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Alexandra [31]
3 years ago
14

As a positively charged object moves toward a negatively charged object, their potential energy increases. As a positively charg

ed object moves toward another positively charged object, their potential energy increases. As a negatively charged object moves toward another negatively charged object, their potential energy increases. Increasing the distance between two electrically charged objects may raise or lower their potential energy. All of the above are true.
Physics
1 answer:
Zielflug [23.3K]3 years ago
4 0

Answer:

As a positively charged object moves toward another positively charged object, their potential energy increases. True

As a negatively charged object moves toward another negatively charged object, their potential energy increases True

Explanation:

When two like charges move towards each other, there is increase in potential energy because the motion is in opposition to the direction if the field. Hence when two negative or two positive charges move towards each other, there is an increase in potential energy.

However, potential energy is inversely related to the distance of separation of the charges.

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Explanation:

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provides some pertinent background for this problem. A pendulum is constructed from a thin, rigid, and uniform rod with a small
gavmur [86]

Answer:

the period of the physical pendulum is 0.498 s

Explanation:

Given the data in the question;

T_{simple = 0.61 s

we know that, the relationship between T and angular frequency is;

T = 2π/ω ---------- let this be equation 1

Also, the angular frequency of physical pendulum is;

ω = √(mgL / I ) ------ let this equation 2

where m is mass of pendulum, L is distance between axis of rotation and the center of gravity of rod and I  is moment of inertia of rod.

Now, moment of inertia of thin uniform rod D is;

I = \frac{1}{3}mD²

since we were not given the length of the rod but rather the period of the simple pendulum, lets combine this three equations.

we substitute equation 2 into equation 1

we have;

T = 2π/ω OR T = 2π/√(mgL/I) OR T = 2π√(I/mgL)

so we can use I = \frac{1}{3}mD² for moment of inertia of the rod

Since center of gravity of the uniform rod lies at the center of rod

so that L =  \frac{1}{2}D.

now, substituting these equations, the period becomes;

T = 2π/√(I/mgL) OR T = 2\pi \sqrt{\frac{\frac{1}{3}mD^2 }{mg(\frac{1}{2})D } } OR T = 2π√(2D/3g )  ----- equation 3

length of rod D is still unknown, so from equation 1 and 2 ( period of pendulum ),

we have;

ω_{simple = 2π/T_{simple OR  ω_{simple = √(g/D) OR  ω_{simple = 2π√( D/g )  

so we simple solve for D/g and insert into equation 3

so we have;

T = √(2/3) × T_{simple

we substitute in value of T_{simple

T = √(2/3) × 0.61 s

T = 0.498 s

Therefore, the period of the physical pendulum is 0.498 s

 

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3 years ago
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sveticcg [70]

Answer:

<h2>2.22 kPa</h2>

Explanation:

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<h3>2.22 kPa</h3>

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