A 2.0 g ball with a +0.05 C charge is moved from the negative plate to the positive plate, and then is released. The potential difference between the plates is 12.0V. When the ball strikes the (–) plate its velocity is most nearly equal to

24.5 m/s

Explanation:

Since the work done by the electric field on the charge equals the kinetic energy of the ball, then

qV = 1/2mv²

and v = √(2qV/m)

where q = charge = + 0.05 C, V = potential difference = 12.0 V, m = mass of ball = 2.0 g = 0.002 kg

Substituting the values into v, we have

v = √(2 × + 0.05 C × 12.0 V/0.002 kg)

= √(1.2 CV/.002 kg)

= √(600 CV/kg)

= 24.5 m/s

hello, i am papaguy.

i have found an answer to your question but use it as information.

information: because weak acids/bases have low concentration of hydrogen ions, the addition of water has a large impact on the ph. 2. a difference in 1 unit is a 10x difference in concentration for example; a liquid with ph of 3 is 10x more acidic than a liquid with a ph of 4. therefore, a liquid with a ph of 3 is stronger.

best of luck!

distance = 150m

displ = root (50 ^2 + 100^2)

root(2500+10000)

root 125000