Each tyre of a car has an area of 100 cm2 in contact with the ground. The car has a mass of 1600 kg. The weight of the car is eq
2 answers:
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Answer:
<em>a. 0.33 Amp</em>
<em>b. 2.4 Volt</em>
<em>c. 0</em>
<em>d. 2.45 Amp</em>
<em>e. Infinite</em>
<em>f. Series is safer</em>
Explanation:
<u>Series Connection of Resistors</u>
When two or more resistors are connected in series, the current through each one of them is the same, and the voltage divides depending on the particular value of each resistance. If all the resistances are equal, then the voltage is equally divided.
a. The string of 50 bulbs is connected to a 120 VAC outlet and consumes 40 W. The power of a circuit is given by
Solving for I
Since all the bulbs are connected in series the current is the same for all of them.
b. The voltage is equally divided, so each bulb has 120/50= 2.4 V
c. If one of the bulbs burns out and its resistance becomes infinite, then the series circuit is open and no current flows through it, neither through the rest of the bulbs. The typical case of the whole string going out.
d. If one of the bulbs short circuits, the resistance of that bulb is zero and the voltage is distributed by the 49 remaining bulbs. Thus the new current is
e. If the bulbs were connected in parallel, all of them would have the same voltage, and the total current will be equally divided among them. In that case, a short circuit in one of the bulbs will cause a parallel short, theoretically producing an infinite current and making the short circuit protection blow up.
f. The condition described above makes the strings be made of series-connected bulbs which is safer than the parallel circuit. If a single bulb shorts, the entire string goes out in a series connection, but the breaker would trigger disconnection of the house circuit if it's a parallel connection. That is why we must deal with unusable strings instead of burning cables.
This is a Fraunhofer single slit experiment, where the light passing through the slit produces an interference pattern on the screen, and where the dark bands (minima of diffraction) are located at a distance of
from the center of the pattern. In the formula, m is the order of the minimum,
the wavelenght, 
the distance of the screen from the slit and 
the width of the slit.
In our problem, the distance of the first-order band (m=1) is
. The distance of the screen is D=86 cm while the wavelength is 
. Using these data and re-arranging the formula, we can find a, the width of the slit:
from the center of the pattern. In the formula, m is the order of the minimum,
In our problem, the distance of the first-order band (m=1) is