Answer:
16.9000000000000001 J
Explanation:
From the given information:
Let the initial kinetic energy from point A be 
= 1.9000000000000001 J
and the final kinetic energy from point B be 
= ???
The charge particle Q = 6 mC = 6 × 10⁻³ C
The change in the electric potential from point B to A;
i.e. V_B - V_A = -2.5 × 10³ V
According to the work-energy theorem:
-Q × ΔV = ΔK
V=(40km/hr)(hr/3600s)(1000000mm/km)
v=11111.1mm/s
v=d/t
d=vt
d=(11111.1mm/s)(5s)
d=55555mm
d=5.56x10^4mm
Weight doesn't really mean much as it just means gravity the bigger a space object is the more force it has to pull on something since the moon is smaller than the earth then it has less gravity and then less weight on scales.
The weightiness of the added
water displaced is equivalent to the joined weight of the two extra people who come
to be into the boat:
<span>m water g = 2 x 690 N</span>
<span> =
1,380 N</span>
<span>
</span>
The mass of the water displace
is then
<span>m water g = 1,380 N</span>
<span> = 1,380 N / 9.8 m/s^2</span>
<span> = 141 kg</span>
<span>
</span>
Compute the calculation for
density for the volume of water displace and practice this outcome for the mass
of the water displace to get the answer:
<span>p water = mass of water / volume of water</span>
<span>
</span>
<span>volume of water = mass of water / p water</span>
<span> = 141 kg / 1000 kg /m^3 eliminate
kilogram</span>
<span> = 0.14 m^3 the additional volume
of water that is displaced</span>
