Answer:
ω = 0.467 rad/s
Explanation:
given,
tangential force exerted by the person = 37.7 N
radius of merry-go-round = 2.75 m
mass of merry-go-round = 144 Kg
angle = 33.2°
moment of inertia
I = 544.5 kg.m²
torque = force x radius
τ = 37.7 x 2.75
τ = 103.675 N.m
angular acceleration
α = 0.190 rad/s²
now ,
distance =
d = 0.579 rad
we know,
using equation of rotational motion
t = 2.46 s
angular speed
ω = α x t
ω = 0.19 x 2.46
ω = 0.467 rad/s
Answer:
L=0.654 m
Explanation:
<u>Concepts and Principles </u>
1- The speed of sound in air is expressed as a function of the temperature of air as follows:
v=(331 m/s)√(1+T_C/273°C) (1)
where 331 m/s is the speed of sound in air at temperature 0°C and Tc is the temperature of air in Celsius.
<u>Standing Wave Patterns in Pipes: </u>
A pipe open at both ends can have standing wave patterns with resonant frequencies:
f=v/λ=nv/2L n=1,2,3.........
where v is the speed of sound in air.
<u>Given Data </u>
f_1 (fundamental frequency of the flute) = 262 Hz
T (temperature of the air) = 20°C
The flute is open at both ends.
<u>Required Data </u>
We are asked to determine the length of the tube.
<u>Solution</u><u> </u>
The speed of sound in air at temperature T = 20°C is found from Equation (1):
v=(331 m/s)√(1+T_C/273°C)
=342.91 m/s
The fundamental frequency of the flute is found by substituting n = 1 into Equation (2):
f=v/2L
Solve for L:
L=v/2f_1
L=0.654 m
