Explanation:
A projectile (Cannon ball) is launched at an angle to the horizontal and rises up to a peak while moving horizontally. When it reaches the peak, the projectile starts to fall.
Answer:
0.94 m³/s
Explanation:
From the question given above, the following data were obtained:
Air flow (in ft³/min) = 2×10³ ft³/min
Air flow (in m³/s) =.?
Next, we shall convert 2×10³ ft³/min to m³/min. This can be obtained as follow:
35.315 ft³/min = 1 m³/min
Therefore,
2×10³ ft³/min = 2×10³ ft³/min × 1 m³/min / 35.315 ft³/min
2×10³ ft³/min = 56.63 m³/min
Finally, we shall convert 56.63 m³/min to m³/s. This can be obtained as follow:
1 m³/min = 1/60 m³/s
Therefore,
56.63 m³/min = 56.63 m³/min × 1/60 m³/s ÷ 1 m³/min
56.63 m³/min = 0.94 m³/s
Thus, 2×10³ ft³/minis equivalent to 0.94 m³/s.
To develop this problem, it is necessary to apply the concepts related to the description of the movement through the kinematic trajectory equations, which include displacement, velocity and acceleration.
The trajectory equation from the motion kinematic equations is given by
Where,
a = acceleration
t = time
= Initial velocity
= initial position
In addition to this we know that speed, speed is the change of position in relation to time. So
x = Displacement
t = time
With the data we have we can find the time as well
With the equation of motion and considering that we have no initial position, that the initial velocity is also zero then and that the acceleration is gravity,
Therefore the vertical distance that the ball drops as it moves from the pitcher to the catcher is 1.46m.
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