The angular frequency of the wave is determined as 75.4 rad/s.
<h3>
What is wave function?</h3>
A wave function is a mathematical equation for the motion of the wave.
y(x, t) = A sin(kx + ωt + Φ)
where;
- ω is angular speed
- k is angular wavenumber
- Φ is phase angle
<h3>What is angular frequency?</h3>
The angular frequency is the angular displacement of any wave element per unit of time or the rate of change of the waveform phase.
<h3>Angular frequency</h3>
ω = 2πf
ω = 2π(12)
ω = 75.4 rad/s
Thus, the angular frequency of the wave is determined as 75.4 rad/s.
Learn more about angular frequency here: brainly.com/question/3654452
#SPJ4
Explanation:
Speed of the marathon runner, v = 9.51 mi/hr
Distance covered by the runner, d = 26.220 mile
Let t is the time taken by the marathon runner. We know that the speed of the runner is given by total distance divided by total time taken. Mathematically, it is given by :
t = 2.75 hours
Since, 1 hour = 60 minutes
t = 165 minutes
Since, 1 minute = 60 seconds
t = 9900 seconds
Hence, this is the required solution.
<h2>Answer:</h2>
<u>Turning a magnet very quickly would be BEST used to create an electric current</u>
<h2>Explanation:</h2>
In Electromagnetic waves electric field produces magnetic field and vice versa. A moving magnet can produce electric current. Dynamo is the best example for it. In dynamo armature is rotated between the magnets which results in the development of electric field and hence an electric current is produced in it.
Answer:
the SI unit of momentum is :- kg.ms-1
and we know that,
kinetic energy = 1/2 mv2
E=p2/2m
p=(2Em)1/2
so the derived units are (J.kg)1/2
Explanation:
(a) 5.66 m/s
The flow rate of the water in the pipe is given by
where
Q is the flow rate
A is the cross-sectional area of the pipe
v is the speed of the water
Here we have
the radius of the pipe is
r = 0.260 m
So the cross-sectional area is
So we can re-arrange the equation to find the speed of the water:
(b) 0.326 m
The flow rate along the pipe is conserved, so we can write:
where we have
and where 
is the cross-sectional area of the pipe at the second point.
Solving for A2,
And finally we can find the radius of the pipe at that point:
