Recent content by Dorothy Weglend

Ah, thanks arildno. And I also mistyped in my first post, I will correct that now. Thanks again for all your help.

I'm not sure why you changed the variables here. It still seems to me the areas must be equal. The original equations, y=x^n and y=x^1/n. the second one is equivalent to y^2 = x. Therefore the distance from the x-axis to the curve y=x^2 must be the same as the distance from the y-axis to the...

But how is that possible, if they are mirror images, the area above x^1/n must equal the area below x^n. Otherwise I don't see how they can sum to 1. What am I missing?

Thank you both, so much! It must be a mistake in the text. Now the problem seems solvable. To answer your question, it seems to me that the geometric argument would go something like this: the curve x^n and x^1/n are mirror images of each other, reflected about the line y=x (as they are...

Homework Statement By appealing to geometric evidence show that \int_0^8x^n\,dx + \int_0^1 x^{1/n}\,dx = 1 for n a positive integer. Homework Equations Fundamental theorem of calculus, power rule for integration. The Attempt at a Solution I integrated. For the first integral, I...

It sounds like something is missing, I agree. Without knowing what is stopping the coaster, then it's hard to say what the force is. Is it supposed to stop suddenly at B? That would involve talking about the impulse. Or if there is friction on the horizontal track, and it slows gradually...

Think of this in terms of work and energy, instead of forces. Dorothy

Consider a spring with 5 coils and F = mg hanging from the spring which causes a spring extenstion of 5 cm. That means that each coil will move 1 cm. Now double the number of coils while keeping the material exactly the same. Hang the same mass from the spring. Each coil will still move 1...

Hi Gyroscope, Yes, sorry. I think your solution is correct. Dorothy

I don't think you can use a scalar here. My experience is that a shadow "accelerates". When he is underneath the candle, there will be practically no shadow. When he is farther away, the shadow will be much larger. So it accelerates, so to speak. Dorothy

There are more coils with a longer spring, that means it should stretch farther given the same force, and the spring constant will be proportionally less. So you can use this to get an expression for the k of the longer spring. For part b, that's just a straightforward application of the 2nd...

If the bungee cord obeys Hooke's law, that means it is essentially a spring. You should be able to figure out the spring constant from the data given (body weight stretches a 5.00 m length by 1.30 m). I haven't tried to solve it, but it seems to me you would need this for part (a) also. Dorothy

If you can answer (b), that will help you answer (c). Dorothy

Assuming you are pushing parallel to the incline, the force is just F , the force of gravity that you are pushing against is the component of gravity along the incline, or F_g(\sin \theta). Dorothy

Good. Right. t = D/v Also good. Right again, t = D/v Why did you ignore the distance in your final calculation? Dorothy