Answer:
Explanation:
Δ
- Δ
is the difference in velocity before and after a given time. 
is the acceleration of the object during this time. 
is time
is another way to write this equation.
- The Δ symbol represents "the difference between the initial and final values of a magnitude or vector", so Δ
- I rearranged this equation to solve for , but this is a step that you don't need to take, it's just good to get in the habit of doing this.
- Plug in the given values. Note that our final velocity is
, because the car travels until at <em>rest</em>.
![a=\frac{v_f-v_i}{t}\\a=\frac{(0)-[(17.1\frac{miles}{hour} )(\frac{hour}{3600s})(\frac{1609.34m}{mile})]}{9.7s}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7Bv_f-v_i%7D%7Bt%7D%5C%5Ca%3D%5Cfrac%7B%280%29-%5B%2817.1%5Cfrac%7Bmiles%7D%7Bhour%7D%20%29%28%5Cfrac%7Bhour%7D%7B3600s%7D%29%28%5Cfrac%7B1609.34m%7D%7Bmile%7D%29%5D%7D%7B9.7s%7D)
- Our initial velocity is in mph, something not in standard units, so if not changed, you will get an incorrect answer. What you need to do is cancel out the units your prior value had using division and multiplication, and at the same time multiply and divide the correct numbers and units into your equation. Or look up a converter.
![a=\frac{(0)-[(17.1\frac{miles}{hour} )(\frac{hour}{3600s})(\frac{1609.34m}{mile})]}{9.7s}\\a=\frac{0m/s-7.6m/s}{9.7s} \\a=\frac{-7.6m/s}{9.7s}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B%280%29-%5B%2817.1%5Cfrac%7Bmiles%7D%7Bhour%7D%20%29%28%5Cfrac%7Bhour%7D%7B3600s%7D%29%28%5Cfrac%7B1609.34m%7D%7Bmile%7D%29%5D%7D%7B9.7s%7D%5C%5Ca%3D%5Cfrac%7B0m%2Fs-7.6m%2Fs%7D%7B9.7s%7D%20%5C%5Ca%3D%5Cfrac%7B-7.6m%2Fs%7D%7B9.7s%7D)
- if you converted correctly, your answer for
will be ≅ 
. - Now divide. Notice that the units for acceleration are
or <em>meters per second, per second</em>.
- Our final answer is <em>negative </em>because the car is <em>slowing down</em>. Do not square this answer as the square symbol only applies to the units, not the magnitude.
Answer:
a. +10.9μC
b. 0.600N and downward
Explanation:
To determine the magnitude of the charge, we use the force rule that exist between two charges which us expressed as
F=(kq₁q₂)/r²
since q₁=-0.55μC and the force it applied on the charge above it is upward,we can conclude that the second charge is +ve, hence we calculate its magnitude as
q₂=Fr²/kq₁
q₂=(0.6N*0.3²)/(9*10⁹*0.55*10⁻⁶)
q₂=0.054/4950
q₂=1.09*10⁻⁵c
q₂=10.9μC.
Hence the second charge is +10.9μC
b. From the rule of charges which state that like charges repel and unlike charges attract, we can conclude that the two above charges will attract since they are unlike charges. Hence the direction of the force will be downward into the second charge and the magnitude of the force will remain the same as 0.600N
