When I went through with the math, the answer I came upon was:
<span>6.67 X 10^14 </span>
<span>Here is how I did it: First of all we need to know the equation. </span>
<span>c=nu X lamda </span>
<span>(speed of light) = (frequency)(wavelength) </span>
<span>(3.0 X 10^8 m/s) = (frequency)(450nm) </span>
<span>We want the answer in meters so we need to convert 450nm to meters. </span>
<span>450nm= 4.5 X 10^ -7 m </span>
<span>(3.0 X 10^8 m/s) = (frequency)(4.5 X 10^ -7 m) </span>
<span>Divide the speed of light by the wavelength. </span>
<span>(3.0 X 10^8m/s) / (4.5 X 10^ -7m) =6.67 X 10^ 14 per second or s- </span>
<span>Answer: 6.67 X 10^14 s- hope this helps</span>
To find:
The equation to find the period of oscillation.
Explanation:
The period of oscillation of a pendulum is directly proportional to the square root of the length of the pendulum and inversely proportional to the square root of the acceleration due to gravity.
Thus the period of a pendulum is given by the equation,
Where L is the length of the pendulum and g is the acceleration due to gravity.
On substituting the values of the length of the pendulum and the acceleration due to gravity at the point where the period of the pendulum is being measured, the above equation yields the value of the period of the pendulum.
Final answer:
The period of oscillation of a pendulum can be calculated using the equation,
You should check A, and D.
The formula that will be used in this problem is E = q/ 4pi*r^2 z where z is the elctric charge constant equal to 8.854 *10 ^-12. The magnitude using r equal to 0.0525 m and q equal to -22.3 *10^-6 C is equal to -22.3 *10^-6/ 4pi*(0.0525)^2 *8.854 *10 ^-12 or equal to -7.272 *10 ^7. The magnitude 5 cm outside the surface is -22.3 *10^-6<span>/ 4pi*(0.0525+0.05)^2 *8.854 *10 ^-12 equal to -1.908 *10^7.
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