Answer: Both cannonballs will hit the ground at the same time.
Explanation:
Suppose that a given object is on the air. The only force acting on the object (if we ignore air friction and such) will be the gravitational force.
then the acceleration equation is only on the vertical axis, and can be written as:
a(t) = -(9.8 m/s^2)
Now, to get the vertical velocity equation, we need to integrate over time.
v(t) = -(9.8 m/s^2)*t + v0
Where v0 is the initial velocity of the object in the vertical axis.
if the object is dropped (or it only has initial velocity on the horizontal axis) then v0 = 0m/s
and:
v(t) = -(9.8 m/s^2)*t
Now, if two objects are initially at the same height (both cannonballs start 1 m above the ground)
And both objects have the same vertical velocity, we can conclude that both objects will hit the ground at the same time.
You can notice that the fact that one ball is fired horizontally and the other is only dropped does not affect this, because we only analyze the vertical problem, not the horizontal one. (This is something useful to remember, we can separate the vertical and horizontal movement in these type of problems)
Answer:
The depth is 5.15 m.
Explanation:
Lets take the depth of the pool = h m
The atmospheric pressure ,P = 101235 N/m²
The area of the top = A m²
The area of the bottom = a m²
Given that A= 1.5 a
The force on the top of the pool = P A
The total pressure on the bottom = P + ρ g h
ρ =Density of the water = 1000 kg/m³
The total pressure at the bottom of the pool = (P + ρ g h) a
The bottom and the top force is same
(P + ρ g h) a = P A
P a +ρ g h a = P A
ρ g h a = P A - P a
h=5.15 m
The depth is 5.15 m.
I think the correct answer from the choices listed above is the first option. In order for a person to "see" an object, light waves pass through the cornea. The cornea is the transparent layer forming at the front of the eye. Hope this answers the question. Have a nice day.
Explanation:
It is given that,
Mass of the woman, m₁ = 52 kg
Angular velocity, 
Mass of disk, m₂ = 118 kg
Radius of the disk, r = 3.9 m
The moment of inertia of woman which is standing at the rim of a large disk is :
I₁ = 790.92 kg-m²
The moment of inertia of of the disk about an axis through its center is given by :
I₂ =897.39 kg-m²
Total moment of inertia of the system is given by :
I = 1688.31 kg-m²
The angular momentum of the system is :
So, the total angular momentum of the system is 4980.5 kg-m²/s. Hence, this is the required solution.
