If a ball dropped from a tower reaches the ground after 3.5 seconds, what is the height of the tower? Given: g = –9.8 meters/sec
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Answer:
Yes, the paths of the two particles cross.
Location of path intersection = ( 1 , 2 , 3)
Explanation:
In order to find the point of intersection, we need to set both locations equal to one another. It should be noted however, that the time for each particle can vary as we are finding the point where the <u>paths</u> meet, not the point where the particles meet themselves.
So, we can name the time of the first particle , and the time of the second particle
.
Setting the locations equal, we get the following equations to solve for and
:
Equation 1
Equation 2
Equation 3
Solving these three equations simultaneously we get:
2 seconds
4 seconds
Since, we have an answer for when the trajectories cross, we know for a fact that they indeed do cross.
The point of crossing can be found by using the value of or
in the location matrices. Doing this for the first particle we get:
Location of path intersection = ( -1 + 2 , 4 - 2 , -1 + 2(2) )
Location of path intersection = ( 1 , 2 , 3)
Answer:
KE = m
Explanation:
In the generation of energy from hydroelectric power station, the motion of water, and the turbines are paramount. The falling flowing water turns the blades of the turbine, which in-turn causes the movement of a coil within a strong magnetic field.
The motion of the coil which cuts the strong magnetic field induces current. Thus, the system generates electrical energy.
The equation that links kinetic energy (KE), mass (m) and speed (v) can be expressed as:
KE = m
(a) 5.66 m/s
The flow rate of the water in the pipe is given by
where
Q is the flow rate
A is the cross-sectional area of the pipe
v is the speed of the water
Here we have
the radius of the pipe is
r = 0.260 m
So the cross-sectional area is
So we can re-arrange the equation to find the speed of the water:
(b) 0.326 m
The flow rate along the pipe is conserved, so we can write:
where we have
and where is the cross-sectional area of the pipe at the second point.
Solving for A2,
And finally we can find the radius of the pipe at that point: