Answer:
Final momentum after a head on collision is -2kgm/s
Explanation:
One ball moves to the right and the other moves opposite and momentum is a vector quantity so that considering the direction
Initial momenta are P₁=2x3=6kgm/s P₂=4x(-2)=-8kgm/s
Final momentum is the vector sum of P(final)= 6-8= -2 kgm/s
Answer:
1.06 secs
Explanation:
Initial speed of sled, u = 8.4 m/s
Final speed of sled, v = 5.8 m/s
Coefficient of kinetic friction, μ = 0.25
Using the impulse momentum theory, we know that the impulse applied to the sled is equal to change in momentum of the sled:
FΔt = mv - mu
where m = mass of the object
Δt = time interval
F = force applied
The force applied on the sled is the frictional force, which is given as:
F = -μmg
where g = acceleration due to gravity
Therefore:
-μmgΔt = mv - mu
-μmgΔt = m(v - u)
-μgΔt = v - u
Making Δt subject of formula:
Δt = (v - u) / -μg
Δt = (5.8 - 8.4) / (-0.25 * 9.8)
Δt = -2.6/ -2.45
Δt = 1.06 secs
It took the sled 1.06 secs to travel from A to B.
Answer:
The maximum energy stored in the combination is 0.0466Joules
Explanation:
The question is incomplete. Here is the complete question.
Three capacitors C1-11.7 μF, C2 21.0 μF, and C3 = 28.8 μF are connected in series. To avoid breakdown of the capacitors, the maximum potential difference to which any of them can be individually charged is 125 V. Determine the maximum energy stored in the series combination.
Energy stored in a capacitor is expressed as E = 1/2CtV² where
Ct is the total effective capacitance
V is the supply voltage
Since the capacitors are connected in series.
1/Ct = 1/C1+1/C2+1/C3
Given C1 = 11.7 μF, C2 = 21.0 μF, and C3 = 28.8 μF
1/Ct = 1/11.7 + 1/21.0 + 1/28.8
1/Ct = 0.0855+0.0476+0.0347
1/Ct = 0.1678
Ct = 1/0.1678
Ct = 5.96μF
Ct = 5.96×10^-6F
Since V = 125V
E = 1/2(5.96×10^-6)(125)²
E = 0.0466Joules
