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
. . . 'protect' its domestic steel industry, by
increasing the price of imported steel.
Answer:
μ = 0.725
Explanation:
This problem refers to Newton's second law.
F = ma
Let's write the equations on each axis
Y Axis
N-W = 0
N = W
N = mg
X axis
F-fr = ma
With the body not started moving its acceleration is zero
F-fr = 0
F = fr
The friction force equation is
fr = μ N
fr = μ m g
Let's replace and calculate
F = μ m g
μ = F / mg
μ = 321 /45.2 9.8
μ = 0.725
Answer:
a. 3.039cm
b.magnetic field is 
Explanation:
Direction of the solenoid magnetic field is along the axis of the solenoid. and magnetic field due to the wire perpendicular to that due to the solenoid.. Magnetic field at r is given by:
Angle of net magnetic field from axial direction is given by:
,
Field due to solenoid:
Field due to wire:
Therefore, r:
Hence, the radial distance is 3.039cm
b.The magnetic field strength is given by:
Hence, the magnetic field is
Answer:
Transition metal.
Explanation:
Reason;in periodic table gold is third element in eleventh column. So its classified as TRANSITION METAL. Gold contain 79 protons, 79 electrons and 118 neutrons and gold is the most abundant isotope.
Hope i helped you..
