As capacitor was discharging, The charge on the plate got reversed and the motion of charge is opposite to the flow of current.
The charging contemporary asymptotically processes 0 as the capacitor becomes charged up to the battery voltage.
The capacitor is completely charged when the voltage of the electricity supply is equal to that at the capacitor terminals. that is referred to as capacitor charging; and the charging segment is over when modern-day stops flowing thru the electrical circuit.
A capacitor can be slowly charged to the important voltage and then discharged quick to provide the power wanted. it's far even viable to charge several capacitors to a positive voltage and then discharge them in any such way as to get extra voltage out of the gadget than became installed.
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The process which involves plants to cause weathering to a seed that falls into a crack in a rock therefore, breaking apart the rock is called Mechanical weathering.
<h3>
Mechanical weathering</h3>
This is defined as the set of weathering processes that break apart rocks into particles (sediment) through physical processes. The most common form of mechanical weathering is the freeze-thaw cycle. Water seeps into holes and cracks in rocks then the water freezes and expands, therefore making the holes larger eventually making the rock split apart.
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The element of art that creates edges of shapes and visual cues for the viewer in terms of both motion and space is line. The correct option among all the options that are given in the question is the third option or option "c". The lines have the ability to create a sense of motion when drawn in spaces.
The edge length of the unit cell at the given atomic mass and density of the molybdenum is 314.2 pm.
<h3>Volume of molybdenum</h3>
V = (zm/ρN)
where;
- z is 2 for cubic unit cell
- m is mass of the molybdenum
- ρ is density of the molybdenum
V = (2 x 95.96) / (10.28 x 6.02 x 10²³)
V = 3.10 x 10⁻²³ cm³
<h3>Edge length of the unit cell</h3>
a³ = V
a = (V)^¹/₃
a = ( 3.10 x 10⁻²³)^¹/₃
a = 3.142 x 10⁻⁸ cm
a = 3.142 x 10⁻¹⁰ m
a = 314.2 x 10⁻¹² m
a = 314.2 pm
Thus, the edge length of the unit cell at the given atomic mass and density of the molybdenum is 314.2 pm.
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During a climb UP the mountain, gravity does NO work on the climber.
Actually, it's more correct to say that gravity does NEGATIVE work
on him. The climber has to DO the positive work to haul himself up.
Work = (mass) x (gravity) x (height) .
For the guy in this problem:
Work = (67 kg) x (9.8 m/s²) x (3,500 meters)
= 2,298,100 joules.
If he eats no candy bars on the way, and completely depends on
his stored body fat for the energy, then he'll burn off
(2,298,100 joules) / (3.8 x 10⁷ joules/kg)
= 0.06 kg of fat.
That's only about 2.1 ounces. We KNOW he'll lose more weight than that,
climbing 11,000 feet. That's because climbing is pretty inefficient.
In addition to the potential energy you have to give your body weight,
you also have to expend energy breathing, digesting, metabolizing,
and sweating.
