Answer:
β2= β1+10*f
Explanation:
comparing β2 and β1, it is said that β2 is increased by a factor of f.
for each factor of f, there is a 10*f dB increase.
therefore if the β1 is increases by an intensity of factor f
the new intensity would be β1+ 10*f
Answer:
The tone really matters and if there are any exclamation marks also.
Explanation:
Answer:
lowest frequency = 535.93 Hz
distance between adjacent anti nodes is 4.25 cm
Explanation:
given data
length L = 32 cm = 0.32 m
to find out
frequency and distance between adjacent anti nodes
solution
we consider here speed of sound through air at room temperature 20 degree is approximately v = 343 m/s
so
lowest frequency will be =
..............1
put here value in equation 1
lowest frequency will be =
lowest frequency = 535.93 Hz
and
we have given highest frequency f = 4000Hz
so
wavelength =
..............2
put here value
wavelength =
wavelength = 0.08575 m
so distance =
..............3
distance =
distance = 0.0425 m
so distance between adjacent anti nodes is 4.25 cm
Answer: 17.83 AU
Explanation:
According to Kepler’s Third Law of Planetary motion <em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”. </em>
(1)
Talking in general, this law states a relation between the <u>orbital period</u>
of a body (moon, planet, satellite, comet) orbiting a greater body in space with the <u>size</u>
of its orbit.
However, if
is measured in <u>years</u>, and
is measured in <u>astronomical units</u> (equivalent to the distance between the Sun and the Earth:
), equation (1) becomes:
(2)
This means that now both sides of the equation are equal.
Knowing
and isolating
from (2):
(3)
(4)
Finally:
(5)