Answer:
it is 2.2 m
Explanation:
because he goes back 2.2 m so 4.4 minus 2.2 equals 2.2
The fragment of an asteroid or any interplanetary material is known as A. METEROID
Answer:
481.76 J/mol
133.33 K
Explanation:
= Avogadro's number = 
Change in enthalpy is given by

Entropy is given by

Latent heat of fusion is given by

The latent heat of fusion is 481.76 J/mol
Melting point is given by

Melting occurs at 133.33 K
The waste products of a nuclear fission power plant can best be described as radioactive waste.
These are the by-products from the processes carried out that produce nuclear energy. This type of waste is highly dangerous. A lot of attention has to be paid to the collection and disposal of this waste as it must not reach any near by water bodies for example. It can be deadly for life.
A 5.00 A current runs through a 12 gauge copper wire (diameter 2.05 mm) and through a light bulb. Copper has 8.5*10^28 free electrons per cubic metre.
a) How many electrons pass through the light bulb each second?
b) What is the current density in the wire? (answer in A/m^2)
<span>c) At what speed does a typical electron pass by any given point in the wire? (answer in m/s)
</span>a) 5.0 A = 5.0 C/s
. Number of electrons in 5.0C = 5.0 / 1.60^-19 = 3.125^19
. 5.0 A ►= 3.125^19 electrons/s
b) A/m² = 5.0 / π(1.025^-3 m)² .. .. ►= 1.52^6 A/m²
c) Charge density (q/m³) = 8.50^28 e/m³ x 1.60^-19 = 1.36^10 C/m³
(q/m³)(m²)(m/s) = q/s (current i in C/s [A])
(m²) = Area
(m/s) = mean drift speed
(q/m³)(A)(v) = i
v = i.[(q/m³)A]ˉ¹
<span>v = 5.0 [1.36^10 * π(1.025^-3 m)²]ˉ¹.. .. ►v = 1.10^-4 m/s</span>