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Sever21 [200]
3 years ago
5

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?

Physics
2 answers:
spin [16.1K]3 years ago
3 0

Answer:

2.5 mm

Explanation:

It's just big enough for red and blue to fall through but small enough that yellow will stay in

beks73 [17]3 years ago
3 0

Answer:b

Explanation:

Because it look like it

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Seven days after a waxing gibbous moon, the moon will be waning gibbous, and at some point during that seven days, it will have been Full.
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estion: Why is it important to use vector quantities and not just scalar quantities to describe the motion of an object? Vector
Vera_Pavlovna [14]

Answer:

Vector quantities are important in the study of motion. Some examples of vector quantities include force, velocity, acceleration, displacement, and momentum. The difference between a scalar and vector is that a vector quantity has a direction and a magnitude, while a scalar has only a magnitude. Vector, in physics, a quantity that has both magnitude and direction. It is typically represented by an arrow whose direction is the same as that of the quantity and whose length is proportional to the quantity's magnitude. A quantity which does not depend on direction is called a scalar quantity. Vector quantities have two characteristics, a magnitude and a direction. The resulting motion of the aircraft in terms of displacement, velocity, and acceleration are also vector quantities. A vector quantity is different to a scalar quantity because a quantity that has magnitude but no particular direction is described as scalar. A quantity that has magnitude and acts in a particular direction is described as vector.

Explanation:

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3 years ago
How does a fuse work?
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5 0
3 years ago
A horizontal force in used to pull a 5 kilogram cart at a constant speed of 5 meters per second across the floor as shown in the
ddd [48]

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

F_{net} = 0

So here we will have

F_{net} = F_{ap} - F_f = 0

here we know that

F_f = 10 N

so we will have

F_{ap} - 10 = 0

F_{ap} = 10 N

so here applied force on handle will be

<em>b. 10 N</em>

5 0
3 years ago
A penny rides on top of a piston as it undergoes vertical simple harmonic motion with an amplitude of 4.0cm. If the frequency is
BlackZzzverrR [31]

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

6 0
3 years ago
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