The gravitational force between two objects is given by:
where
G is the gravitational constant
m1 and m2 are the masses of the two objects
r is their separation
In this problem, the first object has a mass of
, while the second "object" is the Earth, with mass
. The distance of the object from the Earth's center is
; if we substitute these numbers into the equation, we find the force of gravity exerted by the Earth on the mass of 0.60 kg:
Answer:
The resistance is 0.124 ohm.
Explanation:
It is common for domestic electrical installations to use copper wire with a diameter of 2.05 mm. Determine the resistance of such a wire with a length of 24.0 m.
diameter, d = 2.05 mm
radius, r = 1.025 mm
Length, L = 24 m
resistivity of copper = 1.7 x 10^-8 ohm m
Let the resistance is R.
A.) is chemical, B.) is physical, C.) is physical, D.) is chemical, E.) is physical, F.) is physical, G.) is physical, and H.) is chemical.
Answer:
It would not be possible the cohesion among water molecules by the polar covalent bonding.
Well, to understand this in a better way, let's begin by explaining that water is special due to its properties, which makes this fluid useful for many purposes and for the existence of life.
In this sense, one of the main properties of water is cohesion (molecular cohesion), which is the attraction of molecules to others of the same type. So, water molecule (
) has 2 hydrogen atoms attached to 1 oxygen atom and can stick to itself through hydrogen bonds.
How is this possible?
By the polar covalent bonding, a process in which electrons are shared unequally between atoms, due to the unequal distribution of electrons between atoms of different elements. In other words: slightly positive and slightly negative charges appear in different parts of the molecule.
Now, it can be said that a water molecule has a negative side (oxygen) and a positive side (hydrogen). This is how the oxygen atom tends to monopolize more electrons and keeps them away from hydrogen. Thanks to this polarity, water molecules can stick together.
