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lina2011 [118]
3 years ago
15

Tungsten is used to make the filament of electric bulbs. Give reason to justify the statement.

Physics
1 answer:
Anuta_ua [19.1K]3 years ago
3 0

Answer:

Tungsten is used to make the filament of electric bulbs. This is because tungsten has a high melting point. This prevents the filament from melting as current flows through, along with an increasing temperature.

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The winning time for a 500.0 mile circular race track was 3 hours and
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Explanation:

We have,

Distance traveled in a circular track is 500 miles

The winning time was 3 hours and  13 minutes. It means time is 3.217 hours.

The driver's average speed is given by total distance divided by total time taken. Its formula can be written as :

v=\dfrac{d}{t}\\\\v=\dfrac{500\ miles}{3.217\ h}\\\\v=155.42\ mph

At the end of the race, the driver reaches the point form where he has started. It means the displacement of the driver is equal to 0. Hence, driver's average velocity is equal to 0.

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Which two staments explain how a cell's parts help it get nutrients
Goryan [66]

Answer:

I think it's A and D.

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The 5 major unit processes include chemical coagulation, flocculation, sedimentation, filtration, and disinfection (described below). There are chemicals added to the water as it enters the various treatment processes.
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2 years ago
A battery with an emf of 12.0 V shows a terminal voltage of 11.7 V when operating in a circuit with two lightbulbs, each rated a
wariber [46]
<h2>Answer:</h2>

0.46Ω

<h2>Explanation:</h2>

The electromotive force (E) in the circuit is related to the terminal voltage(V), of the circuit and the internal resistance (r) of the battery as follows;

E = V + Ir                      --------------------(a)

Where;

I = current flowing through the circuit

But;

V = I x Rₓ                    ---------------------(b)

Where;

Rₓ = effective or total resistance in the circuit.

<em>First, let's calculate the effective resistance in the circuit:</em>

The effective resistance (Rₓ) in the circuit is the one due to the resistances in the two lightbulbs.

Let;

R₁ = resistance in the first bulb

R₂ = resistance in the second bulb

Since the two bulbs are both rated at 4.0W ( at 12.0V), their resistance values (R₁ and R₂) are the same and will be given by the power formula;

P = \frac{V^{2} }{R}

=> R = \frac{V^{2} }{P}             -------------------(ii)

Where;

P = Power of the bulb

V = voltage across the bulb

R = resistance of the bulb

To get R₁, equation (ii) can be written as;

R₁ = \frac{V^{2} }{P}    --------------------------------(iii)

Where;

V = 12.0V

P = 4.0W

Substitute these values into equation (iii) as follows;

R₁ = \frac{12.0^{2} }{4}

R₁ = \frac{144}{4}

R₁ = 36Ω

Following the same approach, to get R₂, equation (ii) can be written as;

R₂ = \frac{V^{2} }{P}    --------------------------------(iv)

Where;

V = 12.0V

P = 4.0W

Substitute these values into equation (iv) as follows;

R₂ = \frac{12.0^{2} }{4}

R₂ = \frac{144}{4}

R₂ = 36Ω

Now, since the bulbs are connected in parallel, the effective resistance (Rₓ) is given by;

\frac{1}{R_{X} } = \frac{1}{R_1} + \frac{1}{R_2}       -----------------(v)

Substitute the values of R₁ and R₂ into equation (v) as follows;

\frac{1}{R_X} = \frac{1}{36} + \frac{1}{36}

\frac{1}{R_X} = \frac{2}{36}

Rₓ = \frac{36}{2}

Rₓ = 18Ω

The effective resistance (Rₓ) is therefore, 18Ω

<em>Now calculate the current I, flowing in the circuit:</em>

Substitute the values of V = 11.7V and Rₓ = 18Ω into equation (b) as follows;

11.7 = I x 18

I = \frac{11.7}{18}

I = 0.65A

<em>Now calculate the battery's internal resistance:</em>

Substitute the values of E = 12.0, V = 11.7V and I = 0.65A  into equation (a) as follows;

12.0 = 11.7 + 0.65r

0.65r = 12.0 - 11.7

0.65r = 0.3

r = \frac{0.3}{0.65}

r = 0.46Ω

Therefore, the internal resistance of the battery is 0.46Ω

5 0
4 years ago
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son4ous [18]

Answer:

Explanation:

I think the answer is statement no 3.

Hope it helps.

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3 years ago
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