Answer:
The force constant for each elastic band is 77.93 N/m
Explanation:
Hooke's law of a spring or an elastic band gives the relation between elastic force (Fe) and stretching (x), the magnitude of that force is:
(1)
With k, the elastic force constant. The three elastic bands support the child’s weight (W) and maintain him at rest, so by Newton’s second law for one of the elastic bands:
(2)
(3)
Using (1) on (3):
(4)
Answer:
L= 0.059 mH
Explanation:
Given that
R = 855 Ω and C = 6.25 μF
V= 84 V
Frequency
ω = 51900 1/s
We know that
L=Inductance
C=Capacitance
ω =angular Frequency
ω² L C =1
(51900)² x L x 6.25 x 10⁻⁶ = 1
L= 5.99 x 10⁻⁵ H
L= 0.059 mH
The characteristics of the speed of the traveling waves allows to find the result for the tension in the string is:
T = 10 N
The speed of a wave on a string is given by the relationship.
v =
Where v es the velocty, t is the tension ang μ is the lineal density.
They indicate that the length of the string is L = 2.28 m and the pulse makes 4 trips in a time of t = 0.849 s, since the speed of the pulse in the string is constant, we can use the uniform motion ratio, where the distance traveled e 4 L
v =
v =
v =
v = 10.7 m / s
Let's find the linear density of the string, which is the length of the mass divided by its mass.
μ =
μ = 8.77 10⁻² kg / m
The tension is:
T = v² μ
Let's calculate
T = 10.7² 8.77 10⁻²
T = 1 0 N
In conclusion using the characteristics of the velocity of the traveling waves we can find the result for the tension in the string is:
T = 10 N
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