcd = dθ/dt
we have
⇒ 0 = tan θ*aB + vB*Sec²θ*ωcd
ωcd = - Sin (2θ)*aB/(2*vB)
If
v = ωcd*L
where v = vA*Cos θ ⇒ ωcd = v/L = vA*Cos θ/L
⇒ vA*Cos θ/L = - Sin (2θ)*aB/(2*vB)
⇒ aB = - vA*vB/((Sin θ)*L)
We plug the known values into the equation
aB = - (0.23 m/s)*(0.5418 m/s)/(L*Sin 23°)
⇒ aB = - (0.3189/L) m/s²
Finally we obtain the angular velocity of CD as follows
ωcd = vA*Cos θ/L
⇒ ωcd = 0.23 m/s*Cos 23°/L
⇒ ωcd = (0.2117/L) rad/s
Place the permanent magnet into an AC magnetic field and slowly reduce the field to 0 and the magnet will be temporarily demagnetized.
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Cloud droplets collide to form larger droplets in a process called coalescence.