CaptainBlood

OK, thanks, I did it.

Two cars start moving form the same place. One goes north at a rate of 50 mi/hr, while the other heads east at a rate of 30 mi/hr. At what rate is the distance between the two cars changing exactly two hours later? I thought that the equation for the velocity vectors is: r(x) = 30xi + 50yj and thus distance d two hours later is given by: d = √(30x² + 50x² ) and dd/dt = ∂x/dt + ∂y/dt and after solving the equation I just plug in 2 for x and y to get what time the distance changes at the two hour mark. Is this right?

Ahoy, OK, thanks 4/5 I can't believe I didn't see that. Arrrhhhhhhh

If the product of two numbers if 5, and the sum of these same two numbers is 4, what's the sum of the reciprocals of these two numbers? I tried: let's say we have two numbers a and b, then a*b = 5 a = 5/b since their product is 4, we get b + 5/b = 4 or b^2 -4b + 5 = 0 but this has no real solutions just imaginary because this function is > 0 for all b I can't factor it and the quadratic formula doesn't work since 4^2 < 4*1*5 also the derivative is 2b -4 = 0 so b = 2 is the minimum since second derivative is positive but that doesn't help either. What do I do?

You can't just arbitrarily remove numbers (3 in your case) in the factored part of the equation and plugging them into another equation. Just think that if you remove a number from factored part of the equation, you are affecting the whole factored part of the equation with that number, in other words that number has an effect on all parts of the factored part of the equation.

Let f(x - 2) = x^2 + 2 Find f(3) Do I just sub 3 for x and get f(1) = 11? or do I do it this way: x - 2 = 3 x = 5 , therefore f(5) = (5)^2 + 2 = 27? or just let x = 5, then f(5 - 2) = f(3) = 5^2 + 2 = 27?

Personally I find the above line hilarious especially given the fact that he asked this question several times on multiple forums.

Thanks Doc. P.S. I guess I felt that I cheated because the problem seemed too simple, also I don't know how to include the free body diagram so everybody can see it that's why I tried to explain the best I could. I'd appreciate if someone could tell me how I could do that.

OK, thanks imatfaal.

Right, v, v0 and x is all we have given, didn't really think this one through, sorry.

First, since you know velocity solve v = v0 + at for time. Then, use x = x0 + v0t + 1/2 at^2 solve this for acceleration and the figure out what it is considering the gravity is 9.8 m/s^2.

I have 3.43 GPA at the community college, because in the beginning I didn't care about my GPA and got a C in Biology and an F in Chem I and a couple of B's, but I retook Chem I and got all A's since then. Will my low GPA affect my chances to get to medical school? I am planning to get all A's and at the 4-year University, but I'm not sure if they will count my GPA form community college.

Don't worry about it I solved it already.

Ok, I did it. Thanks a lot for your help I really appreciate it.

Great, thank you very much DrRocket, I gotta run now but I'll try to solve this a little later. This problem is given in the first semester Physics course and unfortunately before we learned about rotational motion besides the ar = v^2/r , I'm undergrad.

I'm sorry, I don't understand what is t and what is n in ωt = 2n pi. Also why are you using ωt in x = r cos (ωt) instead of theta?

An object is moving counterclockwise in a circle of radius r at constant speed v. The center of the circle is at the origin of rectangular coordinates (x, y), and at t = 0 the particle is at (r, 0). If the angular frequency is given by ω = v/r, show that x'' + ω^2 r = 0 and y'' + ω^2 r = 0 Attempted Solution:If particle is at (r,0) then r = x we know: ar = v2/r = ω^2 since velocity is constant ar = so x'' has to equal 0 thus x'' + ω2 r = 0 the same argument can be applied to y'' + ω^2 r = 0 proving the second statement Is this correct? Is there a better way of showing the two statements are true?

Unfortunately, what everybody is saying is exactly right, if you want to learn about astrophysics you need to dedicate your life to it. That means becoming a student for 5 years undergrad and 4 years graduate, full time. Only then do you have a shot at a relevant job. So if you really want it, is it is feasible, don't let anyone tell you it isn't. But it means quitting your job, studying full time while getting financial aid, and then when you get to graduate school, they'll pay you just enough to survive, but still enough to continue your studies. Make sure you do well in undergrad or they won't take you to graduate school. Work your butt off and eat noodles in the mean time and don't worry time flies, especially when you're studying. Oh yeah, don't fall in love or you'll never get there, unless she's rich. Good Luck.

Astrophysics. First learn physics, then you'll be ready to learn astrophysics.

Try to do e^(e^2lnx) = e^(ln(e^(2x^3)) then you should get e^2 = 2x^3 and solve for x

There are four masses hanging by a rope from the ceiling in the simplest arrangement possible, mass 4 is attached by the rope to mass three right above it, mass three is attached by a rope to mass 2 right above it, mass 2 is attached by the rope to mass 1 right above it and mass one is attached by the rope to the ceiling. So the masses are hanging vertically from the ceiling attached by the rope. Two of the tensions and three of the masses have been measured. We know: T1 T2 m1 m2 m3 Show that the fourth mass can be expressed as m4 = (m1T2/T1 - T2) - m2 - m3 Solution: We know that m4g + m3g + m2g = T2 so m4 = (T2/g) - m2 - m3 since multiplying the first term by m1/m1 is the same as multiplying the term by one, we get m4 = (m1T2/m1g) - m2 - m3 using the fact that T1 - T2 = m1g and substituting this equation in the denominator we get m4 = (m1T2/T1 - T2) - m2 - m3 QED Is this right? Did I answer the question properly? Just seems like I cheated. If you can point me in the direction of a better answer I'd greatly appreciate it.

because y gives me the equation for the parabolic trajectory of the ball kicked at angle theta I made a mistake earlier in my algebra, but I took a look at my derivation again and it worked out beautifully.

I know how to solve quadratic equations, it's just that when I try to solve this quadratic equation, the answer that I get is wrong and completely confusing. Here's the equation that I'm trying to solve y = y = (tan θ) x - g/(2(v0) (sin θ)^2) x^2 which describes the parabolic trajectory of a particle--in this case a soccer ball. Since theta is given to us and y = 0 when the ball hits the ground I was able to solve the quadratic now, but I don't know I tried to use the value I get for x0 , when I solved the quadratic, and plug that into the equation x0 + xp = d which is (cosθ sinθ (2v0)/g) - x0/v0cosθ = d and i get -1 = (gd / 2v0 sin θ0) - v0 cos θ0 which is close to what I'm trying to show but still not quite there.

A soccer player kicks a ball to his teammate, who is a distance d away. Even though the kick launches the ball with speed v0 and angle θ0 , the teammate knows it will not travel far enough to reach him before it lands. So as soon as the ball is kicked, the teammate begins running toward the ball. If he is to meet the ball just before it hits the ground, show that his average speed must be vp = (gd / 2v0 sin θ0) - v0 cos θ0 where g is the acceleration due to gravity. Neglect air resistance. I understand that t0 = tp and that the distance that the ball travels depends on the angle at which it is kicked. I solved for y to describe the trajectory of the ball in terms of angle theta and I know that xp = vp*t , then my plan was to solve the trajectory equation for x and use the fact that x0 + xp = d but I can't solve the trajectory equation for x since it quadratic. What do I do?

Agent 007 has just stolen some blueprints from Dr. No, whose henchmen are now in vigorous pursuit of Mr. Bond. He sees his best chance of escape: a wooden boxcar. In a moment, James is in the car and moving along a straight railroad track at speed v1 . A sniper fires a bullet (initial speed v2 ) at it from a high- powered rifle. The bullet passes through both length- wise walls of the car, its entrance and exit holes being exactly opposite each other as viewed from within the car. Assume that the bullet is not deflected upon en- tering the car, but that its speed decreases by 20%. Show that the direction, relative to the track, from which the bullet is fired is given by θ = π − arccos ( 5v1 / 4v2 ) (Why don’t you need to know the width of the box- car?) From the information given, we know that the triangle that is being formed by the trajectories of the v1 and (.8) v2 is a right triangle in which sin theta = 5v1 / 4v2. We also know that the trajectory of the bullet before it reaches the boxcar is given by sin theta = v1/v2, so the combined trajectory is the vector sum of v2 and (.8)v2 which would give us the angle theta that we are looking for, but then I don't know how to proceed. Then I thought that that is not right because in the problem it states that the trajectory of the bullet didn't change, so while the bullet traveled at (.8)v2 speed the cart traveled enough distance for the bullet to hit the other wall at the middle point. That means that when the bullet is traveling along a hypotenuse with the speed (.8)v2 the cart is traveling up at speed v1 until they meet, thus the width of the cart doesn't matter. Does what I'm saying make sense? How should I approach this problem?