The formula to use is: I = (<span> ΔV / R )
Once you solve for R, your new formula would be: R= (</span><span> ΔV / I )
Plug in your values to get: R = (1.5V / .75A )
Finally, R = 2</span><span>Ω</span>
In order to solve this question, we will need to know the E-field of a capacitor
E = sigma/epsilon naught
To find sigma (charge density), we will need to divide the total charge by the area
sigma = (.576 x 10^-9)/(4.952x10^-2)^2 = 2.34 x 10^-7
E = 2.34 x 10^-7/ (8.85x10^-12) = 26541.04 V/m
<span>The quick way to visualize this is to compare isometric and perspective drawings of a cube. In the isometric drawing, all the edges of the cube would be parallel in sets of four. In the perspective drawing, the edges would taper towards one or more vanishing points.
The isometric drawing, being easier to construct, perserving all scales and dimensions, is the preferred method for mechanical drawings, and are practical for use in the shop. The perspective drawing, which are trickier to draw properly, and does not preserve scales and dimensions, is the preferred method for architectural drawings, because they illustrate what the eye actually sees.</span>
Hi Pupil Here is Your Answer ::
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Balanced Forces =》 If the resultant of several forces acting on a body is zero, the forces are said to be Balanced Forces.
Or
When two forces of equal magnitudes but acting in opposite direction on an object simultaneously then the object continuous in its state of rest all uniform motion in a straight line such forces acting on the object are known as Balanced Forces.
Unbalance Forces =》 If the resultant of several forces acting on a body is not zero, the forces are said to be Unbalanced Forces.
Or
When two forces of unequal magnitudes act in opposite direction on an object simultaneously then object moves in the direction of the force with larger magnitude forces. These force acting on the object are known as Unbalanced Forces.
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Hope this helps . . . . ..c
Answer:
A ball thrown into the air has the most potential energy when it has reached the highest point above the ground before it begins descending. If we consider the vertical motion only beginning when the ball leaves the thrower’s hand, the ball is exchanging kinetic energy for (gravitational) potential energy. When all of the kinetic energy has been transformed, the ball begins falling, and exchanges it’s potential energy back into kinetic energy. If you ignore air resistance, the ball will land with as much energy as it began with.
