Answer: 17.83 AU
Explanation:
According to Kepler’s Third Law of Planetary motion <em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”. </em>
(1)
Talking in general, this law states a relation between the <u>orbital period</u>
of a body (moon, planet, satellite, comet) orbiting a greater body in space with the <u>size</u>
of its orbit.
However, if
is measured in <u>years</u>, and
is measured in <u>astronomical units</u> (equivalent to the distance between the Sun and the Earth:
), equation (1) becomes:
(2)
This means that now both sides of the equation are equal.
Knowing
and isolating
from (2):
(3)
(4)
Finally:
(5)
Answer:
a = (v2 - v1) / t
From A to B (8 - 4) m/s / 1 s = 4 m / s^2
From A to D ( 7 - 4) m/s / 5 s = .6 m / s^2
Note these equations hold for "uniform" values
They say nothing about the acceleration at intermediate points - the equation just says that his average speed increased from 4 m/s to 7 m/s during a 5 sec period
Answer:
the required revolution per hour is 28.6849
Explanation:
Given the data in the question;
we know that the expression for the linear acceleration in terms of angular velocity is;
= rω²
ω² =
/ r
ω = √(
/ r )
where r is the radius of the cylinder
ω is the angular velocity
given that; the centripetal acceleration equal to the acceleration of gravity a
= g = 9.8 m/s²
so, given that, diameter = 4.86 miles = 4.86 × 1609 = 7819.74 m
Radius r = Diameter / 2 = 7819.74 m / 2 = 3909.87 m
so we substitute
ω = √( 9.8 m/s² / 3909.87 m )
ω = √0.002506477 s²
ω = 0.0500647 ≈ 0.05 rad/s
we know that; 1 rad/s = 9.5493 revolution per minute
ω = 0.05 × 9.5493 RPM
ω = 0.478082 RPM
1 rpm = 60 rph
so
ω = 0.478082 × 60
ω = 28.6849 revolutions per hour
Therefore, the required revolution per hour is 28.6849
Sorry that you got your answer late but the answer is 0.035m
It Increases. I just took a quiz with the same question.