Mars, which has a radius of 3.4 × 106 m and a mass of 6.4 × 1023 kg, orbits the Sun, which has a mass of 2.0 × 1030 kg at a dist
2 answers:
The correct answer is: Revolution (The tangential speed of Mars' revolution is greater than that of Mars' rotation).
Explanation:
Given data:
Radius-of-Mars = r = m
Distance-between-Sun-and-Mars = d = m
Now let us first calculate the Mars' rotation:
One complete rotation of Mars = 2πr = 2π() =
m
Tangential speed of Mars' rotation (by taking 24hours 37 minutes in seconds (88620s)—as Mars takes that many seconds to complete one rotation) =
Now let's calculate the Mars' revolution (around Sun):
One complete revolution of Mars around Sun = 2πd = 2π() =
m
Tangential speed of Mars' revolution (by taking 687 days in seconds (59356800s)—since Mars takes 687days to complete one revolution)=
So we can see that the tagential speed of:
Revolution > Rotation
>
Hence, the correct answer is Revolution.
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Answer:
a) F_net = 6.48 10⁻¹⁸ ( ), b) x = 0.15 m
Explanation:
a) In this problem we use that the electric force is a vector, that charges of different signs attract and charges of the same sign repel.
The electric force is given by Coulomb's law
F =
Since when we have the two negative charges they repel each other and when we fear one negative and the other positive attract each other, the forces point towards the same side, which is why they must be added.
F_net= ∑ F = F₁ + F₂
let's locate a reference system in the load that is on the left side, the distances are
left side - electron r₁ = x
right side -electron r₂ = d-x
let's call the charge of the electron (q) and the fixed charge that has equal magnitude Q
we substitute
F_net = k q Q ( )
F _net = kqQ ( )
let's substitute the values
F_net = 9 10⁹ 1.6 10⁻¹⁹ 4.50 10⁻⁹ ( )
F_net = 6.48 10⁻¹⁸ ( )
now we can substitute the value of x from 0.05 m to 0.25 m, the easiest way to do this is in a spreadsheet, in the table the values of the distance (x) and the net force are given
x (m) F (N)
0.05 27.0 10-16
0.10 8.10 10-16
0.15 5.76 10-16
0.20 8.10 10-16
0.25 27.0 10-16
b) in the adjoint we can see a graph of the force against the distance, it can be seen that it has the shape of a parabola with a minimum close to x = 0.15 m
