We shall consider two properties:
1. Temperature difference
2. Thermal conductivity of the material
Use a cylindrical rod of a given material (say steel) which is insulated around its circumference.
One end of the rod is dipped in a large reservoir of water at 100 deg.C and the other end is dipped in water (with known volume) at 40 deg. C. The cold water if stored in a cylinder which is insulated on all sides. A thermometer reads the temperature of the cold water as a function of time.
This experiment will show that
(a) heat flows from a region of high temperature to a region of lower temperature.
(b) The thermal energy of a body increases when heat is added to it, and its temperature will rise.
(c) The thermal conductivity of water determines how quickly its temperature will rise. If mercury replaces water in the cold cylinder, its temperature will rise at a different rate because its thermal conductivity is different.
Answer:
Light waves slow down as they travel from gas to solid.
Explanation:
The speed of light is often quoted with reference to a vacuum. Light travels fastest in a vacuum and in a gas.
When light travels through other media such as solids or liquids, the speed of light is decreased due to absorption and scattering of photons by molecules as well as remissions.
Hence the speed of light decreases from gas to solid.
Answer:
Explanation:
The Coulomb's law states that the magnitude of the electrostatic force between two charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them:
In this case, we have 
:
Answer:
W₂= 10000 N
Explanation:
Pascal´s Principle can be applied in the hydraulic press:
If we apply a small force (F1) on a small area piston A1, then, a pressure (P) is generated that is transmitted equally to all the particles of the liquid until it reaches a larger area piston and therefore a force (F2) can be exerted that is proportional to the area (A2) of the piston:
Pressure is defined as the force (F) applied per unit area (A)
P=F/A (N/m²)
P1=P2
Equation (1)
Data
W₁ = weight sits on the small piston
F₁ = W₁= 500 N
A₁ = 2.0 cm²
A₂ = 40 cm²
Calculation of the weight (W₂) can the large piston support
We replace data in the equation (1)
F₂ = 10000 N
W₂= F₂= 10000 N
Answer:
I'm guessing it would remain a mystery to us all
