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Alexxandr [17]
3 years ago
13

The Importance of Assuming Responsibilities and Self-Esteem in Acquiring the Accuracy and Precision of Measurements.

Physics
1 answer:
saveliy_v [14]3 years ago
3 0
Hey good job yeaaa YEAAA GREAT JOB WOOOHOOO
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A boom box, stereo set.
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3 years ago
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Consider the classic problem with holiday lights, one little bulb goes out and the whole string goes out. First consider a strin
Flura [38]

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

P=V.I

Solving for I

\displaystyle I=\frac{P}{V}=\frac{40}{120}=0.33\ Amp

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

\displaystyle I=\frac{V}{R}=\frac{120}{49}=2.45\ A

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.

6 0
3 years ago
Um corpo de massa igual a 2kg move-se com velocidade constante num plano horizontal sem atrito, conforme a figura. Em seguida, e
castortr0y [4]
Only answering to ask a question
4 0
3 years ago
Any push or pull on an object is Question 5 options: Acceleration Velocity Force Inertia
Kobotan [32]
This would be force. Acceleration is increasing the speed in an object and velocity is how fast an object is going. Also, inertia basically says that an object will stay at rest or in motion unless an outside force acts on it. So, for example, a ball will stay in the air unless gravity acts on it and pulls it down. By definition, force is any action, unopposed (or by itself without any other forces that would do the opposite) will change the motion of an object, so this definitely makes the most sense for the question. Hope this helps!
8 0
3 years ago
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When the cylinder is displaced slightly along its vertical axis it will oscillate about its equilibrium position with a frequenc
Nesterboy [21]

Answer:

w = √[g /L (½ r²/L2 + 2/3 ) ]

When the mass of the cylinder changes if its external dimensions do not change the angular velocity DOES NOT CHANGE

Explanation:

We can simulate this system as a physical pendulum, which is a pendulum with a distributed mass, in this case the angular velocity is

          w² = mg d / I

In this case, the distance d to the pivot point of half the length (L) of the cylinder, which we consider long and narrow

         d = L / 2

The moment of inertia of a cylinder with respect to an axis at the end we can use the parallel axes theorem, it is approximately equal to that of a long bar plus the moment of inertia of the center of mass of the cylinder, this is tabulated

        I = ¼ m r2 + ⅓ m L2

        I = m (¼ r2 + ⅓ L2)

now let's use the concept of density to calculate the mass of the system

        ρ = m / V

        m = ρ V

the volume of a cylinder is

         V = π r² L

          m =  ρ π r² L

let's substitute

        w² = m g (L / 2) / m (¼ r² + ⅓ L²)

        w² = g L / (½ r² + 2/3 L²)

        L >> r

         w = √[g /L (½ r²/L2 + 2/3 ) ]

When the mass of the cylinder changes if its external dimensions do not change the angular velocity DOES NOT CHANGE

4 0
3 years ago
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