Respuesta :
Answer;
= 90600 m/s...
Explanation;
velocity of ball = 96 mi/hr = 42.916 m/s...
velocity of bat = -56mi/hr = -25.034m/s.. ( note the negativity here..)
time = 0.75 millisec = 0.75 * 10^-3 sec
acceleration = (velball - velbat) / time
= (42.916 - (-25.034)) / (0.75 * 10^-3)
=90600 m/s
Answer:
The magnitude of the average acceleration of the ball during the contact with the bat is 90666m/s^2
Explanation:
Average acceleration (a) is defined as the change in velocity divided by the time it took:
[tex]a=\frac{V_{2}-V_{1} }{t}[/tex] (eq. 1)
V2: final velocity
V1: initial velocity
t: time the change in velocity took
In this example, the ball traveled at 96mi/h in a direction and changed its velocity to 56mi/h in the opposite direction because of the contact with the bat.
In this case, we need to express all in meters and seconds. To achieve this use some conversion factors:
[tex]V_{1} =96mi/h=96mi/h*(\frac{0.45m/s}{1mi/h} )=43m/s\\V_{2} =-56mi/h=-56mi/h*(\frac{0.45m/s}{1mi/h} )=-25m/s\\t=0.75ms=0.75ms*(\frac{1s}{1000ms} )=7.5*10^{-4} s[/tex]
Note there's a negative sign beside 56 mi/hr because it goes in the opposite direction to 96 mi/hr.
Plugging in these values in eq. 1:
[tex]a=\frac{-25m/s-43m/s }{7.5*10^{-4} s}=90666m/s^2[/tex]
