In general, impulse = F x time=mxv
if J is the impulse, J=<span>Ft, J=mv, and J=p because p=mv
possible formulas are
</span><span>J = Ft, J = mv and J = p</span>
For a point charge, how does the potential vary with distance from the point charge, r?
a constant
b. r.
c. 1/r.
d. 
.
e. 
.
Answer:
The correct option is C
Explanation:
Generally for a point charge the electric potential is mathematically represented as
Here we can deduce that the electric potential varies inversely with the distance i.e
So
Given:
u = 6.5 m/s, initial velocity
a = 1.5 m/s², acceleration
s = 100.0 m, displacement
Let v = the velocity attained after the 100 m displacement.
Use the formula
v² = u² + 2as
v² = (6.5 m/s)² + 2*(1.5 m/s²)*(100 m) = 342.25 (m/s)²
v = 18.5 m/s
Answer: 18.5 m/s
Given:
m = 555 g, the mass of water in the calorimeter
ΔT = 39.5 - 20.5 = 19 °C, temperature change
c = 4.18 J/(°C-g), specific heat of water
Assume that all generated heat goes into heating the water.
Then the energy released is
Q = mcΔT
= (555 g)*(4.18 J/(°C-g)*(19 °C)
= 44,078.1 J
= 44,100 J (approximately)
Answer: 44,100 J
