Answer: The time required for the impluse passing through each other is approximately 0.18seconds
Explanation:
Given:
Length,L = 50m
M/L = 0.020kg/m
FA = 5.7×10^2N
FB = 2.5×10^2N
The sum of distance travelled by each pulse must be 50m since each pulse started from opposite ends.
Ca(t) + CB(t) = 50
Where CA and CB are the velocities of the wire A and B
t = 50/ (CA + CB)
But C = Sqrt(FL/M)
Substituting gives:
t = 50/ (Sqrt( FAL/M) + Sqrt(FBL/M))
t = 50/(Sqrt 5.7×10^2/0.02) + (Sqrt(2.5×10^2/0.02))
t = 50 / (168.62 + 111.83)
t = 50/280.15
t = 0.18 seconds
A point at which parts of an artificial structure are joined. Hope this helps
Answer:
true
Explanation:
dbgkdodocofkci ifkdcl k kfododocp v
W = ∫ (x from 0.1 to +oo) F dx
= ∫ (x from 0.1 to +oo) A e^(-kx) dx
= A/k x [ - e^(-kx) ](between 0.1 and +oo)
= A/k x [ 0 + e^(-k * 0.1) ]
<span>
= A/k x e^(-k/10) </span>
