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Is it? If the ball is in your system then you must account for the upward force F applied on it.

The gravitational force and normal force from the earth do not cancel out though. There is a reaction force from the Earth equal in magnitude to the force applied to the ball by the man. This force is not cancelled out by the force of the cart pushing on the Earth because the Earth is not in...

Say that you take the Ball, Man, Cart system. The forces between the ball and the man, and the man and the cart are all action-reaction pairs and thus cancel out. But the cart also pushes on something outside of the system: the Earth. The Earth pushes back with a reaction force. Is this force...

Whoops.

Perhaps I'm misunderstanding the question, but if a skier goes off a jump with a known velocity, then their path through space is determined. Isn't it clear that he can only land in one particular spot on the ramp?

I think it's safe to assume that the useful energy in this case is the light being emitted from the bulb. Which would mean that 51 J of light energy is emitted. The other energy doesn't "remain in the circuit" it is lost to heat.

Not quite. The horizontal part of the triangle is her resultant velocity,or the vector sum of her speed with respect to the water and the current flow with respect to the shore. Her velocity relative to the water should be inclined at some angle with the horizontal, making it the hypotenuse of...

This is correct. I think you may be overthinking this. If she ends up 15.36 m downstream, that's how far she will have to walk to reach the market. Your intuition is right and you are correct that this is basically a trigonometry problem. Knowing that her speed relative to the water is 2.5...

Take the two equations I've quoted, and set them equal to each other. The stones have the same displacement: 1/2at1^2 = 1/2at2^2 + vi * t2 Now sub in a = 9.8 m/s2, t2 = t1 - 1.50 s, vi = 24.0 m/s, and leave t1 as the unknown. The squared terms can be subtracted out and you are left with a...

The equation for stone number one is just d = 1/2at1^2 Stone 2: d = 1/2at2^2 + v0 * t2 Remember that t2 is t1 - 1.50 and that the displacements of the two stones are equal.

No problem. :smile: If you need some help analyzing its motion, when it enters the magnetic field, just post again. :wink:

Well, when the electron is accelerated through two plates with a potential difference of 26000V, the electron has lost 26000V of electric potential. What does this tell you about the electric potential energy of the electron at the moment it exits the plates? In turn, what does that tell you...

Yes, "d" would be the correct response. Your head tends to want to continue in its current state of motion, so you percieve it being jerked back when the car accelerates. Your right about the shoe with the snow. And the apple hanging in the tree has essentially no "push or pull" or outside...

OK, so I'm getting the velocities as 1.0 * 108 m/s and 5.0 * 107 m/s. I'm pretty confident about this but the textbook only gives one answer (5.3 *7 10) m/s. There should be two velocities as an answer, so I think that must be an error. Is my answer correct?

Homework Statement Two electrons separated by a large distance are fired directly at each other. The closest approach in this head-on collision is 4.0 * 10-14 m. One electron starts with twice the speed of the other. Assuming there is no deflection from the original path, determine the initial...

Nothing. I posted the question verbatim, and there was no diagram either.

Homework Statement Two carts equipped with spring bumpers on an air track have an elastic collision. The 253-g cart has an initial velocity of 1.80 m/s [N]. The 232-g cart is initially stationary. What is the velocity of each cart after the collision? Homework Equations I know because cart 2...

How did you arrive at that? E.G. which equation did you use?

The force exerted by the balloon on the instruments is 98N, but what is the net force after accounting for gravity?

You'd need to use sine for the vertical and cosine for the horizontal. The sine of the angle is y / the hypotenuse, so y = sin71.2 * h. Similarly, x = cos71.2 * h. Hope that helps.

Well, this is kind of an odd question. I would think that a weighted average could work. E.g. multiply the first speed by 1/4, the second by 1/3, and the third by 5/12. Then add these numbers together. See if that gives you the right answer, I'm not entirely sure.

Just use the Pythagorean theorem, then use some right angle trig to get the direction angle.

For your first problem you need to find the total displacement (magnitude and direction) for the motion. Draw one vector pointing to the east and another to the south. Then draw a vector pointing from the tail of the first vector to the tip of the last vector. You need the length of this line...

I've essentially already shown him that: any of my equations can be rearranged to the form t (s or p) = ?

This won't work if you've got 3 unknowns. And your second equation still doesn't make sense. dp = vp (tp) ds = vs (tp -78.8) or: ds = vs (ts) dp = (vs + 78.8) Now you've got two equations that accurately describe the situation. You need to find the values for those two variables that work...

Your first equation is wrong. The distance that wave s travels will be given by multiplying the velocity of that wave times the time for the movement, not the difference between the two times. Your second equation would make sense if you had vs and tp in it. Remember that the time for the s-wave...

Try making two equations of the form d = vt. :wink:

Correct. Can you make two equations relating the distance travelled by both sound waves to each wave's velocity and the time it took them?

How do you express distance in terms of velocity and time?

You have two unknowns: Distance, and t_p (You know that t_s = t_p + 78.8). What do you do? Think back to some basic algebra. Hint: it has nothing to do with wavelength.