Roberto and Genevieve are trying to separate a mixture of beads using a sifter. The chart below shows the colors and sizes of th
e beads. What size must the holes in the sifter be to separate the red and blue beads from the yellow beads?
2 answers:
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A cart is pulled by horizontal force such that it moves with constant velocity
So here since velocity is constant we can say that its acceleration will be ZERO
now here for zero acceleration we can say
So here we will have
here we know that
so we will have
so here applied force on handle will be
<em>b. 10 N</em>
Answer:
the penny loses contact at the piston's highest point.
f = 2.5 Hz
Explanation:
Concepts and Principles
1- Newton's Second Law: The net force F on a body with mass m is related to the body's acceleration a by
∑F = ma (1)
2- The maximum transverse acceleration of a particle in simple harmonic motion is found in terms of the angular speed w and the amplitude A as follows:
a_max = -w^2A (2)
3- The angular frequency w of a wave is related to the frequency f by:
w = 2π f (3)
Given Data
- The amplitude of the piston is: A = (4.0 cm) ( 1/ 100 cm)= 0.04 m.
- The frequency of oscillation of the piston is steadily increased.
Required Data
<em>In part (a), we are asked to determine the point at which the penny first loses contact with the piston. </em>
<em>In part (b), we are asked to determine the maximum frequency for which the penny just barely remains in place for a full cycle. </em>
Solution
(a)
The free-body diagram in Figure 1 shows the forces acting on the penny; mi is the gravitational force exerted by the Earth on the penny andrt is the normal contact force exerted by the piston on the penny.
figure 1 is attached
Apply Newton's second law from Equation (1) in the vertical direction to the penny:
∑F_y -mg= ma
Solve for n=m(g+a) The penny loses contact with the surface of the oscillating piston when the normal force n exerted by the piston is zero. So
0 = m(g + a)
a = —g
Therefore, the penny loses contact with the piston when the piston starts accelerating downwards. The piston first acceleratesdownward at its highest point and hence the penny loses contact at the piston's highest point.
(b)
The maximum acceleration of the penny at the highest point of the piston is found from Equation (2):
a = —w^2A
where a = —g at the highest point. So
g = w^2A
Solve for w:
w =√g/A
Substitute for w from Equation (3):
2πf = √g/A
Solve for f :
f = 1/2π√g/A
Substitute numerical values:
f = 1/2π√9.8 m/s^2/0.04
f = 2.5 Hz
