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joja [24]
3 years ago
13

Kepler’s third law states that for all objects orbiting a given body, the cube of the semimajor axis (A) is proportional to the

square of the orbital period (P). For each of our planets orbiting the Sun, the relationship between the orbital period and semimajor axis can be represented by the equation P2 = kA3, where k is a constant value for all the planets. Rewrite this equation so that it solves for the constant of proportionality, k.
Physics
1 answer:
GarryVolchara [31]3 years ago
8 0

Explanation:

Kepler’s third law states that for all objects orbiting a given body, the cube of the semimajor axis (A) is proportional to the square of the orbital period (P).

For each of our planets orbiting the Sun, the relationship between the orbital period and semimajor axis can be represented by the equation as:

P^2=kA^3

k is constant of proportionality

It is required to solve the above equation for k

k=\dfrac{P^2}{A^3}

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In si units, the electric field in an electromagnetic wave is described by ey = 104 sin(1.40 107x − ωt). (a) find the amplitude
melamori03 [73]
Answers:
(a) B_o  = 0.3466μT
(b) \lambda = 0.4488μm
(c) f = 6.68 * 10^{14}Hz

Explanation:
Given electric field(in y direction) equation:
E_y = 104sin(1.40 * 10^7 x -\omega t)

(a) The amplitude of electric field is E_o = 104. Hence

The amplitude of magnetic field oscillations is B_o =  \frac{E_o}{c}
Where c = speed of light

Therefore,
B_o =  \frac{104}{3*10^8} = 0.3466μT (Where T is in seconds--signifies the oscillations)

(b) To find the wavelength use:
\frac{2 \pi }{\lambda} = 1.40 * 10^7
\lambda =  \frac{2 \pi}{1.40} * 10^{-7}
\lambda =  0.4488μm

(c) Since c = fλ
=> f = c/λ

Now plug-in the values
f = (3*10^8)/(0.4488*10^-6)
f = 6.68 * 10^{14}Hz


6 0
3 years ago
A square loop of wire lies in the x, y plane with opposite corners at (0, 0) and (a, a). The loop carries a current I directed t
Ket [755]

Answer:

Please refer to the figure.

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shusha [124]

Answer:

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