A standing wave of the third harmonic is induced in a stopped pipe of length 1.2 m. The speed of sound through the air of the pi
1 answer:
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We have that the Number of stitches per sec and he mass of oscillation motion is mathematically given as
a) Nt=25stitches per sec
b) m=2.033e-5kg
<h3>Number of stitches per sec and he mass of oscillation motion</h3>Question Parameters:
This <u>sewing </u>machine is capable of stitching 1,500 stiches in one minute.
If the <em>sewing </em>machine has a spring constant of 0.5 N/m,
Generally the equation for the Number of stitches per sec is mathematically given as
Nt=N/t
Therefore
Nt=1500/60
Nt=25stitches per sec
b)
Generally the equation for the Time t is mathematically given as
Therefore
m=2.033e-5kg
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Answer:
a) T = 2.26 N, b) v = 1.68 m / s
Explanation:
We use Newton's second law
Let's set a reference system where the x-axis is radial and the y-axis is vertical, let's decompose the tension of the string
sin 30 =
cos 30 =
Tₓ = T sin 30
T_y = T cos 30
Y axis
T_y -W = 0
T cos 30 = mg (1)
X axis
Tₓ = m a
they relate it is centripetal
a = v² / r
we substitute
T sin 30 = m (2)
a) we substitute in 1
T =
T =
T = 2.26 N
b) from equation 2
v² =
If we know the length of the string
sin 30 = r / L
r = L sin 30
we substitute
v² =
v² =
For the problem let us take L = 1 m
let's calculate
v =
v = 1.68 m / s