That statement is true
Retinal disparity : space between your eyes that allow binocular vision to create depth perception
Retinal Convergence : Space between your eyes that signal visual moves to the retina
They both will increases as an object get closer to the individual, allowing them acknowledge and observe the existence of the object
Answer:
a) 145.6kgm^2
b) 158.4kg-m^2/s
c) 0.76rads/s
Explanation:
Complete qestion: a) the rotational inertia of the merry-go-round about its axis of rotation
(b) the magnitude of the angular momentum of the child, while running, about the axis of rotation of the merry-go-round and
(c) the angular speed of the merry-go-round and child after the child has jumped on.
a) From I = MK^2
I = (160Kg)(0.91m)^2
I = 145.6kgm^2
b) The magnitude of the angular momentum is given by:
L= r × p The raduis and momentum are perpendicular.
L = r × mc
L = (1.20m)(44.0kg)(3.0m/s)
L = 158.4kg-m^2/s
c) The total moment of inertia comprises of the merry- go - round and the child. the angular speed is given by:
L = Iw
158.4kgm^2/s = [145kgm^2 + ( 44.0kg)(1.20)^2]
w = 158.6/208.96
w = 0.76rad/s
Answer:
The workdone is
Explanation:
From the question we are told that
The potential difference is 
Generally the charge on
is 
Generally the workdone is mathematically represented as

=>
=>
Answer:
v=77.62 m/s
Explanation:
Given that
h= - 300 m
speed of the bird ,u= 5 m/s
Lets take Speed of the berry when it hit the ground = v m/s
we know that ,if object is moving upward
v² = u² - 2 g h
u=Initial speed
v=Final speed
h=Height
Now by putting the values
v² = u² - 2 g h
v² = 5² - 2 x 10 x (-300) ( take g = 10 m/s²)
v² =25 + 20 x 300
v² ==25 + 6000
v² =6025
v=77.62 m/s
Therefore the final speed of the berry will be 77.62 m/s.
The masses of the objects and how much distance there is between them