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solmaris [256]
4 years ago
13

If 28.0 J of work is done by an external force to move a charge from a potential of 8.0 V 8.0 V to a potential of 4.0 V , 4.0 V,

what is the change in electric potential energy?
Physics
1 answer:
jonny [76]4 years ago
5 0

Answer:

Change in electric potential energy is -28.0 J

Explanation:

Electric potential energy is defined as the work is done to move a charge particle from one position to another in space in the presence of other charge particle or electric potential.

                                               OR

Electric potential energy is also equal to the change in the configuration of the charge particles.

Thus,

Change in electric potential energy = - Work Done

According to the problem, Work Done is equal to 28 J. Thus,

Change in electric potential energy = -28 J  

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When a plant is entering the calvin cycle of photosythesis______.
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The Calvin cycle<span> refers to the light-independent reactions in photosynthesis that take place in three key steps. Although the </span>Calvin Cycle<span> is not directly dependent on light, it is indirectly dependent on light since the necessary energy carriers (ATP and NADPH) are products of light-dependent reactions.

So basically it indirectly needs the light, even it's called light-independant reaction.

So the answer is the last one.</span>
5 0
4 years ago
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During the annual shuffleboard competition, Renee gives her puck an initial speed of 8.12 m/s. Once leaving her stick, the puck
shutvik [7]
<h2>Answer:2.65 seconds</h2>

Explanation:

Let a be the acceleration.

Let u be the initial velocity.

Let v be the final velocity.

Let t be the time taken.

As we know from the equations of motion,

v=u+at

Given,v=0\\u=8.12ms^{-1}\\a=-3.06ms^{-2}

0=8.12-3.06t

t=2.65sec

6 0
4 years ago
Why is silicon needed in solar panels?
kodGreya [7K]
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4 0
3 years ago
A close coiled helical spring of round steel wire 10 mm diameter having 10 complete turns with a mean radius of 60 mm is subject
kow [346]

Answer:

The deflection of the spring is 34.56 mm.

Explanation:

Given that,

Diameter = 10 mm

Number of turns = 10

Radius_{mean} = 60\ mm

Diameter_{mean} = 120\ mm

Load = 200 N

We need to calculate the deflection

Using formula of deflection

\delta=\dfrac{8pD^3n}{Cd^4}

Put the value into the formula

\delta=\dfrac{8\times200\times(120)^3\times10}{80\times10^{3}\times10^4}

\delta =34.56\ mm

Hence, The deflection of the spring is 34.56 mm.

4 0
3 years ago
6. The image to the right shows a moment of inertia
Trava [24]

The moment of inertia is the rotational analog of mass, and it is given by

the  product of mass and the square of the distance from the axis.

  • The moment of inertia changes as the position of the weight is changed, which indicates that; statement is incorrect

Reasons:

The weight on each arm that have adjustable positions can be considered as point masses.

The moment of inertia of a point mass is <em>I</em> = m·r²

Where;

m = The mass of the weight

r = The distance (position) from the center to which the weight is adjusted

Therefore;

The moment of inertia, <em>I </em>∝ r²

Which gives;

Doubling the distance from the center of rotation, increases the moment of inertia by factor of 4.

Therefore, the statement contradicts the relationship between the radius of rotation and moment of inertia.

Learn more about moment of inertia here:

brainly.com/question/4454769

7 0
3 years ago
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