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9966 [12]
4 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]4 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:

F = 2123.33N

Explanation:

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\Sigma \tau=0

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Then, you obtain the following formula:

-\tau_l+\tau_p+\tau_{cm}=0          (1)

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τcm: torque produced by the center of mass of the wooden

The torque is given by:

\tau=Fd           (2)

F: force applied

d: distance to the pivot of the torque, in this case, distance to the right support.

You replace the equation (2) into the equation (1) and take into account that the force applied by the person and the center of mass of the wood are the their weight:

-Fd_1+W_pd_2+W_{cm}d_3=0\\\\d_1=6.0m\\\\d_2=2.0m\\\\d_3=3.0m\\\\W_p=(200kg)(9.8m/s^2)=1960N\\\\W_{cm}=(300kg)(9.8m/s^2)=2940N

Where d1, d2 and d3 are distance to the right support.

You solve the equation for F and replace the values of the other parameters:

F=\frac{W_pd_2+W_d_3}{d_1}=\frac{(1960N)(2.0m)+(2940N)(3.0m)}{6.0m}\\\\F=2123.33N

The force applied by the left support is 2123.33 N

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