Recent content by Victoria_235

Good evening, Here is my last attempt. Thank you very much for your help .

Ok. I have tried again. As you said, at first the particle should be moving with the arrow, because of the positive slope, then should reach some point at t1, which is the maximum distance to O. After t1, is moving against the arrow, until t1, which is in the origin O (where the arrow begins)...

In A, I thought speed is increasing from origin to t1, were it stops. From 0 to t1, I guess is going from O, in the trayectory to some point at t2. In fact, I think I don't understand it... because how it is in my mind, in t2 should be again in O (in the trayectory draw).

Ok, I think I have something... It is ok? Here is my attempt. Thank you very much for your help :)

So, I understand that the equations (4), about dinamic motion, would be: \begin{equation} T_1 -m_1 (\vec{g}- \vec{a_0}) =m_1a_1\\ m_2 (\vec{g}- \vec{a_0}) -T_2 =m_2a_2\\ \end{equation} When the lift is going up. Because someone inside the lift would fell the gravity harder if the lift is going...

Homework Statement Two blocks are attached by strings of negligible mass to a pulley with two radii R1=0.06 m and R2=0.08 m. The strings are wrapped around their respective radios so that the masses can move either up or down. The pulley has a moment of inertia I=0.0023 kgm2 , and is supported...

Thank you! I think I understand it. When the graph (S, t) is in the negative part, it is moving against the reference system, I understand that in the opposite direction to that indicated by the arrow in the drawing on the right.

In the figure on the right is represented the trajectory, and what the problem asks is to draw on it the components of the acceleration and the direction of the velocity. Also say if the speed is greater in t1 or in t2. Yes, I upload the original plot. Thank you in advance. Thank you.

Homework Statement For each of the following cases, represent on the trajectory, the velocity and tangential and normal components of the acceleration at times t1 and t2 of a mobile that moves according to the graph of s (arc) versus t (time) on the left. The point O on the trajectory marks the...

Sorry, may be not too much clear in the last image, might here is better.

I have put AP distant as function of radio and radio/2. Because of the cosine theorem, ##c^2 = a² + b^2 - 2abcos\phi##, I wrote ##L^2 = (R/2)^2 + R² - (2R^2)/2 cos\phi## --> then, ##R \sqrt{ ((\frac{5}{4})- cos)\phi)}##, where L = AP. So, analyzing the energy in the points A and B, (see the...

Hello, thank you very much for your response. We could give "AP" as function of R, where R is the radio. With law of cosines, we have that AP = R\sqrt \frac{5}{4} - cos\alpha , where alpha is the angle between OA and AP. This way, we could give potential energy as function of angle. But, what...

Homework Statement Here is the problem. A small ball of mass m moves without friction attached to a cwire. The ball moves under the action of gravity and a spring of constant k whose other end is fixed in the point A, with OA = R / 2. The spring is tension free when the spring is R/2 to...

I am student, I am interested in optics