Answer:
38 cm from q1(right)
Explanation:
Given, q1 = 3q2 , r = 60cm = 0.6 m
Let that point be situated at a distance of 'x' m from q1.
Electric field must be same from both sides to be in equilibrium(where EF is 0).
=> k q1/x² = k q2/(0.6 - x)²
=> q1(0.6 - x)² = q2(x)²
=> 3q2(0.6 - x)² = q2(x)²
=> 3(0.6 - x)² = x²
=> √3(0.6 - x) = ± x
=> 0.6√3 = x(1 + √3)
=> 1.03/2.73 = x
≈ 0.38 m = 38 cm = x
Answer:
B) 12 m
Explanation:
Gravitational potential energy is:
PE = mgh
Given PE = 5997.6 J, and m = 51 kg:
5997.6 J = (51 kg) (9.8 m/s²) h
h = 12 m
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
<span>The amplitude because that controls the height of the wave. Correct answer: Amplitude.</span>
Answer:
a) -2.038 m/s²
b) 40.33 mph
c) 312.5 m
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
Acceleration of the boat is -2.083 m/s² if the boat will stop at 150 m.
Speed of the boat by when it will hit the dock is 18.03 m/s
Converting to mph
Speed of the boat by when it will hit the dock is 40.33 mph
The distance at which the boat will have to start decelerating is 312.5 m
