Answer:

Explanation:
We'll use the momentum equation:

where:
p = momentum
m = mass
v = velocity
Since we're doing the magnitude of momentum of the system, we'll add the mass of the cyclist and the mountain bike together:

Given that, we can now substitute our given values into the momentum equation:

Our final answer is:

Point charges q1=+2.00μC and q2=−2.00μC are placed at adjacent corners of a square for which the length of each side is 5.00 cm.?
Point a is at the center of the square, and point b is at the empty corner closest to q2. Take the electric potential to be zero at a distance far from both charges.
(a) What is the electric potential at point a due to q1 and q2?
(b) What is the electric potential at point b?
(c) A point charge q3 = -6.00 μC moves from point a to point b. How much work is done on q3 by the electric forces exerted by q1 and q2?
Answer:
a) the potential is zero at the center .
Explanation:
a) since the two equal-magnitude and oppositely charged particles are equidistant
b)(b) Electric potential at point b, v = Σ kQ/r
r = 5cm = 0.05m
k = 8.99*10^9 N·m²/C²
Q = -2 microcoulomb
v= (8.99*10^9) * (2*10^-6) * (1/√2m - 1) / 0.0500m
v = -105 324 V
c)workdone = charge * potential
work = -6.00µC * -105324V
work = 0.632 J
Formula: PE = mgh
m = 20 kg, g = 9.8 m/s^2 h= ?
1000 = 20 * 9.8 * h
1000 = 196h
h = 5.10204082
The height is around 5m