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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Approximately 150-200 species.
Answer:
15.7 m
Explanation:
The range (horizontal distance) of the projectile is determined only by its horizontal motion.
The horizontal motion is a motion with constant speed, which is equal to the initial horizontal velocity of the object:

where
v = 12.0 m/s is the initial velocity
is the angle between the direction of v and the horizontal
Substituting,

We know that the projectile hits the ground in a time of
t = 2.08 s
so the horizontal distance covered is
