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9966 [12]
3 years ago
11

A 60-W light bulb radiates electromagnetic waves uniformly in all directions. At a distance of 1.0 m from the bulb, the light in

tensity is I0, the average energy density of the waves is u0, and the rms electric and magnetic field values are E0 and B0, respectively.
1. At 2.0 m from the bulb, what is the light intensity?

2. At 2.0 m from the bulb, what is the rms magnetic field value?

3. At 2.0 m from the bulb, what is the average energy density of the waves?
Physics
1 answer:
krek1111 [17]3 years ago
5 0
1. If we increase the distance to twice it's original value, the light intensity is reduced by one-fourth, the light intensity would be:
I0/4

2. rms magnetic field is inversely proportional to distance, so the new rms magnetic field would be:
B0/2

3. average energy density is inversely proportional to the square of the distance, so the new average energy density is:
E0/4
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Answer:

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The initial kinetic energy imparted to a 0.25 kg bullet is 1066 J. The acceleration of gravity is 9.81 m/s 2 . Neglecting air re
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Answer:

The range of the bullet is 0.435 kilometers.

Explanation:

According to the problem, maximum height is equal to the range of the bullet. That is:

\Delta x = \Delta y

Where:

\Delta x - Range of the bullet, measured in meters.

\Delta y - Maximum height of the bullet, measured in meters.

By the Principle of Energy Conservation, gravitational potential energy reaches its maximum at the expense of the initial kinetic energy. That is to say:

K_{1} = U_{2}

Where:

K_{1} - Kinetic energy at point 1, measured in joules.

U_{1} - Gravitational potential energy at point 2, measured in joules, and:

U_{2} = m\cdot g \cdot \Delta y

Where:

m - Mass of the bullet, measured in kilograms.

g - Gravitational constant, measured in meters per square second.

The maximum height is now cleared:

K_{1} = m\cdot g \cdot \Delta y

\Delta y = \frac{K_{1}}{m\cdot g}

If K_{1} = 1066\,J, m = 0.25\,kg and g = 9.81\,\frac{m}{s^{2}}, the maximum height is now computed:

\Delta y = \frac{1066\,J}{(0.25\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}

\Delta y = 434.791\,m

\Delta y = 0.435\,km

Lastly, the range of the bullet is 0.435 kilometers.

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