Answer:
Orbital period, T = 1.00074 years
Explanation:
It is given that,
Orbital radius of a solar system planet, 
The orbital period of the planet can be calculated using third law of Kepler's. It is as follows :
M is the mass of the sun
T = 31559467.6761 s
T = 1.00074 years
So, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about 1.00074 years.
Answer:
Yes it would be different on Earth and the moon
Answer:
r = 0.02 m
Explanation:
from the question we have :
speed = 1 rps = 1x 60 = 60 rpm
coefficient of friction (μ) = 0.1
acceleration due to gravity (g) = 9.8 m/s^{2}
maximum distance without falling off (r) = ?
to get how far from the center of the disk the coin can be placed without having to slip off we equate the formula for the centrifugal force with the frictional force on the turntable force
mv^2 / r = m x g x μ
v^2 / r = g x μ .......equation 1
where
velocity (v) = angular speed (rads/seconds) x radius
angular speed (rads/seconds) = (\frac{2π}{60} ) x rpm
angular speed (rads/seconds) = (\frac{2 x π}{60} ) x 60 = 6.28 rads/ seconds
now
velocity = 6.28 x r = 6.28 r
now substituting the value of velocity into equation 1
v^2 / r = g x μ
(6.28r)^2 / r = 9.8 x 0.1
39.5 x r = 0.98
r = 0.02 m
To get a uniform field in the central region between the coils, current flows in the same direction in each.
Atomic number 1,symbol H,name hydrogen
