Answer:
Gpe = 15680 Joules
Explanation:
Gravitational potential energy (GPE) is an energy possessed by an object or body due to its position above the earth.
Mathematically, gravitational potential energy is given by the formula;
G.P.E = mgh
Where;
G.P.E represents potential energy measured in Joules.
m represents the mass of an object.
g represents acceleration due to gravity measured in meters per seconds square.
h represents the height measured in meters.
Given the following data;
Mass = 20 kg
Height = 80 m
We know that acceleration due to gravity is equal to 9.8 m/s²
To find the gravitational potential energy;
Gpe = mgh
Gpe = 20 * 80 * 9.8
Gpe = 15680 Joules
I think it would'nt move at all but im not postive
-- <span>The gravitational force that you feel when you stand on the surface
of a planet depends on the planet's mass and size. It has </span><span><span>nothing
to do with the planet's orbit. (</span>Of course,"size" is also related to the
planet's mass, density, and surface area.)
-- One possible cause of deforestation is the removal of trees without
adequate replanting.
-- According to Hubble’s Law, the farther away a galaxy is, the faster
it is moving away from us
-- Electromagnetic energy can be defined as energy that moves at
the speed of light. If you conduct experiments to determine whether
the electromagnetic energy is moving in the form of particles or waves,
you find that it behaves as both.</span>
I think so not take my answer because I think the answer is b
Remember Coulomb's law: the magnitude of the electric force F between two stationary charges q₁ and q₂ over a distance r is
where k ≈ 8,98 × 10⁹ kg•m³/(s²•C²) is Coulomb's constant.
8.1. The diagram is simple, since only two forces are involved. The particle at Q₂ feels a force to the left due to the particle at Q₁ and a downward force due to the particle at Q₃.
8.2. First convert everything to base SI units:
0,02 µC = 0,02 × 10⁻⁶ C = 2 × 10⁻⁸ C
0,03 µC = 3 × 10⁻⁸ C
0,04 µC = 4 × 10⁻⁸ C
300 mm = 300 × 10⁻³ m = 0,3 m
600 mm = 0,6 m
Force due to Q₁ :
Force due to Q₃ :
8.3. The net force on the particle at Q₂ is the vector
Its magnitude is
and makes an angle θ with the positive horizontal axis (pointing to the right) such that
where we subtract 180° because 
terminates in the third quadrant, but the inverse tangent function can only return angles between -90° and 90°. We use the fact that tan(x) has a period of 180° to get the angle that ends in the right quadrant.
