Answer:
1 astronomical unit, or AU, is the average distance from the Earth to the Sun; that's about 150 million km. So, Neptune's average distance from the Sun is 30.1 AU. Its perihelion is 29.8 AU, and it's aphelion is 30.4 AU.
Short Answer: it is 29
Explanation:
sorry if its wrong
Answer:
Δ T = 2.28°C
Explanation:
given,
mass of marble = 100 Kg
height of fall = 200 m
acceleration due to gravity = 9.8 m/s²
C_marble = 860 J/(kg °C)
using conservation of energy
Potential energy = heat energy
Δ T = 2.28°C
Answer:
vi = 4.77 ft/s
Explanation:
Given:
- The radius of the surface R = 1.45 ft
- The Angle at which the the sphere leaves
- Initial velocity vi
- Final velocity vf
Find:
Determine the sphere's initial speed.
Solution:
- Newton's second law of motion in centripetal direction is given as:
m*g*cos(θ) - N = m*v^2 / R
Where, m: mass of sphere
g: Gravitational Acceleration
θ: Angle with the vertical
N: Normal contact force.
- The sphere leaves surface at θ = 34°. The Normal contact is N = 0. Then we have:
m*g*cos(θ) - 0 = m*vf^2 / R
g*cos(θ) = vf^2 / R
vf^2 = R*g*cos(θ)
vf^2 = 1.45*32.2*cos(34)
vf^2 = 38.708 ft/s
- Using conservation of energy for initial release point and point where sphere leaves cylinder:
ΔK.E = ΔP.E
0.5*m* ( vf^2 - vi^2 ) = m*g*(R - R*cos(θ))
( vf^2 - vi^2 ) = 2*g*R*( 1 - cos(θ))
vi^2 = vf^2 - 2*g*R*( 1 - cos(θ))
vi^2 = 38.708 - 2*32.2*1.45*(1-cos(34))
vi^2 = 22.744
vi = 4.77 ft/s
The answer would be D hope it helps and sorry if it is wrong. :)
