Answer:
the can's kinetic energy is 0.42 J
Explanation:
given information:
Mass, m = 460 g = 0.46 kg
diameter, d = 6 cm, so r = d/2 = 6/2 = 3 cm = 0.03 m
velocity, v = 1.1 m/s
the kinetic energy of the can is the total of kinetic energy of the translation and rotational.
KE =
I ω^2 + 
where
I =
and ω = 
thus,
KE =
(
)^2 + 
=
+ 
=
+ 
= 
=
= 0.42 J
The car has an initial velocity
of 23 m/s and a final velocity
of 0 m/s. Recall that for constant acceleration,

The car stops in 7 s, so the acceleration is


Answer:
Explanation:
v = u +at
u = 0
a = 2.3 m /s²
t = 20 s
v = 2.3 x 20
= 46 m /s
Distance covered under acceleration of 2.3 m/s²
s = ut + 1/2 at²
= 0 + .5 x 2.3 x 20²
= 460 m
After that it moves under free fall ie g acts on it downwards .
v² = u² - 2gh , h is height moved by it under free fall
0 = 46² - 2 x 9.8 h
h = 107.96 m
Total height attained
= 460 + 107.96
= 567.96 m
b ) At its highest point ,it stops so its velocity = 0
c ) rocket's acceleration at its highest point = g = 9.8 downwards .
At highest point , it is undergoing free fall so its acceleration = g
I'm not sure what "60 degree horizontal" means.
I'm going to assume that it means a direction aimed 60 degrees
above the horizon and 30 degrees below the zenith.
Now, I'll answer the question that I have invented.
When the shot is fired with speed of 'S' in that direction,
the horizontal component of its velocity is S cos(60) = 0.5 S ,
and the vertical component is S sin(60) = S√3/2 = 0.866 S . (rounded)
-- 0.75 of its kinetic energy is due to its vertical velocity.
That much of its KE gets used up by climbing against gravity.
-- 0.25 of its kinetic energy is due to its horizontal velocity.
That doesn't change.
-- So at the top of its trajectory, its KE is 0.25 of what it had originally.
That's E/4 .