Answer:
frequency is 195.467 Hz
Explanation:
given data
length L = 4.36 m
mass m = 222 g = 0.222 kg
tension T = 60 N
amplitude A = 6.43 mm = 6.43 ×
m
power P = 54 W
to find out
frequency f
solution
first we find here density of string that is
density ( μ )= m/L ................1
μ = 0.222 / 4.36
density μ is 0.050 kg/m
and speed of travelling wave
speed v = √(T/μ) ...............2
speed v = √(60/0.050)
speed v = 34.64 m/s
and we find wavelength by power that is
power = μ×A²×ω²×v / 2 ....................3
here ω is wavelength put value
54 = ( 0.050 ×(6.43 ×
)²×ω²× 34.64 ) / 2
0.050 ×(6.43 ×
)²×ω²× 34.64 = 108
ω² = 108 / 7.160 ×
ω = 1228.16 rad/s
so frequency will be
frequency = ω / 2π
frequency = 1228.16 / 2π
frequency is 195.467 Hz
Latitude, elevation, ocean currents, topography, and prevailing winds. There's probably a few others but these are the most important.
To determine the height of the object given the time, we simply use the given relation between height and time in the problem statement. It is given as:
h = -16t^2 + 127t
We substitute 55 seconds to t and obtain,
h = -16(55)^2 + 127(55)
h = - 41415
Answer:
Different surfaces
<h3>You can see that dull surfaces are good absorbers and emitters of infrared radiation. Shiny surfaces are poor absorbers and emitters (but they are good reflectors of infrared radiation</h3>
Answer:
Choice C is not equivalent to 2.50 miles.
Explanation:
The given data is now converted into feet, inches, kilometers, yards and centimeters:
mi - ft


(Choice A)
mi - in


(Choice B)
mi - km

(Different from Choice C)
mi - yd


(Choice D)
mi - cm


(Choice E)
Choice C is not equivalent to 2.50 miles.