Answer:
(a)
(b)
Explanation:
<u>Electric Circuits</u>
Suppose we have a resistive-only electric circuit. The relation between the current I and the voltage V in a resistance R is given by the Ohm's law:
(a) The electromagnetic force of the battery is and its internal resistance is . Knowing the equivalent resistance of the headlights is , we can compute the current of the circuit by using the Kirchhoffs Voltage Law or KVL:
Solving for i
i=2.28\ A
The potential difference across the headlight bulbs is
(b) If the starter motor is operated, taking an additional 35 Amp from the battery, then the total load current is 2.28 A + 35 A = 37.28 A. Thus the output voltage of the battery, that is the voltage that the bulbs have is
To solve this problem we will apply the concepts related to the Doppler effect. The Doppler effect is the change in the perceived frequency of any wave movement when the emitter, or focus of waves, and the receiver, or observer, move relative to each other. Mathematically it can be described as,
Here,
= Frequency of Source
= Speed of sound
f = Frequency heard before slowing down
f' = Frequency heard after slowing down
v = Speed of the train before slowing down
So if the speed of the train after slowing down will be v/2, we can do a system equation of 2x2 at the two moments, then,
The first equation is,
Now the second expression will be,
Dividing the two expression we have,
Solving for v, we have,
Therefore the speed of the train before and after slowing down is 22.12m/s
Answer:
1.) 274.5v
2.) 206.8v
Explanation:
1.) Given that In one part of the lab activities, students connected a 2.50 µF capacitor to a 746 V power source, whilst connected a second 6.80 µF capacitor to a 562 V source.
The potential difference and charge across EACH capacitor will be
V = Voe
Where Vo = initial voltage
e = natural logarithm = 2.718
For the first capacitor 2.50 µF,
V = Vo × 2.718
746 = Vo × 2.718
Vo = 746/2.718
Vo = 274.5v
To calculate the charge, use the below formula.
Q = CV
Q = 2.5 × 10^-6 × 274.5
Q = 6.86 × 10^-4 C
For the second capacitor 6.80 µF
V = Voe
562 = Vo × 2.718
Vo = 562/2.718
Vo = 206.77v
The charge on it will be
Q = CV
Q = 6.8 × 10^-6 × 206.77
Q = 1.41 × 10^-3 C
B.) Using the formula V = Voe again
165 = Vo × 2.718
Vo = 165 /2.718
Vo = 60.71v
Q = C × 60.71
Q = C
Equation: Mass x Velocity = Momentum
Answer: 93 x 13 = 1,209