EvoN1020v

I'm not sure if you still want help with this but I recommend you to do this:

[math]\int cos^4xdx = \int cos^2x \cdot cos^2x[/math]

You know that [math]cos^2x = 1- sin^2x[/math]

So you have:

[math]\int (1-sin^2x) + \int (1-sin^2x) [/math], then let [math]u=sinx[/math].

Hope this helps you abit.

well the completely wrong answer i got to #2 was 4.98 m/s for the submarine and 12.78 or 8.89(both probably wrong) m/s for the torpedo

The velocity for the sumbmarine is 4.98 m/s which you got it right. ([math]V_s=4.981m/si - 0.00684m/sj[/math]). The speed of the torpedo is [math]V_T=3.413m/sj+14.38m/si[/math] or [math]14.78m/s[/math]. Your answer of 12.78 was close. Can you tell us how you got your calculations? That'd be appreciated.

Today is the day for a new question:

3rd question:

An electromagnetic launcher applies a force to 0.007 kg projectile that follows the illustrated force/time profile. How fast is the initially stationary mass moving at t=0.08s?

Good luck.

[math]cos^2(x)=1-sin^2(x)[/math]

You should know that [math]\int cosx = sinx + C[/math]

You need to know the u-substitution also.

I disagree with you because you don't input any velocity values in the formula of [math]F_f=\mu F_N[/math], therefore it's not dependent.

Yeah I will put up a new physics problem tomorrow. You're welcome to participate dumbman29.

[math]

F = \mu R

[/math]

 

Is the maximum frictional force normally, as 5614 says for a proper friction equation mu is velocity dependent.

You are implying that air resistance is included in the friction forces. For basic friction with air resistance negligble, velocity is not dependent.

Velocity won't affect the object's normal force because velocity is a vector that is independent from the normal force. You should know that normal force is always perpendicular to the object's centre of mass. The normal force of an object differs depending on the slope of the surface.

For instance, an object of 10 kg on a flat surface: 98.1 N

An object of 10 kg on a 45 degrees slope: 69.4 N

In conclusion, velocity does not affect the object's normal force.

Just a reminder: Three days left for the 2nd question before I put up another new question. (It will be easier, I promise you )

It's not lame it's the truth, your question is a 2 body conversation of momentum, I could do it but I don't want to put others who might enjoy it more from doing it.

 

It'd be like going to a primary school and answering all the teachers questions loudly from the back.

Then do you notice any students in the front answering the teacher's question?

TI-83 Plus has been treating me well for 2 years now. ($110 CAN taxes included).

That is the objective of my "Physics F-U-N doom" thread. I give you guys a question, and I give it a week. If nobody can get it, I will give the solution, and give a new question.

If you want to continue this thread, I can give out another question, and that's no problem with me.

That's a lame excuse, Klaynos. Everyone is welcome to contribute and participate.

Ok, I refrained from looking at other posts for the first question, here's what I got:

 

 

There are 3 forces acting on the object: The normal force [math]F_N[/math], the weight vector [math]W[/math], and the frictional static [math]f_s[/math]. Since acceleration only occurs in the x direction, we only consider [math]\sum F_x=ma[/math]. To find [math]f_s[/math], which is the only force acting in the x direction, I convert the mass to kilograms, and multipliy the normal force ( [math]F_N=40.9*9.8=400.82[/math] ) by the frictional coefficient of .9 to get 360.738 Newtons. Setting that equal to m*a and solving for a I get a=8.82m/s^2.

Good job hotcommodity.

b) calculate the impulse.

Do you mind explaining more about that? To determine the velocities of the submarine and the torpedo, you don't need to calculate the impulse. Just use the momentum conversation theory and relative motion theory.

Anyone else tried this question? Or is it too hard for you guys?

This question is related with conversion of momentum. Please leave the answer in i and j direction. (i.e. [math]V_b=2.3m/si -5.6m/sj[/math]). If you got the right answer, I will be really impressed, because this is a challenging university physics question.

2nd question:

A 50,000 kg submarine moving forward at 5 m/s launches a 100 kg torpedo. The ejection mechanism acts on the torpedo for 0.25s, at the end of which the torpedo enters the water with a velocity of 10 m/s relative to the submarine and at 20 degrees elevation to the orginial path of the submarine.

(a) Calculate the velocities of the submarine and torpedo just after ejection is complete.

Good luck.

You know what's funny? Last year in high school, my math teacher refused to teach me the Chain Rule. So I had to do the research myself, and I couldn't understand it, so I got really depressed. ender7x77, just get some sleep, and you'll feel better tomorrow.

In university, (actually around a month ago), I finally learned it, and I was really happy. It just takes time before you can understand - requires a lot of practices, that's all. Don't worry, you're fine.

thanks, you are very helpful.

Anytime.

This is what your originial function and the first derivative should look like:

You should be able to notice the change of slope at x=2.

The second derivative function shape is kinda very wierd, so I'll try it one more time. I can post the graph of the second derivative if you are interested.

Did you use the Chain Rule?

You know that the Quotient Rule is [math]\frac{f'g-fg'}{g^2}[/math].

On the [math]fg'[/math] part, you have to use Chain Rule to find the derivative of [math]g[/math]. ([math]2(x^3+x^2)\cdot3x^2\cdot2x[/math])

If you don't know how to do it, then it's out of your league. I didn't learn Chain Rule in high school myself. This is more of an university expertise. You can just tell your teacher it's an unacceptable question, or you could just leave it alone.

Yes, you're right. I did some research on "concavity", and it said that you have to find the second derivative of the function to find the point of inflection.

My calculation of the second derivative is different than yours. Since [math]f(x)=-20x^4+60x^2+40x[/math] and [math]f(g)=(x^3+x^2)^2[/math], you use the quotient rule.

I found the calculation of the second derivative to be: [math]\frac{240x^{10}+160x^9-880x^8-510x^7-200x^6+200x^5+40x^4}{(x^3+x^2)^4}[/math].

Such a painstaking calculation!! So you put [math]f''(x)=0[/math], and you should be able to find the value of x.

P.S. Is there any derivative tool on the internet so we can check if it's right?

From [math]x^3-3x-2[/math], I used synthetic division, and I got [math](x-2)(x+1)^2[/math], therefore the critical points are: [math]2,-1[/math].

I don't really understand why you had to find the second derivative of the function. Just input 2 and -1, in the original function, and you'll get the local minimum/maximum.

Note: If you want to know if the shape is concave in positive or negative direction, then you use test points between the critical points. Input the test points in the first derivative of the function, and you'll find the shape of the original function.

If you don't understand this, then I can give you some examples.

Evo, this is an international forum 96% of the world uses SI. SI is the official standard of the forum so pst the questions in SI or don't complain if thats what you get the answer in.

I disagree with you on the 96% part. Engineers in Europe use the U.S. metric system as well.

I understand that you guys want it to be in the international metric system - I got no problem with that.

I know how to solve it. You can use synthetic divsion to figure out the equation that the tree gave you: [math]x^3=3x+2[/math]

Therefore: [math]x^3-3x-2=0[/math]. I strongly recommend you to use synthetic division.

Calculation for the 1st question:

Goal: Determine the maximum decleration of the car so the TV won't fall off the roof.

Given: [math]\mu_s=0.9[/math], [math]m_{tv} = 90lb[/math].

Diagram:

Formulate: [math]F_{net}=ma[/math]

Solution:

From the Free Body Diagram, you can gather all the forces into Fnet. Using the provided y and x axes, you can determine if it's in the positive and negative direction. (i_hat for x, and j_hat for y)

[math]-F_fi +N_{tv}j-W_{tv}j = -mai[/math]

Now separate the i and j:

(1) [math]i: [/math][math]-F_f = -ma[/math]

(2) [math]j: [/math][math]N_{tv}-W_{tv}=0[/math]

(2) equation can be rearranged to: [math]N_{tv}=W_{tv}[/math]

Substitute (2) in (1):

[math](-\mu(W_{tv}) = -ma[/math]

[math](-\mu(mg))=-ma[/math]

Insert all the numbers that you have:

[math]-0.9(90lb(32.2ft/s^2) = -(90lb)a[/math]

The resultant acceleration or deceleration in this situtation is [math]29.0 ft/s^2[/math].

I'll have a new problem for Monday.

Video consoles are for enjoyment ONLY.

Very sad, I know.

*A quiet clapping*