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3 years ago

11

Examine the lightbulbs in the circuit below. Write a sentence explaining what would happen if lightbulb A burned out. Repeat thi

s for lightbulbs B, C, and D.

1 answer:

tatiyna3 years ago

8 0

If the lightbulb A in the circuit shown in the image burned out, the path for the current to flow is disrupted because one of its terminals is connected direct to the source. So, there will be no current through the lightbulbs B, C, and D, and they will turn off. Similarly it will happen, if the lightbulb D burned out.

If the lightbulb B burned out the current will continue circulating through the lightbulbs A, C, and D, because lightbulb B is connected in parallel. Similarly it will happen, if the lightbulb C burned out.

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Which property do liquids and gases share?

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C, They change their shapes depending on their containers

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A 100 Ω resistor is connected in series with a 47 µF capacitor and a source whose maximum voltage is 5 V, operating at 100.0 Hz.

Pani-rosa [81]

Answer:

$X_c=-33.86275385\Omega$

$|Z|=105.5778675\Omega$

$I=0.04735841062A$

$\phi=20.78612878\°$

Explanation:

The electrical reactance is defined as:

$X_c=-\frac{1}{2\pi fC}$

Where:

$f=Frequency\\C=Capacitance$

So, replacing the data provided by the problem:

$X_c=\frac{1}{2\pi *100*(47*10^{-6} )} =-33.86275385\Omega$

Now, the impedance can be calculated as:

$Z=R+jX_c$

Where:

$R=Resistance\\X_c= Capacitive\hspace{3}reactance$

Replacing the data:

$Z=100-j33.86275385$

In order to find the magnitude of the impedance we can use the next equation:

$|Z|=\sqrt{(R^2)+(X_c^2)}=\sqrt{(100)^2+(-33.86275385)^2} =105.5778675\Omega$

We can use Ohm's law to find the current:

$V=I*Z\\I=\frac{V}{Z}$

Therefore the current is:

$I=\frac{5}{100-j33.86275385}=0.04485638113+0.01518960593j$

And its magnitude is:

$|I|=\sqrt{(0.04485638113)^2+(0.01518960593)^2} =0.04735841062\Omega$

Finally the phase angle of the current is given by:

$\phi=arctan(\frac{0.01518960593}{0.04485638113})=20.78612878\°$

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3 years ago

What is the formula to calculate the readings on an ammeter?

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3 years ago

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A mine car (mass=390 kg) rolls at a speed of 0.50 m/s on a horizontal track, as the drawing shows. A 250-kg chunk of coal has a

Nady [450]

Answer:

v=0.60 m/s

Explanation:

Given that

m ₁= 390 kg ,u ₁= 0.5 m/s

m₂ = 250 kg ,u₂ = 0.76 m/s

As we know that if there is no any external force on the system the total linear momentum of the system will be conserve.

Pi = Pf

m ₁u ₁+m₂u₂ = (m₂ + m ₁ ) v

Now putting the values in the above equation

390 x 0.5 + 250 x 0.76 = (390 + 250 ) v

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v=0.60 m/s

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4 years ago

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Simora [160]

Answer:

a) 45571 N

b) 22786 N

c) 4557 N

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In other words, the change in momentum, must be equal to the initial one, but with opposite sign.

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Now, just applying the original form of Newton's 2nd Law, we know that this change in momentum must be equal to the impulse needed to stop the person:

$\Delta p = F* \Delta t (2)$

So, as we know the magnitude of Δp from (1) and we have different Δt as givens, we can get the different values of F (in magnitude) required to stop the person for each one of them, as follows:

$F_{1} = \frac{\Delta p}{\Delta t_{1}} = \frac{1595kgm/s}{0.035s} = 45571 N (3)$

$F_{2} = \frac{\Delta p}{\Delta t_{2}} = \frac{1595kgm/s}{0.07s} = 22786 N (4)$

$F_{3} = \frac{\Delta p}{\Delta t_{3}} = \frac{1595kgm/s}{0.35s} = 4557 N (5)$

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