<span>If two wheels are exactly the same but spin at different speeds, wheel b is twice te speed of wheel a, it is possible to find the ratio of the magnitude of radial acceleration at a singular point of the rim on wheel be to the spot is four.</span>
Answer: NNOOOOOOOOOOOOOOOOOOONONONO
Explanation: simple harmonic motion, in physics, repetitive movement back and forth through an equilibrium, or central, position, so that the maximum displacement on one side of this position is equal to the maximum displacement on the other side. The time interval of each complete vibration is the same. The force responsible for the motion is always directed toward the equilibrium position and is directly proportional to the distance from it. That is, F = −kx, where F is the force, x is the displacement, and k is a constant. This relation is called Hooke’s law.
A specific example of a simple harmonic oscillator is the vibration of a mass attached to a vertical spring, the other end of which is fixed in a ceiling. At the maximum displacement −x, the spring is under its greatest tension, which forces the mass upward. At the maximum displacement +x, the spring reaches its greatest compression, which forces the mass back downward again. At either position of maximum displacement, the force is greatest and is directed toward the equilibrium position, the velocity (v) of the mass is zero, its acceleration is at a maximum, and the mass changes direction. At the equilibrium position, the velocity is at its maximum and the acceleration (a) has fallen to zero. Simple harmonic motion is characterized by this changing acceleration that always is directed toward the equilibrium position and is proportional to the displacement from the equilibrium position. Furthermore, the interval of time for each complete vibration is constant and does not depend on the size of the maximum displacement. In some form, therefore, simple harmonic motion is at the heart of timekeeping.
Answer:
Fy=107.2 N
Explanation:
Conceptual analysis
For a right triangle :
sinβ = y/h formula (1)
cosβ = x/h formula (2)
x: side adjacent to the β angle
y: opposite side of the β angle
h: hypotenuse
Known data
h = T = 153.8 N : rope tension
β= 44.2°with the horizontal (x)
Problem development
We apply the formula (1) to calculate Ty : vertical component of the rope force.
sin44.2° = Ty/153.8 N
Ty = (153.8 N ) *(sen44.2°)= 107.2 N directed down
for equilibrium system
Fy= Ty=107.2 N
Fy=107.2 N upward component of the force acting on the stake
<h3>
Answer:</h3>
<h3>
Explanation:</h3>
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λ=500 nm = 500·10⁻⁹ m
c=3·10⁸ m/s
h=6,63·10⁻³⁴ J·s = 4,14·10⁻¹⁵ eV·s
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E - ?
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V=IR
V=15x11
V=165ohms
I don’t quite remember the unit
Ohms law
