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:
6.53 m/s²
Explanation:
Let m₁ = 5 kg and m₂ = 10 kg. The figure is attached and free body diagrams of the objects are also attached.
Both objects (m₁ and m₂) have the same magnitude of acceleration(a). Let g be the acceleration due to gravity = 9.8 m/s². Hence:
T = m₁a (1)
m₂g - T = m₂a (2)
substituting T = m₁a in equation 2:
m₂g - m₁a = m₂a
m₂a + m₁a = m₂g
a(m₁ + m₂) = m₂g
a = m₂g / (m₁ + m₂)
a = (10 kg * 9.8 m/s²) / (10 kg + 5 kg) = 6.53 m/s²
Both objects have an acceleration of 6.53 m/s²
