The valence electrons are the one furthest from the nucleus
(30, 5)
(10, 1)
change of y / change of x
= (30 - 10) / (5 - 1)
= 20 /4
= 5
Answer:
Part a)
Part b)
Explanation:
As we know that mountain climber is at rest so net force on it must be zero
So we will have force balance in X direction
now we will have force balance in Y direction
Part a)
so from above equations we have
Part b)
Now for tension in right string we will have
Answer:
O The particles of the medium move more slowly and there are fewer chances to transfer energy.
Explanation:
Various media are made up of particles. These particles are in constant motion according to the kinetic theory of matter. Recall that temperature has been defined as the average kinetic energy of the particles in a medium. Hence, for any given medium, the velocity of particle motion increases or decreases linearly with temperature.
The speed of particles in any medium increases or decreases as the temperature of the medium increases or decreases as emphasised above. Hence, at low temperature, the velocity of waves set up by the motion of particles in a medium decreases and transfer the wave energy to neighbouring particles occurs more slowly than at high temperatures.
Answer:
serie Ceq=0.678 10⁻⁶ F and the charge Q = 9.49 10⁻⁶ C
Explanation:
Let's calculate all capacity values
a) The equivalent capacitance of series capacitors
1 / Ceq = 1 / C1 + 1 / C2 + 1 / C3 + 1 / C4 + 1 / C5
1 / Ceq = 1 / 1.5 + 1 / 3.3 + 1 / 5.5 + 1 / 6.2 + 1 / 6.2
1 / Ceq = 1 / 1.5 + 1 / 3.3 + 1 / 5.5 + 2 / 6.2
1 / Ceq = 0.666 + 0.3030 +0.1818 +0.3225
1 / Ceq = 1,147
Ceq = 0.678 10⁻⁶ F
b) Let's calculate the total system load
Dv = Q / Ceq
Q = DV Ceq
Q = 14 0.678 10⁻⁶
Q = 9.49 10⁻⁶ C
In a series system the load is constant in all capacitors, therefore, the load in capacitor 5.5 is Q = 9.49 10⁻⁶ C
c) The potential difference
ΔV = Q / C5
ΔV = 9.49 10⁻⁶ / 5.5 10⁻⁶
ΔV = 1,725 V
d) The energy stores is
U = ½ C V²
U = ½ 0.678 10-6 14²
U = 66.4 10⁻⁶ J
e) Parallel system
Ceq = C1 + C2 + C3 + C4 + C5
Ceq = (1.5 +3.3 +5.5 +6.2 +6.2) 10⁻⁶
Ceq = 22.7 10⁻⁶ F
f) In the parallel system the voltage is maintained
Q5 = C5 V
Q5 = 5.5 10⁻⁶ 14
Q5 = 77 10⁻⁶ C
g) The voltage is constant V5 = 14 V
h) Energy stores
U = ½ C V²
U = ½ 22.7 10-6 14²
U = 2.2 10⁻³ J
