Answer:
313.6 m downward
Explanation:
The distance covered by the bullet along the vertical direction can be calculated by using the equation of motion of a projectile along the y-axis.
In fact, we have:
where
y(t) is the vertical position of the projectile at time t
h is the initial height of the projectile
is the initial vertical velocity of the projectile, which is zero since the bullet is fired horizontally
t is the time
a = g = -9.8 m/s^2 is the acceleration due to gravity
We can rewrite the equation as
where the term on the left, 
, represents the vertical displacement of the bullet. Substituting numbers and t = 8 s, we find
So the bullet has travelled 313.6 m downward.
We wouldn’t really have a solid concept of time. We also wouldn’t be able to keep up with the tides
She could look for an online job or a job she can work at from home.
Here is the full question:
The rotational inertia I of any given body of mass M about any given axis is equal to the rotational inertia of an equivalent hoop about that axis, if the hoop has the same mass M and a radius k given by:
The radius k of the equivalent hoop is called the radius of gyration of the given body. Using this formula, find the radius of gyration of (a) a cylinder of radius 1.20 m, (b) a thin spherical shell of radius 1.20 m, and (c) a solid sphere of radius 1.20 m, all rotating about their central axes.
Answer:
a) 0.85 m
b) 0.98 m
c) 0.76 m
Explanation:
Given that: the radius of gyration
So, moment of rotational inertia (I) of a cylinder about it axis = 
k = 0.8455 m
k ≅ 0.85 m
For the spherical shell of radius
(I) = 
k = 0.9797 m
k ≅ 0.98 m
For the solid sphere of radius
(I) = 
k = 0.7560
k ≅ 0.76 m
