Answer:
Wave A
<em>I</em><em> </em><em>hope this</em><em> </em><em>helps</em><em> </em>
Answer:
-8.04 m/s2
Explanation:
To find the answer to this, you have to use the 4th kinematic equation:
You plug into the equation to get:
solve for a to get
-8.04 m/s2
Answer:
P=3.42×10^-6 J/s
Explanation:
From the kinematics of motion with constant acceleration we know that :
vf^2=vi^2+2*a(xf-xi)
Where :
• vf , vi, are the the final and the initial velocity of the electron
• a is the acceleration of the electron
• xf , xi are the final and the initial position of the electron .
Strategy for solving the problem : at first from the given information we calculate the acceleration of the electron.
Givens: vf = 8.4 x 10^7 m/s , vi, = 0 m/s , xf = 0.025 m and xi = 0 m
vf^2 =vi^2+2*a(xf-xi)
vf^2-vi^2=2*a(xf-xi)
2*a(xf-xi)= vf^2-vi^2
a = (vf^2-vi^2)/2(xf-xi)
Pluging known information to get :
a = (vf^2-vi^2)/2(xf-xi)
= 1.411 × 10^17
From the acceleration and the previous Eq. we can calculate the final velocity of the electron but a new position xf = 0.01 m
so,
vf^2 =vi^2+2*a(xf-xi)
vf^2 =5.312× 10^7
From the following Eq. we can calculate the time elapsed in this motion .
xf =xi+vi*t+1/2*a*t
xf =xi+vi*t+1/2*a*t
t=√2(xf-xi)/a
t=3.765×10^-10 s
now we can use the power P Eq.
P=W/Δt => ΔK/Δt
Where: the work done W change the kinetic energy K of the electron ,
ΔK=Kf-Ki=>1/2*m*vf^2-1/2*m*vi^2
P=1/2*m*vf^2-1/2*m*vi^2/Δt
P=3.42×10^-6 J/s
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
