Explanation:
It is given that,
A planet were discovered between the sun and Mercury, with a circular orbit of radius equal to 2/3 of the average orbit radius of Mercury.
Mass of the Sun, 
Radius of Mercury's orbit, 
Radius of discovered planet, 

Let T is the orbital period of such a planet. Using Kepler's third law of planetary motion as :




T = 4135214.625 s
or
T = 47.86 days
So, the orbital period of such a planet is 47.86 days. Hence, this is the required solution.
9.3 x 10⁻⁶N
Explanation:
Given parameters:
Mass 1 = 70kg
Mass 2 = 2000kg
distance = 1m
Unknown:
force between them =
Solution:
The force between the two masses will be a gravitational force of attraction.
F = 
G is universal gravitation constant = 6.67430×10−¹¹ N⋅m²/kg²
r is the distance between the two masses
Substituting the parameters:
F =
= 9.3 x 10⁻⁶N
Learn more:
Universal gravitation constant
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Even though the object is weightless, it would need inertia, I.e, you pushing it or any form of transportation. So you would still have to push that 500kg to just keep it moving in space. Say if it were a planet with less gravitational force, it would be weightless.
Answer:


-0.04194 V
Explanation:
= Number of turns in outer solenoid = 330
= Number of turns in inner solenoid = 22
= Current in inner solenoid = 0.14 A
= Rate of change of current = 1800 A/s
= Vacuum permeability = 
r = Radius = 0.0115 m
Magnetic field is given by

The average magnetic flux through each turn of the inner solenoid is 
Magnetic flux is given by

Mutual inductance is given by

The mutual inductance of the two solenoids is 
Induced emf is given by

The emf induced in the outer solenoid by the changing current inthe inner solenoid is -0.04194 V