Recent content by fro

Here is a slightly different one: Water flows at 12m/s with pressure of 3X10^4 N/m^2. If the pipe widens to twice its original radius, what is the pressure in the wider section? I am trying to plug everything to Bernoulli's principle. Problem is that both the pressure and radius is unknown...

Thanks, I got it.

Liquid flows through a 4cm diameter at 1m/s. There is 2cm diameter restriction i n the line. Find fluid velocity in restriction. p_1+ \rho\cdot g \cdot y_1 + \frac{1}{2}\rho \cdot(v_1)^2 = p_2+ \rho \cdot g\cdot y_2 + \frac{1}{2}\cdot \rho\cdot(v_2)^2 I know I have to use Bernoulli's...

Need some verification. I am trying to drive a 9 inch nail into a log with a 10kg hammer, raising it to a maximum height of 1m and then gain momentum as it is swung down. a. Energy of the hammer immediately prior to striking the nail: 10kg\times10\frac{m}{s^2}\times1m = 100J. b. If 20% of...

Yes, we have been introduced to conservation of energy. However, I am not sure how to apply it in this particular situation. Are you suggesting that KE_i = KE_f + PE_f - 1500J?

So I add on the weight of the sled at the top (25m). Therefore, 60N+\frac{20kg\times10\frac{m}{s^2}\times25m}{25m} = 260N 360N > 260N, so the sled is able to climb up the hill?

Frictional force: \frac{1500J}{25m} = 60N Work done at the bottom: \frac{1}{2}\times20kg\times30^2\frac{m}{s^2} = 9000J Force: \frac{9000J}{25m} = 360N Since 360N is more than the frictional force, the sled should be able to ascend. Am I right on this part?

A 20kg sled has final velocity of 20m/s after it ascended a 25m tall hill. I calculated the initial velocity as 30m/s: v_0 = \sqrt{{20\frac{m}{s^2}^2}+2\times10\frac{m}{s^2}\times25m v_0 = 30\frac{m}{s} Now, if 1500J of work is being done by friction, how can I find the final velocity...

\Delta L = 20{kg} \cdot{m^2} \alpha = \frac{20kg\cdotm^2}{4s} = {5kg} \cdot {m^2} \tau = \frac{\Delta L}{\Delta \omega} \times \alpha \tau = \frac{20kg\cdot m^2}{\Delta \omega} \times {5kg} \cdot {m^2} So, how would I find \Delta \omega?

Sorry, I'm really confused about this. By your equation, angular acceleration should be 5kg*m^2. Not sure how and where to plug it into get the answer.

Maybe I could use L = I*w and T = I*angular acceleration? But I don't know what to do after that.

I have no clue as to how to solve this. Any hints/suggestions will be helpful. Problem: An object's angular momentum changes by 20kg*m^2/s in 4 seconds. What magnitude of torque acted on the object?

Somewhat confused about this problem and need some clarification: Problem: A particle of mass 0.01kg rotates with angular speed of 2 rad/s in a circle of radius 0.8m. What is its angular momentum? I think the answer should be: I = 0.01kg X (0.8m^2) = 0.0064 kg/m^2. w = 2 rad/s...

So you are saying that if I completely stopped pushing the box, then only gravitational force of 8N would be acting on the box? So there would be no external forces acting on the box at all?

If I stopped pushing the box into the wall, should not the applied force, frictional force and the normal force go away? Then only the gravitational force should be acting on the box. I am assuming that frictional force will go away since nothing is pushing the box to the wall anymore. Am I...