Answer:
at d the charge will be 3q and at 3d it will be 9q
Explanation:
for V=Vp-V2d
V=KQ/d=K*6q/2d=3kq/d for potential to 2d at 6q be zero the Vp will equal 3kq/d; hence at d, Q=3q and at 3d, Q=9q
The equation is garbled and the question is missing.
I found this equation for the same statement:
S = - 2.7t ^2 + 30t + 6.5
And one question is: after how many seconds is the ball 12 feet above the moon's surface?
Given that S is the height of the ball, you just have to replace S with 12 and solve for t.
=> 12 = - 2.7 t^2 + 30t + 6.5
=> 2.7t^2 - 30t - 6.5 + 12 = 0
=> 2.7t^2 - 30t + 5.5 = 0
Now you can use the quadratic equation fo find t:
t = { 30 +/- √ [30^2) - 4(2.7)(5.5)] } / (2*2.7)
=> t = 0.186s and t = 10.925 s
Answer: after 0.186 s the ball is at 12 feet over the surface, and again 10.925 s
Final speed after collision with the wall

before collision the speed of ball initially

time taken for the collision

now as per the formula of acceleration we know that

now plug in all values in it


so acceleration is - 5 m/s/s for above situation
Answer:
45
Explanation:
because the equation for speed is distance divided by time! hope that helps gave a nice day!