Answer:
The distance covered by puck A before collision is 
Explanation:
From the question we are told that
The label on the two hockey pucks is A and B
The distance between the two hockey pucks is D 18.0 m
The speed of puck A is 
The speed of puck B is 
The distance covered by puck A is mathematically represented as
=> 
The distance covered by puck B is mathematically represented as
=> 
Since the time take before collision is the same
substituting values
=> 
=>
The general formula for the frequency of the nth-harmonic of the column of air in the tube is given by
where f1 is the fundamental frequency.
In our problem, we have two harmonics, one of order n and the other one of order (n+1) (because it is the next higher harmonic), so their frequencies are
so their difference is
So, the difference between the frequencies of the two harmonics is just the fundamental frequency of the column of air in the tube, which is:
Answer:
1. All electromagnetic waves travelS at 3×10⁸m/s in vacuum
2.They are transverse in nature.
3.they can travel through vacuum.( doesn't require any materials for their transmission.
4.They under go phenomena of diffraction,reflection, refraction, interference and polarization.
HOPE IT HELPED.
PLEASE PICK BRAINLIEST
Answer:
option (c) - 10 j
Explanation:
F = (6 i + 4 j - 2 k) N
r1 = (1.5, 3, -4.5) m = (1.5 i + 3j - 4.5 k) m
r2 = (4, -2.5, - 3) m = (4 i - 2.5 j - 3 k) m
displacement, r = r2 - r1 = ( 2.5 i - 5.5 j + 1.5 k) m
Work done is defined as the dot product of force vector and teh displacement vector.
W = (6 i + 4 j - 2 k) . ( 2.5 i - 5.5 j + 1.5 k)
W = 15 - 22 - 3 = - 10 J
