The planets move eastward against the background of fixed stars with the exception of Venus, Uranus and Pluto moving westward as seen in Earth's sky . This apparent retreating movement is called Retrograde motion. It is illusion created by Earth's movement going by outer planets in their respective orbits.
Answer:
linear density of the string = 4.46 × 10⁻⁴ kg/m
Explanation:
given,
mass of the string = 31.2 g
length of string = 0.7 m
linear density of the string = 
linear density of the string = 
linear density of the string = 44.57 × 10⁻³ kg/m
linear density of the string = 4.46 × 10⁻⁴ kg/m
Answer:
41.4 g
41.4 cm³
1.08695 g/cm³
Explanation:
= Density of water = 1 g/cm³
Mass of water displaced will be the difference of the

Mass of water displaced is 41.4 g
Density is given by

So, volume of bone is 41.4 cm³
Average density of the bird is given by

The average density is 1.08695 g/cm³
Explanation:
The criteria for decision making would be
1. I would fund for the school of young diabetics, for the sole purpose of them leaning and being motivated for a healthy lifestyle.
2. I would also fund for new and improved insulin pumps as old ones cause multiple problems.
Given:
Uniform distributed load with an intensity of W = 50 kN / m on an overhang beam.
We need to determine the maximum shear stress developed in the beam:
τ = F/A
Assuming the area of the beam is 100 m^2 with a length of 10 m.
τ = F/A
τ = W/l
τ = 50kN/m / 10 m
τ = 5kN/m^2
τ = 5000 N/ m^2<span />