if u have 2 cups. 1 with hot water and other at 32 degrees and drop a penny at the same time which fall 1st?
487300
the hot one coz of the increased energey i think
491198
Easy!! we did this in science a while ago, let the pennies soak in pure lemon juice, it works really well.
992192
1. second\n2. first\n3. 4100\n4. mary
454506
100 degrees celcius\n212 degrees Fahrenheit
765294
First the specific heat of water is not needed because both items mixed are water and would have the same specific heat so just use the weight.\n\nso you have 60 * 80 + 40 * 25 = 100 * x    where x is the new temperature in degrees C\n\nIf the items being mixed were different then use the specific heat of each item to calculate the number of joules in each item which will give the total number of joules in the combined item.
744521
One and a half cups \n\nOne cup of water is equal to 8 ounces
278524
Celsius or Centigrade
341668
Water evaporates at 100 degrees, celsius, so the liquid water you touched wouldn't be any hotter than that. The metal, on the other hand, does not. So it would be as hot as the object which is heating it.
602506
1. Water boils- 212 degree F/100 degree C\n2. Normal Body temp- 98.6 degree F/37 degree C\n3. Water Freezing temp- 32 degree F/ 0 degree C
992414
Improve your spelling, unless "frehiet" is sic. If it isn't, and means "fahrenheit", then the answer is F = (9/5 * C) + 32.
481420
first, 1 degree celcius is NOT 32 degrees fahrenheit. ZERO degrees celcius is 32 degrees fahrenheit, but niether here nor there....\n\nnth term will be \n\n(n) + (n-1) + 31. \n\nso if its the first term, then\n(1) + (0) + (31) = 32.....or for the 2nd\n(2) + (2-1) + (31) = 34\n(3) + (3-1) + (31) = 36\n(4) + (4-1) + (31) = 38  .....  and so on. \n\nso for the "42nd" term\n\n(42) + (42-1) + (31) = 114\n\nso for the Nth term\n\n(N) + (N-1) + (31) = .....
780627
WATER BOIL AT:  27*C at 26.739 (pressure(torr)  thru 30*C at 31.824\n pressure(torr)
461815
you do the experiment yourself places a cup of cold and hot water in your freezer, the hot water will freeze first, the\nhotter the water the faster it will freeze
947272
1 penny=2.5 g\n\n100pennies=250g
610262
1 cup of dry corn starch and a 1/2 cup of water
78350
When the metal enters the water, it begins losing heat to the water. At the same time the temperature of the water rises. This process continues until the temperature of the metal and the temperature of the water are equal st which point (30.6oC) no net flow of heat occurs. The heat lost by a gained system is given by (mass)(Specific heat)(change in temp.)First calculate the heat gained by the water, then calculate the specific heat of the metal.\n\nTemp rise in the water - \n   (change in temp) 30.6 - 25.0 = 4.4\n\nHeat gained by water - \n              \n   (185.0)    (4.184J)/(g oC)  (4.4) = 3.4*10 pwr3 J\n\nThe metal dropped into the water must have the same temp as the water (30.6)\n\nTemp drop by the metal - \n   (Change in temp) = 178-30.6 = 147.4\n           \n Specific heat = \n(heat)/(mass)(change in temp)\n\nSpecific heat of the metal =\n(3.4*10 pwr3)\n(36g)(147.4)\n             \n=0.640 J/g oC\n\nhope this helps. good luck
858861
1) The gravitational potential energy of water at the top of a waterfall is first converted to kinetic energy (falling) and then converted to both acoustic (sound) and thermal (temperature) energy when it hits the pool of water at the bottom. The temperature of the water in the pool will be slightly higher than the temperature of the water at the top of the waterfall.\n\n2) The thing to remember for this problem is that the energy in a system before some action = the energy in that system after some action:\n\nP0 + K0 + Th0 = P1 + K1 + Th1\nWhere P is gravitational potential energy (mgh), K is kinetic energy (0.5mv^2), and Th is thermal energy (leftover).\n\n22*9.81*4.5 + 0 + 0 = 0 + 0.5*22*2.5^2 + Th1\nSolve for Th1:\nTh1 = 22*9.81*4.5 - 0.5*22*2.5^2 =  902.44 Joules of thermal energy were generated due to friction.\n\n3) Split this problem into two cases, the first 10m and the second 10m. Find the velocity of the crate after the first 10m:\na = F/m = 200/70 = 2.86 m/s^2\nv = sqrt(2*a*d) = sqrt(2*2.86*10) = 7.56 m/s\n\nFor the second 10m, the force of friction on the crate is:\nFf = 0.3*70*9.81 = 206 N\n\nThe net force on the crate is 6N in the direction opposite of movement. The acceleration for the second 10m is:\na = F/m = 6/70 = 0.086 m/s^2\nvf = sqrt(vi^2 + 2*a*d) = sqrt(7.56^2 + 2*-0.086*10) = 7.45 m/s\n\n5) Since watts (W) are measured in J/s and joules (J) are measured in N*m, a watt is the same as N*m/s. Convert lb to N and ft to m to go from horsepower to watts:\n\n550 ft*lb/s * 0.3048 m/ft * 4.45 N/lb = 746 N*m/x = 746 W\n\nPS - You have in your problem description that horsepower is equal to 764 W but that should be 746 W.\n\nI hope this helps!