Answer:
Chemical energy to electrical energy
Explanation:
In nature, there are several types of energy.
In this example (a flashlight being turned on), we have a conversion of energy from chemical energy to electrical energy. In fact:
- Chemical energy is the energy stored in the chemical bonds of the molecules of the substances used inside the battery. When the chemical reaction inside the battery occurs, this energy is liberated, and it is used to "push" the electrons along the circuit connected to the battery
- Electric energy is the energy associated to the motion of the electrons along the circuit of the flashlight; it is the energy associated to an electric current.
Moreover, in the flashlight the electric energy is then converted into two more types of energy: light energy (since the bulb in the flashlight produces light) and heat energy (because the flashlight also produces heat, so thermal energy).
Answer:
I=VRS=9V90Ω=0.1A
Explanation:
The equivalent resistance is the algebraic sum of the resistances (Equation 10.3. 2): RS=R1+R2+R3+R4+R5=20Ω+20Ω+20Ω+20Ω+10Ω=90Ω. The current through the circuit is the same for each resistor in a series circuit and is equal to the applied voltage divided by the equivalent resistance
From the average speed you can fix an equation:
Average speed = distance / time
You know the average speed = 65.1 kg / h, then
65.1 = distance / total time,
where total time is the time traveling plus 22.0 minutes
Call t the time treavelling and pass 22 minutes to hours:
65.1 = distance / [t + 22/60] ==> distance = [t + 22/60]*65.1
From the constant speed, you can fix a second equation
Constant speed = distance / time traveling
94.5 = distance / t ==> distance = 94.5 * t
The distance is the same in both equations, then you have:
[t +22/60] * 65.1 = 94.5 t
Now you can solve for t.
65.1t + 22*65.1/60 = 94.5t
94.5t - 65.1t = 22*65.1/60
29.4t = 23.87
t = 23.87 / 29.4
t = 0.812 hours
distance = 94.5 km/h * 0.812 h = 76.7 km
Answers: 1) 0.81 hours, 2) 76.7 km
(i) The total capacitance for the circuit is 5 μF.
(ii) The total charge stored in the circuit is 1 x 10⁻⁴ C.
(iii) The charge stored in 3μF capacitor is 6 x 10⁻⁶ C.
<h3>Total capacitance of the circuit</h3>
The total capacitance of the circuit is determined by reolving the series capacitors separate and parallel capacitors separate as well.
<h3>C1 and C2 are in series </h3>
<h3>C1 and C2 are parallel to C3</h3>
<h3>C(123) is series to C5 and C6</h3>
<h3>C7 and C8 are in series</h3>
<h3>Total capaciatnce of the circuit</h3>
Ct + C(78) = 2 μF + 3 μF = 5 μF
<h3 /><h3>Total charge stored in the circuit</h3>
The total charge stored in the capacitor is calculated as follows;
Q = CV
Q = (5 x 10⁻⁶) x (20)
Q = 1 x 10⁻⁴ C
<h3>Charge stored in 3μF capacitor</h3>
Q = (3 x 10⁻⁶) x (20)
Q = 6 x 10⁻⁶ C
Learn more about capacitance of capacitor here: brainly.com/question/13578522
