Answer:
A measured force of (46.5 0.8 N ) would not be in agreement with a theoretically calculated force of (48.4 0.6 N )
Explanation:
From the question we are told that
Measured force is
Calculated force is
Generally the measured force in interval form is
=>
Generally the calculated force in interval form is
=>
Generally looking both interval we see that they do not intersect at any point Hence
A measured force of (46.5 0.8 N ) would not be in agreement with a theoretically calculated force of (48.4 0.6 N )
Answer:
g(h) = g ( 1 - 2(h/R) )
<em>*At first order on h/R*</em>
Explanation:
Hi!
We can derive this expression for distances h small compared to the earth's radius R.
In order to do this, we must expand the newton's law of universal gravitation around r=R
Remember that this law is:
In the present case m1 will be the mass of the earth.
Additionally, if we remember Newton's second law for the mass m2 (with m2 constant):
Therefore, we can see that
With a the acceleration due to the earth's mass.
Now, the taylor series is going to be (at first order in h/R):
a(R) is actually the constant acceleration at sea level
and
Therefore:
Consider that the error in this expresion is quadratic in (h/R), and to consider quadratic correctiosn you must expand the taylor series to the next power:
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For a circular aperture, the first minima (n=1) as an angular separation from the peak of the central maxima given by
Sinθ = 1.22λ / d
Where,
d is the aperture or pupil diameter
d = 4.69 mm = 4.69 × 10^-3m
λ is the wavelength
λ = 545 nm = 545 × 10^-9 m
Then,
Sinθ = 1.22λ / d
Sinθ = 1.22 × 545 × 10^-9 / 4.69 × 10^-3
Sinθ = 1.418 × 10^-4 rad
Then, the head light sources have the same angular separation θ from the eye as the image have inside the eye.
For the headlight
Sinθ ≈ light separation / distantce for the eye
Light separation is give as x = 0.659 m
And let the distance of the eye be D
Then,
Sinθ = x / D
Make D subject of formula
D = x / Sinθ
D = 0.695 / 1.418 × 10^-4
D = 4902.316m
To km, 1km = 1000m
D ≈ 4.9 km