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# Defining a Coordinate System in Torqu...

When choosing which direction of the system will be positive and which will be negative, do we always have to choose the counterclockwise direction to be positive in free body diagrams involving torque(s) applied on pulleys with mass? Will points be taken off on exams if we define the positive direction to be clockwise, if that is w [...]

# Answered: Question regarding directio...

The direction of the force doesn’t have a direct relation with on the direction of velocity. You can have a positive net force while the velocity is positive, negative, and zero; and vice versa. The momentum is always in the same direction of the velocity since ||\vec P=m\vec v||.

# Question regarding direction of force...

Is the force (F) on an object and the velocity of the object (V) and the momentum of the object (P) all in the same direction?

# Answered: Question 3 Quiz Week 9

If you pick downward as positive, then ||v_2|| is negative and ||v_1|| is positive. So your ||\Delta P|| would be negative too — a negative P2 minus a positive P1 gives a total negative ||\Delta P||, which means an upward ||\Delta P||. In this case, d is correct.
In addition, think about the forces the ground applie [...]

# Answered: Is there anyway we can get ...

The example problems in the lecture notes are the best problems to practice. Try them and make sure you can do them without looking at the solutions. You can also imagine how to vary the examples a bit to create new problems to practice.

# Question 3 Quiz Week 9

Question 3 Quiz Week 9
I don’t have the correct answers for this/any quiz.
A basketball of mass m dropped from a certain height hits the ground with a downward speed v(1) and bounces back with an upward speed v(2) leaving the ground. Ignoreair resistance during the process. What is the impulse applied by the ground to the metal ground.
[...]

# Answered: Textbook Page270 Chapter 9

This level of questions won’t be on midterm. Midterm questions will be similar to the examples in my lecture notes and previous quizzes.

# Answered: Is momentum always conserve...

It depends on how you define your system. If you include the external object which exerted the external force in a grander system, then the momentum of that system would be conserved. — that is if you define the whole universe as your system, then all the momentum of all matters inside the universe would sum to a constant and be co [...]

# Is momentum always conserved? Even if...

I understand that there are two situations one about an isolated system in which there is no ext. force and so deltap=0
and one non-isolated system in which there is an external force acting so deltap=Impulse
My question is, is momentum always conserved in both these situations?

# Textbook Page270 Chapter 9

Water Falls without splashing at a rate of 0.25L/s from a height of 2.6m into a 0.75kg bucket on a scale. If the bucket is originally empty, what does the scale read in newtons 3.00s after water starts to accumalte in it.
It would be very helpful if you could tell me is this question of similar difficultly level to questions on midterm? Just [...]

# Answered: HELP! HW10 Problem 6

Take a look at this answer: http://tutor.leiacademy.org/qa/index.php/62/hw-10-question-6?show=65#a65
Use the elastic collision equations there, you will find the final velocities — pay attention to the directions (signs) of the velocities — ||v_{2i}|| is in the negative x direction so you will need to put in a negative sign t [...]

# HELP! HW10 Problem 6

In this question, v1i=v2i (because v=sqrt(2gh), it doesnt matter the mass). So from elastic collision equations , I get v1f=v1i (because v1i=v2i, so the numerator and denominator are the same, m1+m2). Similarly, we get v2f=v2i, so the heights (for both blocks) after collision are exactly the same as the heights the blocks released. [...]

# Answered: Oscillator equation

Now you have ||\omega||, nice.
At equilibrium, the velocity has its maximum magnitude ||v=\omega A||, which gives you A based on the numbers given in the question.
Then, since at t=5.4 ms, the oscillator is at equilibrium, you would have ||x(t=5.4 ms)=0||. Plug in the t, A, and ||\omega||, you can find ||\phi _0||.

# Oscillator equation

An oscillator with period 2.7 ms passes through equilibrium at t = 5.4 ms with velocity v = -6.8 m/s. The equation of the oscillator’s motion is
x(t) = ( ? ) cm cos ( ( ( ? ) /s ) t + ( ? ) )
For the second question mark I did 1/2.7 (the period) to get the [...]

# Answered: Homework 10 Question 7

Use energy conservation, we can find the initial vertical velocity of the water droplets with $$\frac 12mv^2=mgh$$
For the adjacent two water droplets, we find ||v_1|| and ||v_2||, which also gives ||\omega_1=\frac {v_1}{R}|| and ||\omega_2=\frac {v_2}{R}||.
Between the two droplets, the wheel rotates for a complete circle so i [...]

# Homework 10 Question 7

A bicycle is turned upside down while its owner repairs a flat tire on the rear wheel. A friend spins the front wheel, of radius 0.342 m, and observes that drops of water fly off tangentially in an upward direction when the drops are at the same level as the center of the wheel. She measures the height reached by drops moving vertically (see [...]

# Answered: HW 10 Question 9

More Physics Homework Help at LeiAcademy.
The blocks are going to slide with acceleration to the right and down the ramp.
Do free body diagrams for the two blocks and the pulley. Define the positive direction being clock wise for the rotation and towards right and down the ramp.
For block 1, you would have 4 forces with ||N_1=m_1g|| and ||f_ [...]

# Answered: HW 10 Question 6

For this question, you first use energy conservation to find the velocities of each of the blocks before the collision. The collision is elastic, so both kinetic energy and momentum are conserved. The calculations are complicated so you can use the elastic collision equations directly to find their final velocities.
$$v_{1f}=\frac {m_1-m_2}{m [...]

# Answered: Hw 11 Problem 5

For simple harmonic oscillation, you have $$x(t)=Acos(\omega t+\phi _0)$$ $$v(t)=-\omega Asin(\omega t+\phi _0)$$ $$a(t)=-\omega ^2Acos(\omega t+\phi _0)$$
Set ||t=0|| as the initial state and we have: $$x(0)=Acos(\omega 0+\phi _0)$$ $$v(0)=-\omega Asin(\omega 0+\phi _0)$$ $$a(0)=-\omega ^2Acos(\omega 0+\phi _0)$$
To find ||\omega||, divide t [...]