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
The formula we use
here is:
radial acceleration =
ω^2 * R <span>
110,000 * 9.81 m/s^2 = ω^2 * 0.073 m
<span>ω^2 = 110,000 * 9.81 / 0.073
ω = 3844.76 rad/s </span></span>
<span>and since: ω = 2pi*f --> f = ω/(2pi)</span><span>
f = 3844.76 / (2pi) = 611.91 rps = 611.91 * 60 rpm
<span>= 36,714.77 rpm </span></span>
Answer: 0.2 hours
Explanation: In order to solve this question we have to considerer that a recargeable battery can supply 1800 mA in one hour then we have to determine how long could this battery drive current through a long, thin wire of resistance 34 Ω .
Besides, this battery has a voltage of 12 V
so by using the Ohm law we also know that V=R*I,
Fron this we can obtain:
I= V/R= 12 V/ 34 Ω=0.35 A= 350 mA
then considering that this battery can supply 1800 mA in one hour we have this battery can supply 350 mA in x time in the form:
1hour------- 1800 mA
x hour--------350 mA
time= 350/1800= 0.2 hour
It will move in the direction of applied force following the law of newton that is an object is stationary until an external force is applied then it moves in direction of force
Before swinging, T has only potential energy, (no speed)
Ui = mgh
Where h is the vertical displacement of T
From the laws of geometry,
cos45 = (L-h)/L
cos45 = 1-h/L
h/L = 1-cos45
h = L(1-cos45)
Therefore
Ui = mgL(1-cos45)
Proceeding the same way,
Twill raise to aheight of h' due to swing
h' = L(1-cos30)
The PE of T after swing is
Uf = mgh'
Uf = mgL(1-cos30)
Along with the PE , T has some kinetic energy results due to the moment.
Tf = 0.5*mv^2
According to the law of conservation of energy,
Ui = Uf+Tf
mgL(1-cos45) = mgL(1-cos30) + 0.5*mv^2
gL(co30-cos45) = 0.5*v^2
9.8*20*(co30-cos45) = 0.5*V^2
v = 7.89 m/s
<span>The speed f T after swing is 7.89 m/s</span>
