I believe the correct answer would be kimberlite. Diamonds are usually found in pipes 50 to 200 m across made of kimberlite. It is an igneous rock that is known to contain traces of diamonds. It is named base on the town where it was discovered which is Kimberley, South Africa.
Answer:
The pressure exerted by camel feet is <u>2000 N/m²</u>.
Step-by-step explanation:
<h3><u>Solution</u> :</h3>
Here, we have given that ;
- Force applied on camel feet = 4000 N
- Total area of camel feet = 2 m²
We need to find the pressure exerted by camel feet.
As we know that :

Substituting all the given values in the formula to find the pressure exerted by camel feet.

Hence, the pressure exerted by camel feet is 2000 N/m².

For an inelastic collision where coefficient of restitution,e, is equal to 0, the momentum is conserved but not the kinetic energy. So, there is addition or elimination of kinetic energy.
On the otherhand, when e = 1, like for an elastic collision, kinetic energy and momentum is conserved. Thus, the system's kinetic energy is unchanged.
Choice C.
That's when convection stops.
Explanation:
(1). Formula to calculate the potential difference is as follows.
= 
= 
= 
= 
= 38.7 volts
Therefore, magnitude of the potential difference between the two spheres is 38.7 volts.
(2). Now, formula to calculate the energy stored in the capacitor is as follows.
E = 
= 
= 
Thus, the electric-field energy stored in the capacitor is
.