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Answer:
The magnitude of the force is 1.29*10^-3N in the positive x direction
Explanation:
In order to calculate the magnitude and direction of the force, you take into account that the force is the space derivative of the potential enrgy, as follow:
(1)
where:
You replace the expression for U into the equation (1) and solve for F:
![F(x)=-\frac{d}{dx}[Ax^4]=-4Ax^3](https://tex.z-dn.net/?f=F%28x%29%3D-%5Cfrac%7Bd%7D%7Bdx%7D%5BAx%5E4%5D%3D-4Ax%5E3)
(2)
The force on the particle, for x = -0.080m is:
The magnitude of the force is 1.29*10^-3N in the positive x direction
Answer:
Δx = 6.33 x 10⁻³ m = 6.33 mm
Explanation:
We can use the Young's Double Slit Experiment Formula here:
where,
Δx = distance between consecutive dark fringes = width of central bright fringe = ?
λ = wavelength of light = 633 nm = 6.33 x 10⁻⁷ m
L = distance between screen and slit = 3.7 m
d = slit width = 0.37 mm = 3.7 x 10⁻⁴ m
Therefore,
<u>Δx = 6.33 x 10⁻³ m = 6.33 mm</u>
<span>3598 seconds
The orbital period of a satellite is
u=GM
p = sqrt((4*pi/u)*a^3)
Where
p = period
u = standard gravitational parameter which is GM (gravitational constant multiplied by planet mass). This is a much better figure to use than GM because we know u to a higher level of precision than we know either G or M. After all, we can calculate it from observations of satellites. To illustrate the difference, we know GM for Mars to within 7 significant figures. However, we only know G to within 4 digits.
a = semi-major axis of orbit.
Since we haven't been given u, but instead have been given the much more inferior value of M, let's calculate u from the gravitational constant and M. So
u = 6.674x10^-11 m^3/(kg s^2) * 6.485x10^23 kg = 4.3281x10^13 m^3/s^2
The semi-major axis of the orbit is the altitude of the satellite plus the radius of the planet. So
150000 m + 3.396x10^6 m = 3.546x10^6 m
Substitute the known values into the equation for the period. So
p = sqrt((4 * pi / u) * a^3)
p = sqrt((4 * 3.14159 / 4.3281x10^13 m^3/s^2) * (3.546x10^6 m)^3)
p = sqrt((12.56636 / 4.3281x10^13 m^3/s^2) * 4.458782x10^19 m^3)
p = sqrt(2.9034357x10^-13 s^2/m^3 * 4.458782x10^19 m^3)
p = sqrt(1.2945785x10^7 s^2)
p = 3598.025212 s
Rounding to 4 significant figures, gives us 3598 seconds.</span>
The question is incomplete and no options are given but the answer is;
The orbit of Mars is an ellipse,<span> is generally attributed to Kepler.
Kepler's first law states that; "</span><span>The orbits of the planets are ellipses, with the Sun at one focus of the ellipse.
</span>The planet at that point takes after and follow the ellipse in its orbit, which implies that the distance between Earth and Sun remove is continually changing as the planet goes around its orbit.
