Answer:
a). A conservative force permits a two-way conversion between kinetic and potential energies.
TRUE
Because there is no energy loss in presence of conservative forces so energy conversion in two ways are possible.
b). A potential energy function can be specified for a conservative force.
TRUE
negative gradient of potential energy is equal to conservative force

c). A non-conservative force permits a two-way conversion between kinetic and potential energies.
FALSE
here energy is lost against non-conservative forces
d). The work done by a conservative force depends on the path taken.
FALSE
work done by conservative force is independent of path
e). The work done by a non-conservative force depends on the path taken.
TRUE
work done by non conservative forces depends on path.
f). A potential energy function can be specified for a non-conservative force.
FALSE
It is not defined for non conservative forces
Answer:
(3) The period of the satellite is independent of its mass, an increase in the mass of the satellite will not affect its period around the Earth.
(4) he gravitational force between the Sun and Neptune is 6.75 x 10²⁰ N
Explanation:
(3) The period of a satellite is given as;

where;
T is the period of the satellite
M is mass of Earth
r is the radius of the orbit
Thus, the period of the satellite is independent of its mass, an increase in the mass of the satellite will not affect its period around the Earth.
(4)
Given;
mass of the ball, m₁ = 1.99 x 10⁴⁰ kg
mass of Neptune, m₂ = 1.03 x 10²⁶ kg
mass of Sun, m₃ = 1.99 x 10³⁰ kg
distance between the Sun and Neptune, r = 4.5 x 10¹² m
The gravitational force between the Sun and Neptune is calculated as;

Answer:
Applying enough force in one direction to move the object, making kinetic energy.
Explanation:
Simpleness
Answer:
q = - 93.334 nC
Explanation:
GIVEN DATA:
Radius of ring 73 cm
charge on ring 610 nC
ELECTRIC FIELD p FROM CENTRE IS AT 70 CM
E = 2000 N/C
Electric field due tor ring is guiven as
![E = \frac{KQx}{[x^2+ R^2]^{3/2}}](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7BKQx%7D%7B%5Bx%5E2%2B%20R%5E2%5D%5E%7B3%2F2%7D%7D)

E1 = 3714.672 N/C
electric field due to point charge q



now the eelctric charge at point P is
E = E1 + E2
solving for q
q = - 93.334 nC