A 487-Hz tuning fork is sounded together with an out-of-tune guitar string, and a beat frequency of 4 Hz is heard. When the stri
1 answer:
Answer:
Explanation:
Let f is frequency of of the guitar string
Frequency of tuning fork is given 487 Hz
Beat frequency is difference frequencies of guitar string and the tuning fork
It could be
(f - 487Hz)
or
(487Hz-f)
As it is mentioned that as the frequency of guitar string increases the beat frequency decreases
So the beat frequency is given as
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here's the solution,
we know that,
=》
so,
=》
=》
=》
frequency = 6 hertz
1) 3 miles/Hour
The speed is defined as the distance covered divided by the time taken:
where
d = 1.5 mi is the distance
t = 0.5 h is the time taken
Substituting,
2) 1.34 m/s south
Velocity, instead, is a vector, so it has both a magnitude and a direction. We have:
is the displacement in meters
is the time taken in seconds
Substituting,
And the direction of the velocity is the same as the displacement, so it is south.
Answer:
So airplane will be 1324.9453 m apart after 2.9 hour
Explanation:
So if we draw the vectors of a 2d graph we see that the difference in angles is = 83 - 44.3 =
Distance traveled by first plane = 730×2.9 = 2117 m
And distance traveled by second plane = 590×2.9 = 1711 m
We represent these distances as two sides of the triangle, and the distance between the planes as the side opposing the angle 38.7.
Using the law of cosine, representing the distance between the planes, we see that:
d = 1324.9453 m
The electric field produced by a large flat plate with uniform charge density on its surface can be found by using Gauss law, and it is equal to
where
is the charge density
is the vacuum permittivity
We see that the intensity of the electric field does not depend on the distance from the plate. Therefore, the strenght of the electric field at 4 cm from the plate is equal to the strength of the electric field at 2 cm from the plate:
where
We see that the intensity of the electric field does not depend on the distance from the plate. Therefore, the strenght of the electric field at 4 cm from the plate is equal to the strength of the electric field at 2 cm from the plate: