No, the density of diamond and graphite would not be the same
Explanation:
What is density?
Density is the amount of substance per unit volume. It is simply mass divided by volume.
Density is greatly influenced by the way substances are packed.
Substances that are well packed will have lower volume for the same amount of matter than another that is poorly packed.
- The carbon atoms in graphite are poorly packed. They are arranged layers upon layers.
- Diamond carbon atoms have a cross-linked networked pattern. They are well packed.
- For the same mass of matter, graphite will take up more space than diamond.
Since:
Density = 
The one that has a lesser volume will have a higher density.
Therefore diamond will have a higher density.
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Kinetic energy is energy of motion. Pick choice-A, at the top of the swing, where she stops moving & then goes the other way.
Answer:
The person is 187[m] farther and 70° south to east.
Explanation:
We can solve this problem by drawing a sketch of the location of the person and the truck, then we will draw the displacement vectors and finally the length of the vector and the direction of the vector will be measured in order to give the correct indication of where the person will have to move.
First we establish an origin of a coordinate system.
We can see in the attached schema that the red vector is the displacement vector from the last point to where the truck is located.
The length of the vector is 187 [m], and the direction is 70 degrees south to East.
The answer is Carbon. Beyond being the only element listed here that is located in the second row, it is also in the fourteenth column of the table if you count from left to right. Hope this helps!
Answer:
Explanation:
Assuming the pith balls as point charges, we can calculate the repulsive force between them, using Coulomb's law:
We observe that the magnitude of the electric force is directly proportional to the product of the magnitude of both signed charges(
) and inversely proportional to the square of the distance(d) that separates them.
Replacing the given values, where k is the Coulomb constant:
