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Gravitational Potential Energy can be found with the equation:
Therefore the change in GPE is:
We can see that it will change linearly. The percent it will change by will be:
The actual change is (I'll interpret the units MG as Millions of Grams - and since the formula uses kilograms as the unit of mass, I'll adjust the numbers to get the units right):
How much energy is this?
One way to view it is to say that it's the energy required to move a paperclip 1 metre (3 feet).
The following is the original answer, which answered the question regarding gravitational force:
The force of gravity between two masses can be found with:
Since we're changing the distance between the two objects, one way we can express the change is to see that:
And so the change in the two forces is an
The actual numerical change is (I'll interpret the units MG as Millions of Grams - and since the formula uses kilograms as the unit of mass, I'll adjust the numbers to get the units right):
This is a remarkably small amount of force. If you put two snickers bars in your hand (100g), the amount of force pushing down on your hand is roughly 1N. To achieve .005N, eat one of the snickers bars (you don't need it) and take the other one and cut it up into 250 equal pieces. Take one piece from that and put it in your hand. That is roughly .005N.
Second Law of Thermodynamics comes in way of utilising this resource. Efficiency of conversion is so low as to not justify the cost spent in building machineries for this purpose.
Of the various forms of energies, Mechanical energy and Electrical energy are of high quality whereas Thermal energy is of low quality.
While the Energy Conservation Principle (First Law of Thermodynamics) allows one to convert energy from one form to another, the Second Law of Thermodynamics places constraints on the efficiency of this conversion, especially from low quality forms to high quality forms.
The machines that convert thermal energy into mechanical/electrical energies are called Heat Engines. Heat Engines draw thermal energy from a heat reservoir at high temperature (
Second Law of thermodynamics clearly rules out a 100% conversion. The efficiency of the Carnot Engine is related to the temperatures of the heat reservoir (
Looking at this equation it is clear that a 100% efficient engine would require either a heat reservoir of infinite temperature (
It is fair to ask why 100% why not be happy with what you get. Conventional heat engines (automobile engines and conventional nuclear plant turbines) have efficiencies in the range of 25% to 35%. Here the temperature difference between the heat source and heat sink is of the order of
Now if you consider ocean as huge reservoir of thermal energy, the temperature difference between cold ocean bottom and hot ocean surface is of the order of few kelvins and the surface temperature is only about
SORRY about overshooting the word limits.
Millikan had data for the charges on the oil droplet which when sorted was expected to show a linear trend
The charges on the oil droplets are expected to be some integral multiple of a unknown fundamental unit of charge
If we plot
Remember that all entries in the sequence
Millikan's own data is given below. The third column shows that the sequence interval, a rough estimate of
Millikan's own data is reproduced below.
Millikan's Experiement Data Source:
Least Squares Fit: When we have measured data
To find the
Applying this to Millikan's data:
Therefore the least-square fit value of the fundamental unit of charge for Millikan's data, written to 4 significant digits is:
Option ( B)
Let the object
Considering that time count is started when the object is at origin. So we can write the equation for displacement
Here origin is also its equilibrium position and It returns repeatedly to this position after each complete oscillation for 1.6s.
So the velocity
Now it is given that the velocity of the object after passing the equilibrium position by
So we can insert
The electric field inside a good conductor cannot be non-zero. If the electric field has to vanish inside a conductor, it must get polarized such that
But since the conductor is electrically neutral, each of these negative polar charges will have a positive charge of same magnitude associated with them. These would have drifted to the surface of the good conductor. The electric field will continue outside the conductor originating from these outer
By Gauss's law the electric flux through this surface is -
The electric field is spherically symmetric and so will have the same magnitude (
If we imagine a gaussian sphere centered on the central charge such that point
At the point of contact of the cube with the cone, in the absence of friction two forces act on it.
For circular motion required centripetal force
#m#is mass of cube, #v#its velocity in a horizontal circle of radius #r#.
This is provided by the horizontal component of normal reaction. We have the equation
In the state of equilibrium vertical components of all the forces must be zero. Therefore, we get
Dividing (2) by (1) we get
Inserting given values we get
#v=1.17ms^-1#, rounded to one decimal place.
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