<span>If the refrigerator weights 1365 and you are not exerting any vertical force on it, then the normal force is also 1365N. so Fn=1365
Fsf = Static frictional force = (coefficient of static friction) * (Normal force)
So the least for you could exert to move it is equal to the Fsf.
Fsf = (0.49)(1365N)</span><span>
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A wave is a result of the disturbance in the equilibrium state. There are two types of wave, transverse and longitudinal. Transverse wave affects amplitude while longitudinal wave affects the frequency of the wave. As for the transverse wave, the magnitude of the perpendicular disturbance of the wave is directly proportional to the amplitude of the wave. The higher the transverse disturbance the higher the amplitude.
If you are given distance and a period of time, you can calculate
the speed. The velocity of an object is the rate of change of its position with
respect to a frame of reference, and is a function of time. Velocity is
equivalent to a specification of its speed and direction of motion (e.g. 60
km/h to the north).
Extinction of a species is most likely to occur as a result of "<span>environmental changes"
In short, Your Answer would be Option D
Hope this helps!</span>
Answer:
The charge stored in the capacitor will stay the same. However, the electric potential across the two plates will increase. (Assuming that the permittivity of the space between the two plates stays the same.)
Explanation:
The two plates of this capacitor are no longer connected to each other. As a result, there's no way for the charge on one plate to move to the other. 
, the amount of charge stored in this capacitor, will stay the same.
The formula 
relates the electric potential across a capacitor to:
- , the charge stored in the capacitor, and
, the capacitance of this capacitor.
While stays the same, moving the two plates apart could affect the potential 
by changing the capacitance of this capacitor. The formula for the capacitance of a parallel-plate capacitor is:
,
where
is the permittivity of the material between the two plates.
is the area of each of the two plates.
is the distance between the two plates.
Assume that the two plates are separated with vacuum. Moving the two plates apart will not affect the value of . Neither will that change the area of the two plates.
However, as (the distance between the two plates) increases, the value of will become smaller. In other words, moving the two plates of a parallel-plate capacitor apart would reduce its capacitance.
On the other hand, the formula can be rewritten as:
.
The value of (charge stored in this capacitor) stays the same. As the value of becomes smaller, the value of the fraction will become larger. Hence, the electric potential across this capacitor will become larger as the two plates are moved away from one another.
