The lead ball will float with about 17% of its volume above the surface of the mercury.
We know that density is defined as mass per unit volume of a substance. The density of a substance is an intrinsic property which can be used to identify a substance.
Given that Lead is less dense that mercury, we know that lead will float on mercury. Since the density of mercury is 13.6 g/cm3 and that of lead is 11.3 g/cm3, lead ball will float with about 17% of its volume above the surface of the mercury.
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Missing parts;
A spherical ball of lead (density 11.3 g/cm3) is placed in a tub of mercury (density 13.6 g/cm3). Which answer best describes the result?
A.The lead ball will float with about 83% of its volume above the surface of the mercury.
B.The lead ball will float with about 17% of its volume above the surface of the mercury.
C.The lead ball will float with its top exactly even with the surface of the mercury.
D.The lead will sink to the bottom of the mercury.
E.none of the above
Answer:
When two objects collide and stick together, what will happen to their speed, assuming momentum is conserved? They will move at the same velocity as whichever object was fastest initially. They will move at the same velocity of whichever object was slowest initially.
Explanation:
Answer:
Explanation:
The difference of electric potential between two points is given by the formula 
, where <em>d</em> is the distance between them and<em> E</em> the electric field in that region, assuming it's constant.
The electric field formula is 
, where <em>F </em>is the force experimented by a charge <em>q </em>placed in it.
Putting this together we have 
, so we need to obtain the electric force the charged ball is experimenting.
On the second drop, the ball takes more time to reach the ground, this means that the electric force is opposite to its weight <em>W</em>, giving a net force 
. On the first drop only <em>W</em> acts, while on the second drop is <em>N</em> that acts.
Using the equation for accelerated motion (departing from rest) 
, so we can get the accelerations for each drop (1 and 2) and relate them to the forces by writting:
These relate with the forces by Newton's 2nd Law:
Putting all together:
Which means:
And finally we substitute:
Which for our values means:
