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
I believe the answer is energy
By definition,
q = 1.22y/D
Where,
q = min. angle
y = wavelength
D = Aperture diameter = diameter of the antenna
At distance "x" from the antenna,
L =xq = 1.22xy/D
Where, L = Min. distance
But, y =c/f = (3*10^8)/(16*10^9) = 0.01875 m
Substituting;
L = 1.22*5*10^3*0.01875/2.1 = 54.46 m
Answer: Different fuel components boil at different temperatures, allowing them to be separated.
Explanation:
