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1 answer:
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Missing question in the text of the exercise. Found on internet:
"What is the acceleration due to gravity on the surface of the planet?"
Solution:
The gravitational acceleration at Earth's surface is given by:
where
G is the gravitational constant
M is the Earth's mass
r is the Earth's radius
The Earth's mass can be rewritten also as the product between the Earth's density,
, and its volume (the volume of a sphere of radius r):
Now let's call M' the mass of the new planet, r' its radius and d' its density. The acceleration due to gravity on the surface of the new planet is
(1)
so we need to find M' and r'.
The problem says the radius of the new planet is twice the Earth's radius:
(2)
and that its density is 2/3 of Earth's density:
so the mass M' of the new planet is, with respect to the Earth's mass:
(3)
And if we substitute (2) and (3) into (1), we find the gravitational acceleration on the surface of the new planet:
And since
, we find
"What is the acceleration due to gravity on the surface of the planet?"
Solution:
The gravitational acceleration at Earth's surface is given by:
where
G is the gravitational constant
M is the Earth's mass
r is the Earth's radius
The Earth's mass can be rewritten also as the product between the Earth's density,
Now let's call M' the mass of the new planet, r' its radius and d' its density. The acceleration due to gravity on the surface of the new planet is
so we need to find M' and r'.
The problem says the radius of the new planet is twice the Earth's radius:
and that its density is 2/3 of Earth's density:
so the mass M' of the new planet is, with respect to the Earth's mass:
And if we substitute (2) and (3) into (1), we find the gravitational acceleration on the surface of the new planet:
And since
Complete Question:
Two 3.0µC charges lie on the x-axis, one at the origin and the other at 2.0m. A third point is located at 6.0m. What is the potential at this third point relative to infinity? (The value of k is 9.0*10^9 N.m^2/C^2)
Answer:
The potential due to these charges is 11250 V
Explanation:
Potential V is given as;
where;
K is coulomb's constant = 9x10⁹ N.m²/C²
r is the distance of the charge
q is the magnitude of the charge
The first charge located at the origin, is 6.0 m from the third charge; the potential at this point is:
The second charge located at 2.0 m, is 4.0 m from the third charge; the potential at this point is:
Total potential due to this charges = 4500 V + 6750 V = 11250 V