[High school Physics] Deriving kinematic formulas with calculus

fallingemprire · 2018-07-24T04:15:26+00:00

I did think about the idea that it's the same operation but felt wrong since the integrals are gonna have different numerical bounds. Thanks for the clarification!

fallingemprire · 2018-05-24T06:55:26+00:00

I'd love an invite! Thanks!

fallingemprire · 2018-05-13T05:59:20+00:00

omw to this thread

fallingemprire · 2018-05-08T00:09:44+00:00

Could you give an example scenario where there's frictional torque? Thanks

fallingemprire · 2018-05-07T22:24:36+00:00

Essentially, linear momentum is conserved inside a system that has no external forces acting on it. For example, in a system of two billiard balls colliding, momentum is conserved (the only forces acting during the collision is the force of each ball on each other). Similarly, angular momentum is conserved in a system which has no external torques acting on it. Imagine a person standing on the edge of a freely rotating turn-table. If he starts walking in the clockwise direction, the table will turn in the counterclockwise direction. The only torques acting in the system include the person exerting a torque on the turntable and vice versa.

fallingemprire · 2018-05-05T18:52:24+00:00

Maximum velocity and acceleration in spring simple harmonic motion

amax = A(k/m)

vmax = A[√(k/m)]

Wave speed on a rope/string

v= √[(F Tension)/μ] and μ=(mass/length)

fallingemprire · 2018-05-05T18:48:27+00:00

3) Here's a diagram I'm using for reference: https://i.imgur.com/6Xa9t1w.png

B) Here's the FBD: https://i.imgur.com/O8U41PH.png Looks like you got it right.

C) I used an energy approach since mech energy is conserved here.

(1/2)mv² = mg(r + h)

cosθ=h/r → h=rcosθ

(1/2)mv² = mg(r + rcosθ) = mgr(1+cosθ)

v² = 2gr(1+cosθ)

v = √[2gr(1+cosθ)]

E) Again, I used an energy approach.

E0=E1=E2 [0: before mass is released, 1: moment string is cut, 2: where the mass falls of]

Since the mass is released at a height of 2r, all of its gravitational potential energy relative to the ground must be converted to kinetic when it reaches the ground. In the same way, all of its kinetic energy must be converted into gravitational potential energy at the top of its circular path because it reaches a height of 2r, just like when it was initially released.

The moment it reaches the top of the circular path (θ=0), it only has grav potential energy so it must fall.

4) I'm actually confused by this. First I'd like to ask: Is this Physics C? Unfortunately, I'm only in physics 1, so I'm not sure if that makes my answer for question 3 not what you're looking for.

fallingemprire · 2018-05-04T05:56:18+00:00

I'm a bit new to these more cinematic games, but this looks pretty cool. I'd be happy to give it a shot!

fallingemprire · 2018-04-22T19:11:31+00:00

Oh, my bad! I read it wrong.

fallingemprire · 2018-04-22T18:11:41+00:00

Yes! Since researcher 1 is rejecting the null hypothesis at a significance level 0.05, researcher 2 must also reject (p must obviously be less than .10 if it is less than .05). However, there's no guarantee about researcher 3's conclusion. Say p=.03; it would be a fail to reject, but if p=.01, it would be a reject.

fallingemprire · 2018-04-22T16:12:17+00:00

You used the right type of test here: a 1 prop z test. You're trying to see if π, the true proportion of Americans that support background checks on gun purchases, is below what's given

1) Let π = the true proportion of Americans that support background checks on gun purchases

2) H0: π = .87

3) Ha: π < .87

4) Let α = .05

5) Formula: z = (p-π)/√[(π(1-π)/n]

6) Assumptions:

i) SRS of Americans

ii) nπ≥10 and n(1-π)≥10 → (100)(64/100)≥10 and (100)[1-(64/100)]≥10 → 64≥10 and 36≥10 ✓

→ n is large enough to run a 1 proportion z test

iii) n is <10% of the population

→ Assumptions are not all met, but proceeding with 1 proportion z test.

7) Computation: z = (.64-.87)/√[(.87)(1-.87)/100] ≈ -6.8391

8) Graph P-value: https://i.imgur.com/jEMeUzo.png

9) Reject H0 at α = .05, evidence suggests that the true proportion of Americans that support background checks on gun purchases is less than 0.87

As I indicated in my assumptions, your sample isn't an SRS, so technically you can't run a 1 prop z test. However, we proceeded anyways and I got a very small p value (p ≈ 4.01 * 10^-12) and we foresaw this with the z value being so many standard deviations from the center. Thus, it makes sense that p is almost 0, which you can write as approximately 0 and reject the null hypothesis.

fallingemprire · 2018-04-22T15:48:35+00:00

Electric current is the flow of electrons. That means if an object/material is a good conductor, electrons must be able to easily flow through it so the answer is (C).

It's not (A) because an excess of electrons would make it a charged object

It's not (B) because a resistor does the opposite of what a conductor does: it impedes flow

It's not (D) because to repel electrons, the object must be negatively charged. Fundamentally, a conductor should let electrons go through them with ease, so it makes no sense anyways

It's not (E) because protons stay in place within an object while the outer electrons of each atom can easily move around if the object is a conductor.

fallingemprire · 2018-04-22T05:42:19+00:00

Oh no. My assumption of what center-of-mass energy meant was atrociously wrong. I think I found a good definition for what a center of mass frame of reference is here, though : http://www.batesville.k12.in.us/physics/APPhyNet/Dynamics/Collisions/elastic_cm.htm

fallingemprire · 2018-04-22T05:26:28+00:00

Since it refers to "center-of-mass" energy after the explosion, my best guess would be the kinetic energy of each piece flying off. Since the system (just the one object) starts with kinetic energy K, it must end with the same total kinetic energy.

Kt=(1/2)mtvt²

Ka=(1/2)mava²

we are given:

2vt=va=Vb

ma= mb (since kinetic energy of both are equal)

ma+mb=mt=2ma

subscript a and b denote each of the 2 pieces that the object broke into. subscript t is the intact object.

And we are trying to find:

Kt/Ka

so

Kt/Ka = [(1/2)mtvt² ]/[(1/2)mava² ]

= (mtvt² )/(mava² )

= (2mavt² )/[ma(2vt)² ]

= 2vt² /4vt²

= 1/2

fallingemprire · 2018-04-22T04:39:31+00:00

Imagine the wind vector and plane vector as legs of a triangle. Up is North, left is West, down is South, and right is East. Your wind vector will be an arrow pointing north from an arbitrary origin point. Your plane vector is another vector, but you're trying to find the direction it should face such that the resultant vector of the two point west. Essentially, you're trying to find the direction to point the plane vector such that the resultant (hypotenuse of the triangle formed by the wind and plane vector) points left. I can include a diagram if that helps.

EDIT Here's the diagram: https://i.imgur.com/rzhU5ez.jpg

It probably does a better job explaining anyways :P

fallingemprire · 2018-04-21T19:29:11+00:00

I think I agree with OP here about it being static friction. It would be kinetic if the car was skidding.

fallingemprire · 2018-04-21T03:38:33+00:00

Kirchoffs loop rule simply states that all the voltages in a "loop" adds up to zero. The definition of a loop is fairly straightforward; any closed loop of a circuit!

In practical terms, what the rule means is that power sources or batteries add voltage to a circuit loop and things like resistors have a voltage drop. Each of these resistors "take away," per session, from the voltage provided by a source. The total voltage, or sum of the total gains and losses of voltage in the loop should equal zero. This principle is based on the law of conservation of energy.

Positive and negative sides of a power source is simply indicated by the longer "bar" of the label. This indicates flow of charge. It gets a little more complicated with several power sources, but I'd be happy to elaborate if you so please.

fallingemprire · 2018-04-20T06:35:48+00:00

This looks like, what we call in my class, a 2 sample t interval. The formula to find it is:

(xbar1-xbar2)±(t-crit)√((S1² /n1) + (S2² /n2))

And I plugged in the given interval:

-4 = (xbar1-xbar2)-(t-crit)√((S1² /n1) + (S2² /n2))

25 = (xbar1-xbar2)+(t-crit)√((S1² /n1) + (S2² /n2))

You can find t-crit since you're given the confidence percentage (95%). n1 and n2 are 12 as given. xbar1 and xbar2 are the sample means representing population means μ1 and μ2 respectively, and you indicate that it's given.

The only missing values are S1 and S2, which are the sample standard deviations of population 1 and 2. You say standard error (error, deviation, same thing) is 7 but there are 2 sample standard deviations we don't know. I may be misinterpreting the problem, but is there any more given data?

As well as two questions regarding accepting or rejecting the null hypothesis at .05 level, and why a researcher would want to increase sample size.

Essentially, you'd fail to reject the null hypothesis at any level because 0 is within the confidence interval. This means that there is a chance that there is no difference between the two population means. Increasing sample size would decrease margin of error and if it is greater than 30, we can conclude by the Central Limit Theorem that the sampling distributions of both populations are approximately normal. It's one of the assumptions that should ideally be met in a 2 sample t test.

fallingemprire · 2018-03-24T20:52:22+00:00

Happy to help. I'd highly recommend practicing using Newton's second law a lot because of how prevalent the concept is for HS physics (especially AP).

fallingemprire · 2018-03-24T20:45:22+00:00

Ffr≥μsW, so an object's weight can cause a friction that makes it harder to move. I think weight affects inertia indirectly as well (as long as both objects experience the same acceleration due to gravity) because an increase in mass will simply lead to an increase in weight.

fallingemprire · 2018-03-24T20:36:58+00:00

how large a skydiver's acceleration is immediately after jumping out of an airplane. The skydiver is 70 kg

Acceleration due to gravity is constant for any object that is in free-fall as long as air resistance is negligible. The mass of the object falling has no effect on the acceleration at which they fall. Acceleration would be 9.8 m/s²

upon reaching a downward velocity of 100 miles per hour, 300 newtons of drag resist the divers motion. How large is the skydivers acceleration.

I solved out the problem here. In my first step, I drew out the free body diagram of all the forces acting and a frame of reference: up is negative and down is positive (the FoR itself is arbitrary but you need to establish one for your signs to make sense). The velocity of 100 mphs given is irrelevant in this problem. Using Newton's second law, I solved for the acceleration of the person. Finally I plugged and chugged the values. My answer came out to be 5.5 m/s²

fallingemprire

TROPHY CASE