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Olegator
10 days ago
6

A hard rubber rod with an electric potential energy of 5.2 × 10–3 J has a charge of 4.0 µC at the tip. What is the electric pote

ntial at the tip? Round your answer to one decimal place. × 103 V What is the electric potential if the charge at the tip changes to 2.0 µC? Round your answer to one decimal place. × 103 V
Physics
2 answers:
Keith_Richards [2.2K]10 days ago
8 0
1) The electric potential energy can be defined as the product of the electric potential and the associated charge:
U=q V
where
q refers to the charge
V denotes the electric potential

In this scenario, the charge on the rod is q=4.0 \mu C = 4.0 \cdot 10^{-6}C, and the potential energy is U=5.2 \cdot 10^{-3} J, thus we may rearrange the earlier formula to find the electric potential at the tip:
V= \frac{U}{q}= \frac{5.2 \cdot 10^{-3}J}{4.0 \cdot 10^{-6} C}=1300 V=1.3 \cdot 10^3 V

2) Using this same formula, if the charge changes to q=2.0 \mu C = 2.0 \cdot 10^{-6} C, the resulting electric potential will be:
V= \frac{U}{q}= \frac{5.2 \cdot 10^{-3} J}{2.0 \cdot 10^{-6}C}=2600 V = 2.6 \cdot 10^{3}V
Ostrovityanka [2.2K]10 days ago
3 0

Clarification:

Considering that

Electric potential energy, U=5.2\times 10^{-3}\ J

Charge on a rubber rod, q=4\ \mu C=4\times 10^{-6}\ C

The relationship between electric potential and electric potential energy is established by:

U=qV

CASE 1

Thus,

V=\dfrac{U}{q}

V=\dfrac{5.2\times 10^{-3}\ J}{4\times 10^{-6}\ C}

V=1300\ V

or

V=1.3\times 10^3\ V

CASE 2

In instances where the charge is q=2\ \mu C=2\times 10^{-6}\ C

V=\dfrac{U}{q}

V=\dfrac{5.2\times 10^{-3}\ J}{2\times 10^{-6}\ C}

V=2600\ V

or

V=2.6\times 10^3\ V

Consequently, this presents the necessary solution.

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Answer:

Explanation:

Amount of gold deposited = 0.5 g

Gold's molar mass = 197 g/mol

Time duration, t = 6 hours

= 6 × 3600

= 12600 s

Calculation of moles: mass/molar mass

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= 0.00254 mole

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Faraday's constant = 9.65 x 10^4 C mol-1

Charge, Q = 96500 × 0.00254

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Thus, I = 244.924/12600

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8 days ago
An object is attached to a hanging unstretched ideal and massless spring and slowly lowered to its equilibrium position, a dista
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Answer:

        h = 12.8 cm

Explanation:

The initial parameters are as follows:

distance = 6.4 cm

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        weight = spring force

         mg = ky... equation 1

  • potential energy stored in a stretched spring = work done by the spring

        mgh = 0.5 x k x h^{2}....equation 2

  • Substituting from equation 1 into equation 2

                kyh =  0.5 x k x h^{2}

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6 0
26 days ago
1)After catching the ball, Sarah throws it back to Julie. However, Sarah throws it too hard so it is over Julie's head when it r
Softa [2035]

Answer:

1)

v_{oy}=11.29\ m/s

2)

y=7.39\ m

Explanation:

Projectile Motion

When an object is projected near the surface of the Earth at an angle \theta to the horizontal, it follows a trajectory known as a parabola. The only force acting on it (ignoring wind resistance) is gravity, affecting the vertical axis.

The height of a projectile can be calculated using

\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}

where y_o represents the initial height from ground level, v_{oy} is the vertical component of the initial velocity, and t is the elapsed time.

The vertical speed component is expressed as

v_y=v_{oy}-gt

1) To proceed, we will determine the initial vertical velocity component since we lack sufficient data to calculate the absolute value of v_o.

The peak height is attained when v_y=0, which allows us to compute the time to reach that height.

v_{oy}-gt_m=0

Solving for t_m

\displaystyle t_m=\frac{v_{oy}}{g}

Thus, the maximum height reached is

\displaystyle y_m=y_o+\frac{v_{oy}^2}{2g}

We know this value is equal to 8 meters

\displaystyle y_o+\frac{v_{oy}^2}{2g}=8

Continuing with the calculations for v_{oy}

\displaystyle v_{oy}=\sqrt{2g(8-y_o)}

Substituting known values yields

\displaystyle v_{oy}=\sqrt{2(9.8)(8-1.5)}

\displaystyle v_{oy}=11.29\ m/s

2) At t=1.505 seconds, the ball is positioned above Julie’s head; we can calculate

\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}

\displaystyle y=1.5+(11.29)(1.505)-\frac{9.8(1.505)^2}{2}

\displaystyle y=1.5\ m+16,991\ m-11.098\ m

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1 month ago
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Answer:

Explanation:

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Thus, it becomes:

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λ = 2 m.

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As he moves closer to one of the speakers, his proximity to that speaker increases while the distance to the other speaker decreases, creating a path difference in the sound waves reaching his ears.

If he walks 0.5 m toward one speaker, the created path difference becomes:

0.5 x 2 = 1 m.

This path difference equals λ / 2, leading to destructive interference, resulting in minimal sound being audible.

As he continues walking a full 1 m, the created path difference totals 2 m.

This corresponds to a path difference of λ, causing constructive interference and maximum sound perception.

Finally, if he moves an additional 1.5 m, the resulting path difference increases to 3 m.

Thus, we arrive at a path difference of 3 λ / 2, producing destructive interference once more, leading to minimum sound being perceived again.

In summary, the man begins at a maximum intensity point, moves to minimum intensity, then back to a maximum, and ultimately ends at another minimum sound intensity position.

3 0
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