The question for this problem would be the minimum headphone delay, in ms, that will cancel this noise.
The 200 Hz. period = (1/200) = 0.005 sec. It will need to be delayed by 1/2, so 0.005/2, that is = 0.0025 sec. So converting sec to ms, will give us the delay of:Delay = 2.5 ms.
Answer:
THE FIRST ONE YOU SHOULD TELL HIM AND THE LAST ONE YOU SHOUDENT DO BECAUSE HE WILL DO IT AGAIN AND EXPECT OTHERS TO CLEAN UP AFTER HIM
Explanation:
Answer:
46.4 s
Explanation:
5 minutes = 60 * 5 = 300 seconds
Let g = 9.8 m/s2. And 
be the slope of the road, s be the distance of the road, a be the acceleration generated by Rob, 3a/4 is the acceleration generated by Jim . Both of their motions are subjected to parallel component of the gravitational acceleration 
Rob equation of motion can be modeled as s = a_Rt_R^2/2 = a300^2/2 = 45000a[/tex]
Jim equation of motion is 
As both of them cover the same distance
So Jim should start 346.4 – 300 = 46.4 seconds earlier than Rob in other to reach the end at the same time
Answer:
ω = √(2T / (mL))
Explanation:
(a) Draw a free body diagram of the mass. There are two tension forces, one pulling down and left, the other pulling down and right.
The x-components of the tension forces cancel each other out, so the net force is in the y direction:
∑F = -2T sin θ, where θ is the angle from the horizontal.
For small angles, sin θ ≈ tan θ.
∑F = -2T tan θ
∑F = -2T (Δy / L)
(b) For a spring, the restoring force is F = -kx, and the frequency is ω = √(k/m). (This is derived by solving a second order differential equation.)
In this case, k = 2T/L, so the frequency is:
ω = √((2T/L) / m)
ω = √(2T / (mL))
Answer:
v₂ = v/1.5= 0.667 v
Explanation:
For this exercise we will use the conservation of the moment, for this we will define a system formed by the two students and the cars, for this isolated system the forces during the contact are internal, therefore the moment conserves.
Initial moment before pushing
p₀ = 0
Final moment after they have been pushed
= m₁ v₁ + m₂ v₂
p₀ =
0 = m₁ v₁ + m₂ v₂
m₁ v₁ = - m₂ v₂
Let's replace
M (-v) = -1.5M v₂
v₂ = v / 1.5
v₂ = 0.667 v
