I believe the answer is potential energy if i remember correctly.
The equation that relates distance, velocities, acceleration, and time is,
d = V₀t + 0.5gt²
where d is distance,
V₀ is the initial velocity,
t is time, and
g is the acceleration due to gravity (equal to 9.8 m/s²)
(1) Dropped rock,
(3 x 10² m ) = 0(t) + 0.5(9.8 m/s²)(t²)
The value of t from this equation is 24.73 s
(2) Thrown rock with V₀ = 26 m/s
(3 x 10² m) = (26)(t) + 0.5(9.8 m/s²)(t²)
The value of t from the equation is 5.61 s
The difference between the tim,
difference = 24.73 s - 5.61 s
difference = 19.12 s
<em>ANSWER: 19.12 s</em>
<span>Assuming that the momenta of the two pieces are equal: when they have equal velocities, then
the masses of the two pieces are also equal.
Since there is no force from outside of the system, the center of mass moves on with the same velocity as before the equation. So the two pieces must fly at the side side of the mass center, i.e., they must always be at 90° to the side of the mass center. Otherwise it would not be the mass center, respectively the pieces would not have equal velocities.
This is only possible, when the angle of their velocity with the initial direction is 60°.
Because, cos (60°) = 1/2 = v/(2v).</span>
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 .
Answer:
the knee extensors must exert 15.87 N
Explanation:
Given the data in the question;
mass m = 4.5 kg
radius of gyration k = 23 cm = 0.23 m
angle ∅ = 30°
∝ = 1 rad/s²
distance of 3 cm from the axis of rotation at the knee r = 3 cm = 0.03 m
using the expression;
ζ = I∝
ζ = mk²∝
we substitute
ζ = 4.5 × (0.23)² × 1
ζ = 0.23805 N-m
so
from; ζ = rFsin∅
F = ζ / rsin∅
we substitute
F = 0.23805 / (0.03 × sin( 30 ° )
F = 0.23805 / (0.03 × 0.5)
F F = 0.23805 / 0.015
F = 15.87 N
Therefore, the knee extensors must exert 15.87 N
