Answer:
the final potential energy of this system is 3U0/10
Explanation:
We are given
charge at left end and another test charge at point p
Potential energy is given by = 
where k is electrostatics constant = 
Q1 = first charge , Q2= test charge
R= distance between charges
potential at point p
U0 = k*Q1*Q2 /3 ⇒ kq1q2 = 3U0 ..............1
now the test charge moves to point R
using Pytahgoreou theorem
R(distance) = 
= 10
New Potential energy
U1 = kq1*q2 / 10
substituting kq1q2 = 3U0 from 1
U1 = 3U0/10
So this is the final potential energy of this system.
A ) v = v o + a t ( the acceleration will be negative )
9.50 = 16.0 + a * 1.2
a * 1.2 = -16.0 + 9.50
a * 1.2 = - 6.5
a = - 6.5 : 1.2
a = - 5.4167 m/s²
F = m * a = 950 kg * 5.4167 m/s²
F = 5,145.8 N ( the average force exerted on a car during braking )
b ) d = v o - a t² / 2
d = 16.0 * 1.2 - ( 5.4167 * 1.2² / 2 ) =
= 19.20 - 3.90 = 15.30 m
Answer: final Velocity v = 10.2m/s
Explanation:
Final speed v(t) is given as
v(t) = u + at .......1
Where; u = the initial speed
a = acceleration
t = time taken
The total distance travelled d is given as
d = ut + 1/2(at^2)
Given
d = 5.0m
u = 2.0m
a = g = 10m/s2 (acceleration due to gravity)
Substituting into the equation above we have
5 = 2t + 5t^2
5t^2 +2t -5 = 0
Applying the quadratic formula. We have;
t = 0.82s & t = -1.22s
t cannot be negative
t = 0.82s
From equation 1 above
v = 2.0m/s + 10(0.82)m/s
v = 10.2m/s
15+3=18km/hour
Think about it like this. The boat is going 15 faster than the river, and the river is going 3 faster than the bank, so the boat is going 18 faster than the river bank
