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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D.Power has a time component while energy does not. This is because power is the RATE at which work is performed.
Given required solution
M=10kg W=? W=Fd
v=5.0m/s F=mg
t=2.40s =10*10=100N
S=VT
=5m/s*2.4s
=12m
so W=12*100
W=1200J
First we write the corresponding kinematics equations:
a = -g
v = -g * t + vo
y = -g * ((t ^ 2) / 2) + vo * t + yo
Substituting the values:
y = - (9.81) * (((0.50) ^ 2) / 2) + (19) * (0.50) + (0) = 8.27m
answer:
the displacement at the time of 0.50s is 8.27m
Answer:
Explanation:
The charges will repel each other and go away with increasing velocity , their kinetic energy coming from their potential energy .
Their potential energy at distance d
= kq₁q₂ / d
= 9 x 10⁹ x 36 x 10⁻¹² / 2 x 10⁻² J
= 16.2 J
Their total kinetic energy will be equal to this potential energy.
2 x 1/2 x mv² = 16.2
= 3 x 10⁻⁶ v² = 16.2
v = 5.4 x 10⁶
v = 2.32 x 10³ m/s
When masses are different , total P.E, will be divided between them as follows
K E of 3 μ = (16.2 / 30+3) x 30
= 14.73 J
1/2 X 3 X 10⁻⁶ v₁² = 14.73
v₁ = 3.13 x 10³
K E of 30 μ = (16.2 / 30+3) x 3
= 1.47 J
1/2 x 30 x 10⁻⁶ x v₂² = 1.47
v₂ = .313 x 10³ m/s
