What units are associated with unit vectors i , j and k

1 answer:

4 0

First of all, not to get confused with terminology, unit vectors are called unit vectors because they have a unit length. This has absolutely nothing to do with physical units.

Now, if you are talking about physical units like m, m/s or N. Then it depends on what physical vector quantity you’re talking about. If it’s the displacement vector then the i,j,k will have units of length i.e. meters (m). If it’s the force vector then the i,j,k will have units of force i.e. newtons (N).

But if you are talking about vectors in math, then they are unitless. So, we just pretend the physical unit is 1.

Again stressing that this has nothing to do with their name “unit vectors”. They are called that because the unit, AKA the number “1”, is the value of their magnitude/length.

They have magnitude 1, bearing the unit of physical quantity they represent.

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A 10-kg cart moving at 5 m/s collides with a 5-kg cart at rest and causes it to move 10 m/s. Which principle explains the result

Over [174]

B) law of conservation of momentum

It states that the total momentum of a system before impact is the same as the total momentum of the system after impact.

In this case total momentum before impact:

10kg*5m/s + 5kg * 0m/s = 50 kg m/s

After Impact:

10kg*0m/s + 5kg*10m/s = 50 kg m/s

You can see the momentum before and after impact is same as 50 kg m/s

Of course we assumed that the first cart stopped after the impact, and there are no energy losses.

7 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

In part (a), we are asked to determine the point at which the penny first loses contact with the piston.

In part (b), we are asked to determine the maximum frequency for which the penny just barely remains in place for a full cycle.

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: