Complete Question
Planet D has a semi-major axis = 60 AU and an orbital period of 18.164 days. A piece of rocky debris in space has a semi major axis of 45.0 AU. What is its orbital period?
Answer:
The value is
Explanation:
From the question we are told that
The semi - major axis of the rocky debris
The semi - major axis of Planet D is
The orbital period of planet D is
Generally from Kepler third law
Here T is the orbital period while a is the semi major axis
So
=>
=>
=>
Average acceleration = (change in speed) / (time for the change) .
Change in speed = (ending speed) - (beginning speed)
= (9.89 miles/hour) - (2.35 yards/second) = 26,839.2 ft/hr
Acceleration = (26,839.2 ft/hr) / (4.67 days) = 2,873.58 inch/hour²
Answer:
(a) 104 N
(b) 52 N
Explanation:
Given Data
Angle of inclination of the ramp: 20°
F makes an angle of 30° with the ramp
The component of F parallel to the ramp is Fx = 90 N.
The component of F perpendicular to the ramp is Fy.
(a)
Let the +x-direction be up the incline and the +y-direction by the perpendicular to the surface of the incline.
Resolve F into its x-component from Pythagorean theorem:
Fx=Fcos30°
Solve for F:
F= Fx/cos30°
Substitute for Fx from given data:
Fx=90 N/cos30°
=104 N
(b) Resolve r into its y-component from Pythagorean theorem:
Fy = Fsin 30°
Substitute for F from part (a):
Fy = (104 N) (sin 30°)
= 52 N