We have that the magnitude of the gravitational force is mathematically given as
f=6.377N
<h3>
Force</h3>
Question Parameters:
Earth exerts a 100 N gravitational force on a metal box.
(Mass of the earth is 6e24 kg and radius of the earth is 6.4e6m.)
Generally the equation for the Gravitational mForce is mathematically given as

f=6.377N
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Answer:
16.4287
Explanation:
The force and displacement are related by Hooke's law:
F = kΔx
The period of oscillation of a spring/mass system is:
T = 2π√(m/k)
First, find the value of k:
F = kΔx
78 N = k (98 m)
k = 0.796 N/m
Next, find the mass of the unknown weight.
F = kΔx
m (9.8 m/s²) = (0.796 N/m) (67 m)
m = 5.44 kg
Finally, find the period.
T = 2π√(m/k)
T = 2π√(5.44 kg / 0.796 N/m)
T = 16.4287 s
Answer:
a) R₁ = 14.1 Ω, b) R₂ = 19.9 Ω
Explanation:
For this exercise we must use ohm's law remembering that in a series circuit the equivalent resistance is the sum of the resistances
all resistors connected
V = i (R₁ + R₂)
with R₁ connected
V = (i + 0.5) R₁
with R₂ connected
V = (i + 0.25) R₂
We have a system of three equations with three unknowns for which we can solve it
We substitute the last two equations in the first
V = i (
)
1 = i (
)
1 = i (
) =
i² + 0.75 i + 0.125 = 2i² + 0.75 i
i² - 0.125 = 0
i = √0.125
i = 0.35355 A
with the second equation we look for R1
R₁ =
R₁ = 12 /( 0.35355 +0.5)
R₁ = 14.1 Ω
with the third equation we look for R2
R₂ =
R₂ =
R₂ = 19.9 Ω
This question can be solved from the Kepler's law of planetary motion.
As per this law the square of time period of a planet is proportional to the cube of semi major axis.
Mathematically it can be written as 
⇒
Here K is the proportionality constant.
If
and
are the orbital periods of the planets and
and
are the distance of the planets from the sun, then Kepler's law can be written as-

⇒ 
Here we are asked to calculate the the distance of Saturn from sun.It can solved by comparing it with earth.
Let the distance from sun and orbital period of Saturn is denoted as
and
respectively.
Let the distance from sun and orbital period of earth is denoted as
and
respectively.
we are given that
we know that
1 AU and
1 year.
1 AU is the mean distance of earth from the sun which is equal to 150 million kilometre.
Hence distance of Saturn from sun is calculated as -
From Kepler's law as mentioned above-

=![[1 ]^{3} *\frac{[29.46]^{2} }{[1]^{2} } AU](https://tex.z-dn.net/?f=%5B1%20%5D%5E%7B3%7D%20%2A%5Cfrac%7B%5B29.46%5D%5E%7B2%7D%20%7D%7B%5B1%5D%5E%7B2%7D%20%7D%20AU)

⇒![R_{1} =\sqrt[3]{867.8916}](https://tex.z-dn.net/?f=R_%7B1%7D%20%3D%5Csqrt%5B3%5D%7B867.8916%7D)
=9.5386 AU [ans]