Answer:
V=203 ft/s
Explanation:
answer is available in word document named "attach 1" due to some technical error in maths equation. please find the attached document
The force between the two point charge when they are separated by 18 cm is 3 N
<h3>How do I determine the force when they are 18 cm apart?</h3>
Coulomb's law states as follow:
F = Kq₁q₂ / r²
Cross multiply
Fr² = Kq₁q₂
Kq₁q₂ => constant
F₁r₁² = F₂r₂²
Where
- F₁ and F₂ are the initial and new force
- r₁ and r₂ are the initial and new distance apart
With the above formula, we can obtain the force between the two point charge when they are 18 cm apart. Details below:
- Initial distance apart (r₁) = 6 cm
- Initial force of attraction (F₁) = 27 N
- New distance apart (r₂) = 18 cm
- New force of attraction (F₂) =?
F₁r₁² = F₂r₂²
27 × 6² = F₂ × 18²
972 = F₂ × 324
Divide both side by 324
F₂ = 927 / 324
F₂ = 3 N
Thus, the force when they are 18 cm apart is 3 N
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Answer:
ummm I didn't understand the question
Answer:
Explanation:
Force of gravity = GMm/r^2 = ma
a being the acceleration due to gravity at some distance r from the center of the Earth. I'll use 6400 km for the radius of the Earth so r = 6734 km or 6734000 meters
a = GM/r^2
plugging in G = 6.67 x 10^-11
M is the mass of the Earth = 6 x 10^24
and r is from above
a = 8.825 m/s^2 = 0.9g
so 90% the acceleration of gravity on the surface.