Answer:
See the answer below
Explanation:
The optimal conditions for high biodiversity seem to be a <u>warm temperature</u> and <u>wet climates</u>.
<em>The tropical areas of the world have the highest biodiversity and are characterized by an average annual temperature of above 18 </em>
<em> and annual precipitation of 262 cm. The areas are referred to as the world's biodiversity hotspots. </em>
Consequently, it follows logically that the optimal conditions for high biodiversity would be a warm temperature of above 18
and wet environment with annual precipitation of not less than 262 cm.
The variation in temperature and precipitation across biomes can thus be said to be responsible for the variation in the level of biodiversity in them.
I believe Intangibility is the answer! :P I hope this helps!
<span>Range = 88.5 Km/h - 94.5 Km/h</span><span>
</span>
-- Momentum is (mass) x (speed).
Object B has 1.5 times as much momentum as Object A has.
-- Kinetic energy is (1/2) x (mass) x (speed) .
Object B has 1.5 times as much kinetic energy as Object A has.
-- If they would both stop long enough to get on the scale,
Object B would weigh 1.5 times as much as Object A does.
Answer: The electron number density (the number of electrons per unit volume) in the wire is
.
Explanation:
Given: Current = 5.0 A
Area = 
Density = 2.7
, Molar mass = 27 g
The electron density is calculated as follows.

where,
= density
M = molar mass
= Avogadro's number
Substitute the values into above formula as follows.

Thus, we can conclude that the electron number density (the number of electrons per unit volume) in the wire is
.