Recent content by furtivefelon

There is one problem in the problem set i don't know how to approach it.. Problem: While an artificial diffraction grating is used to study a spectrum of a light beam, electromagnetic waves with wavelengths much less than the wavelength of the visible light are used to study the natural...

After some more thinking, i figured that since the ball bounces and at one moment, it deforms as it hits the ground. Since the internal pressure increases because of the smaller volumne, the pressure increases accordingly, and temperature increases as well.. However, i don't know how to...

hi everyone! I have the following problem: A volley-ball of mass m=200g and volume V=8L is inflated ith the overpresure of delta P = 0.2x10^5 Pa. A ball was thrown upwards to the height of h=20m, fell down, hit the ground, and bounced at approximately the same height. The temperature of...

mmm.. just noticed that my other thread appear to not work for some reason.. anyhow, here is a resubmit: Problem: A heat engine is a decive that uses heat to perform work. the Carnot ideal heat engine runs through a cycle, which consists of a isothermal, adiabatic, again isothermal, and...

I still don't quite understand how to calibrate the scale to measure the acceleration.. I do imagine however that the bubble will move backwards as the tube accelerate forward (however, according to lightgrav, it will accelerate forward??).. Though, is there a reliable way of calibrate the scale...

hi, this is the last question I'm having trouble with.. the question is: Fine sand is sprinkled on the horizontal membrane, which vibrates in the vertical direction with frequency f = 500 Hz. The sand particulates bounce at the height of 3 mm over the level of the position of equilibrium of...

i get that part, but then, the question asks, "How many watts contains the "Wonderland's watt"?".. How would i find the wattage of the "wonderland watt"? Don't i need at least the initial temperature of the water before i can find the "wonderland watt"?

So to calculate joules per second, i can use Q=cm(delta T)/delta t right? I know from density of water that 1 liter = 1kg of water.. and 20 minutes = 1200 seconds, therefore, Q=(4.186 J/g*C)(1000g)(2)/1200s = 6.97Watt.. Though, if I'm right so far, i don't really know the delta T is when the...

mmm.. i don't quite understand how that equation will help me, that is, finding the difference in energy..

sorry about that, i wrote it in response to Pengwuino, and after reading your post, i understood everything :D I didn't make the connection that both are energy transfers, now i did, both quantity makes sense now :D

Now looking at the physics book I've got, i see that they give two equations for calculating heat, one is cm(delta T), the second one is c(delta T).. I'm not sure about the practical differences between "specific heat capacity" and "heat capacity".. The only mass i know from the question is...

mmm.. i got that notion after looking at the power equation, P = W/delta t, as there isn't exactly any relationship that i can find between power and heat.. Further, as work and heat has the same unit, and you can change temperature in a body by doing work to it (such as compressing air), also...

there is one thing i don't understand while in my reading.. The unit of Work is Joules, and the unit of Heat is joules as well.. If those two are interchangeable, then this problem becomes quite easy.. what exactly is the difference(if any) exists between work and heat?

erm.. since we only started learning past Newton machanics, i have to do my own research, just started reading on Thermodynamics (should've done that a long time ago).. Just wondering, what does IOW mean in ur first post? and thanks a lot for responding :smile: i really appreciate it :smile:

that's the exact wording of the question given to me.. However, from my understanding, that the watt inscribed on the cover of the heater isn't the "real" watt, rather, a factory specific watt, so I'm guessing they want me to find the conversion between this "wonderland" watt and normal watt...