maximum speed of cheetah is
speed of gazelle is given as
Now the relative speed of Cheetah with respect to Gazelle
now the relative distance between Cheetah and Gazelle is given initially as "d"
now the time taken by Cheetah to catch the Gazelle is given as
so by rearranging the terms we can say
so above is the relation between all given variable
Explanation:
A wavefront is the long edge that moves, for example, the crest or the trough. Each point on the wavefront emits a semicircular wave that moves at the propagation speed v. These are drawn at a time t later, so that they have moved a distance s = vt.
Answer:
0.83 m/s
Explanation:
FIrst of all, we have to find the time of flight, i.e. the time the baseball needs to reach the ground. This can be done by using the equation for the vertical motion:
where
h is the initial height
u = 0 is the initial vertical velocity
g = 9.8 m/s^2 is the acceleration of gravity
t is the time
Substituting h = 1.8 m and solving for t,
We know that the horizontal distance travelled by the ball is
d = 0.5 m
Therefore, we can find the horizontal velocity (which is constant during the whole motion):
The component of the crate's weight that is parallel to the ramp is the only force that acts in the direction of the crate's displacement. This component has a magnitude of
<em>F</em> = <em>mg</em> sin(20.0°) = (15.0 kg) (9.81 m/s^2) sin(20.0°) ≈ 50.3 N
Then the work done by this force on the crate as it slides down the ramp is
<em>W</em> = <em>F d</em> = (50.3 N) (2.0 m) ≈ 101 J
The work-energy theorem says that the total work done on the crate is equal to the change in its kinetic energy. Since it starts at rest, its initial kinetic energy is 0, so
<em>W</em> = <em>K</em> = 1/2 <em>mv</em> ^2
Solve for <em>v</em> :
<em>v</em> = √(2<em>W</em>/<em>m</em>) = √(2 (101 J) / (2.0 m)) ≈ 10.0 m/s
Answer:
law of gravity
Explanation:
cause the ball was still moving
