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dusya [7]
2 years ago
13

An ordinary flashlight battery has a potential difference of 1.2 V between its positive and negative terminals. How much work mu

st you do to transport an electron from the positive terminal to negative terminal
Physics
1 answer:
Maru [420]2 years ago
4 0

The work done to transport an electron from the positive to the negative terminal is 1.92×10⁻¹⁹ J.

Given:

Potential difference, V = 1.2 V

Charge on an electron, e = 1.6 × 10⁻¹⁹ C

Calculation:

We know that the work done to transport an electron from the positive to the negative terminal is given as:

W.D = (Charge on electron)×(Potential difference)

       = e × V

       = (1.6 × 10⁻¹⁹ C)×(1.2 V)

       = 1.92 × 10⁻¹⁹ J

Therefore, the work done in bringing the charge from the positive terminal to the negative terminal is 1.92 × 10⁻¹⁹ J.

Learn more about work done on a charge here:

<u>brainly.com/question/13946889</u>

#SPJ4

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As a pendulum swings from its highest to lowest position, what happens to its kinetic and potential energy?
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The potential energy decreases while the kinetic energy increases.



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Find the length of a pendulum that oscillates with a frequency of 0.16 hz. the acceleration due to gravity is 9.81 m/s 2 . answe
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The period of the pendulum is the reciprocal of the frequency:
T= \frac{1}{f}= \frac{1}{0.16 Hz}=6.25 s

The period of the pendulum is given by
T=2 \pi  \frac{L}{g}
where L is the length of the pendulum, and g the acceleration of gravity. By re-arranging the formula and using the value of T we found before, we can  calculate the length of the pendulum L:
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7 0
2 years ago
A thin rod of length 0.64 m and mass 120 g is suspended freely from one end. It is pulled to one side and then allowed to swing
valina [46]

Answer:

1. Kinetic Energy = 0.0161 Joules

2. Height = 0.0137m

Explanation:

Given

Length of Rod, l = 0.64m

Mass, m = 120g = 0.12kg

Angular speed, w = 1.40 rad/s

a.

Calculating the Rod's kinetic energy

This is calculated by

Kinetic Energy = ½Iw²

Where I = rotational inertia of the rod about an axis.

This is calculated as follows;

I = Icm + mh²

I = ImL² + m(L/2)²

I = 1/12 * 0.12 * 0.64² + 0.12 * (0.64/2)²

I = 0.016384 kgm²

By substituton

KE = ½Iw² becomes

KE = ½ * 0.016384 * 1.40²

KE = 0.01605632J

KE = 0.0161 Joules

2. Using the total conservation of momentum;

K + U = Kf + V

Where K = Initial Kinetic Energy of the rod at lowest point.

U = Initial gravitational potential energy of the rod at lowest point

Kf = Final Kinetic Energy of the rod at maximum height = 0 J

V = Final gravitational potential energy of the rod at maximum height

So, K + U = Kf + V become

K + U = 0 + V

K + U = V

K = V - U = mgh

substitute 0.01605632J for K

0.01605632J = mgh

h = 0.01605632J/mg

h = 0.01605632J/(0.12 * 9.8)

h = 0.013653333333333

h = 0.0137m

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If the mean velocity adjacent to the top of a wing of 1.8 m chord is 40 m/s and that adjacent to the bottom of the wing is 31 m/
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