Recent content by phucnv87

So what you've done for this problem? You just use the Newton's second law \vec{F} = m\vec{a} to solve these two problems :)

You should use the Newton's second law in this form F = m\frac{dv}{dt} Then you find v as a function of time t. Remember that dx = vdt. Calculate \int_0^{20}dx to find the answer.

Consider motion starting from rest over distance x along the incline: Mgx\sin\theta = \frac{1}{2}Mv^2 + 2(\frac{1}{2}mR^2)(\frac{v}{R})^2 2Mgx\sin\theta = (M+2m)v^2 On the other hand v^2 = 2ax 2Mgx\sin\theta = (M+2m)2ax a = \frac{Mg}{M+2m}\sin\theta

The equation must be M=Fd=I\ddot{\theta}[/color]

The first question, I think you're right. For the second one, find the angle \theta=\frac{1}{2}at^2 The revolutions are N=\frac{\theta}{2\pi}[/color]

Using the law of conservation of momentum[/color]

Centrifugal force is in an opposite direction to the centripental force \vec{F_C}=-\vec{F_r}[/color]

\Delta E=\frac{hc}{\lambda_0}(1-\frac{\lambda_0}{\lambda}) 1-\frac{\lambda_0}{\lambda}=\frac{\lambda-\lambda_0}{\lambda}=\frac{\frac{\lambda-\lambda_0}{\lambda_0}}{1+\frac{\lambda-\lambda_0}{\lambda_0}} Consider \frac{\Delta \lambda}{\lambda_0}=\frac{\lambda-\lambda_0}{\lambda_0} And we...

\large F_{\text{friction}=\mu mg\cos\theta=(\tan\theta-\frac{a}{g\cos\theta})mg\cos\theta=...

The answer is tan\theta=\mu[/color]

It stand for the acceleration of a point at the top of the cylinder.[/color]

Consider \mu=0.45, k=0.32, M=28kg, m_0=1.35kg, m=m_{Sand} a) (m_0+m)g=\mu Mg b) (m_0+m)g-kMg=(M+m_0+m)a[/color]

If all of the springs are on a line, the constants is \frac{1}{k_1}+\frac{1}{2}+...+\frac{1}{k_n} And if all of the springs are parallel each other, the constant is k=k_1+k_2+...+k_n[/color]

That's right, and the equivalent constant is \frac{1}{k}=\frac{1}{k_1}+\frac{1}{k_2}+...+\frac{1}{k_n}[/color]

No, no. If the force apply to the 1^{st} spring, the equation will be F=k_1\Delta l_1[/color]