The equilibrium temperature of aluminium and water is 33.2°C
We know that specific heat of aluminium is 0.9 J/gm-K, and that of water is 1 J/gm-K
Now we can calculate the equilibrium temperature
(mc∆T)_aluminium=(mc∆T)_water
15.7*0.9*(53.2-T)=32.5*1*(T-24.5)
T=33.2°C
Sound waves....................................
Given the following information we have 20 watermelons from mark and 10 fishes from kim therefore we add the longitude of Walmart to the latitude of sams club and end up with a total of 1,000 dish soaps then we convert that into inches which leaves us at 20,000,000 inches of cats then multiply that number to 10 giraffes and we get
1.989 × 10^30 kg and therefore the mass of the sun is 1.989 × 10^30 kg.
Answers are:
(1) KE = 1 kg m^2/s^2
(2) KE = 2 kg m^2/s^2
(3) KE = 3 kg m^2/s^2
(4) KE = 4 kg m^2/s^2
Explanation:
(1) Given mass = 0.125 kg
speed = 4 m/s
Since Kinetic energy = (1/2)*m*(v^2)
Plug in the values:
Hence:
KE = (1/2) * 0.125 * (16)
KE = 1 kg m^2/s^2
(2) Given mass = 0.250 kg
speed = 4 m/s
Since Kinetic energy = (1/2)*m*(v^2)
Plug in the values:
Hence:
KE = (1/2) * 0.250 * (16)
KE = 2 kg m^2/s^2
(3) Given mass = 0.375 kg
speed = 4 m/s
Since Kinetic energy = (1/2)*m*(v^2)
Plug in the values:
Hence:
KE = (1/2) * 0.375 * (16)
KE = 3 kg m^2/s^2
(4) Given mass = 0.500 kg
speed = 4 m/s
Since Kinetic energy = (1/2)*m*(v^2)
Plug in the values:
Hence:
KE = (1/2) * 0.5 * (16)
KE = 4 kg m^2/s^2