When temperature and pressure are applied to sedimentary and igneous rock?
1 answer:
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Answer:
capacitance = 2.242 × F
charge = 1.345 × C
electric field = 600 V/m
work done = 8.07 × J
Explanation:
given data
battery c = 12 V
diameter = 10 in
distance d = 2 cm = 2× m
to find out
capacitance , charge on plate, electric field and work done
solution
we know radius is diameter / 2
so radius r = 10 / 2 = 5 in = 0.127 m
and capacitance formula that is
capacitance = A∈/d
put here all value
capacitance = πr² ∈/d
capacitance = π(0.127)² ×8.85 × /2×
capacitance = 2.242 × F
and
charge on plate is express as
charge = capacitance × c
we know here 2 plate so on 1 plate c is 6
charge = 2.242 × × 6
charge = 1.345 × C
and
electric field is express as
electric field = c / d
electric field = 12 / 2×
electric field = 600 V/m
and
work done is express as
work done = 1/2 × charge × C
put here value
work done = 1/2 × (1.345 × ) (12)
work done = 8.07 × J
Answer:
The high of the ramp is 2.81[m]
Explanation:
This is a problem where it applies energy conservation, that is part of the potential energy as it descends the block is transformed into kinetic energy.
If the bottom of the ramp is taken as a potential energy reference point, this point will have a potential energy value equal to zero.
We can find the mass of the box using the kinetic energy and the speed of the box at the bottom of the ramp.
Now applying the energy conservation theorem which tells us that the initial kinetic energy plus the work done and the potential energy is equal to the final kinetic energy of the body, we propose the following equation.
And therefore