Answer:
The intensity is ![I = 0.0003053 \mu W/cm^2](https://tex.z-dn.net/?f=I%20%3D%200.0003053%20%5Cmu%20%20W%2Fcm%5E2)
Explanation:
From the question we are told
The frequency of the electromagnetic wave is
The peak value of the electric field is ![E_o = 2.30 mV/m = \frac{2.30}{1000 } = 2.30 *10^{-3} V/m](https://tex.z-dn.net/?f=E_o%20%3D%202.30%20mV%2Fm%20%3D%20%5Cfrac%7B2.30%7D%7B1000%20%7D%20%3D%202.30%20%2A10%5E%7B-3%7D%20V%2Fm)
Generally the intensity of this wave is mathematically represented as
![I = c * \frac{1}{2} * \epsilon_o E^2_o](https://tex.z-dn.net/?f=I%20%3D%20c%20%20%2A%20%5Cfrac%7B1%7D%7B2%7D%20%2A%20%5Cepsilon_o%20E%5E2_o)
Where c is the speed of light with value ![c = 3 *10^8 m/s](https://tex.z-dn.net/?f=c%20%3D%203%20%2A10%5E8%20m%2Fs)
is the permittivity of free space with value ![\epsilon _o = 8.85 *10^{-12} C^2 /Nm^2](https://tex.z-dn.net/?f=%5Cepsilon%20_o%20%20%3D%208.85%20%2A10%5E%7B-12%7D%20C%5E2%20%2FNm%5E2)
Substituting values into equation for intensity
![I = 3.0 *10^8 * 0.5 * 8.85 *10^{-12} * 2.30*10^{-3}](https://tex.z-dn.net/?f=I%20%3D%203.0%20%2A10%5E8%20%20%2A%200.5%20%2A%208.85%20%2A10%5E%7B-12%7D%20%2A%202.30%2A10%5E%7B-3%7D)
![I = 3.053 *10^{-6} W/m^2](https://tex.z-dn.net/?f=I%20%3D%203.053%20%2A10%5E%7B-6%7D%20W%2Fm%5E2)
Converting to
we divide by 10,000
![I = \frac{3.053 *10^{-6}}{10000} W/cm^2](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7B3.053%20%2A10%5E%7B-6%7D%7D%7B10000%7D%20W%2Fcm%5E2)
![= 3.053 *10^{-10} W/cm^2](https://tex.z-dn.net/?f=%3D%203.053%20%2A10%5E%7B-10%7D%20W%2Fcm%5E2)
![= 0.0003053 *10^{-6} W/cm^2](https://tex.z-dn.net/?f=%3D%200.0003053%20%2A10%5E%7B-6%7D%20W%2Fcm%5E2)
Answer:
0.56 m/s
Explanation:
The speed of the head at the end of the interval in each case is the area under the acceleration curve. Then the difference in speeds is the difference in areas.
We can find the area geometrically, using formulas for the area of a triangle and of a trapezoid.
A = 1/2bh . . . . area of a triangle
A = 1/2(b1 +b2)h . . . . area of a trapezoid
If h(t) is the acceleration at time t for a helmeted head, the area under that curve will be (in units of mm/s) ...
Vh = 1/2(h(3)·3) +1/2(h(3) +h(4))·1 +1/2(h(4) +h(6))·2 +1/2(h(6))·1
Vh = 1/2(4h(3) +3h(4) +3h(6)) = 1/2(4·40 +3·40 +3·80) = 260 . . . mm/s
If b(t) is the acceleration for a bare head, the area under that curve in the same units is ...
Vb = 1/2(b(2)·2 +1/2(b(2) +b(4))·2 +1/2(b(4) +b(6))·2 +1/2(b(6)·1)
Vb = 1/2(4b(2) +4b(4) +3b(6)) = 1/2(4·120 +4·140 +3·200) = 820 . . . mm/s
Then the difference in speed between the bare head and the helmeted head is ... (0.820 -0.260) m/s = 0.560 m/s
When the object is at the focal point the angular magnification is 2.94.
Angular magnification:
The ratio of the angle subtended at the eye by the image formed by an optical instrument to that subtended at the eye by the object when not viewed through the instrument.
Here we have to find the angular magnification when the object is at the focal point.
Focal length = 6.00 cm
Formula to calculate angular magnification:
Angular magnification = 25/f
= 25/ 8.5
= 2.94
Therefore the angular magnification of this thin lens is 2.94
To know more about angular magnification refer:: brainly.com/question/28325488
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Answer:
light scattering by particles in a colloid or in a very fine suspension
Answer:
The center of the Milky Way most likely contains a supermassive black hole.
Explanation:
Because it is an eleptical galaxy, it has a little rotation to it but not enough to flatten out so the center will contain a supermassive black hole.