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3 years ago

A train travels 72 kilometers in 1 hours, and then 64 kilometers in 1 hours. What is its average speed?

1 answer:

5 0

To find the average of something you must first add everything together and then divide it by the total amount of numbers you added together.

First add 72 and 64 together

72+64=136

Now divide 136 by 2

136/2=68

So the average speed of the train is 68 kilometers

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A plane is heading south with a velocity of 150 kilometers/hour. It experiences a tailwind with a velocity of 20 kilometers/hour

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The tailwind speed would actually add up to the speed of the plane.

It would blow from behind the tail of the plane.

Resultant Velocity = 20 km/h + 150 km/h = 170 km/h

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3 years ago

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3 years ago

Earthquake information would most likely be shown on which type of map?

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Information about Earthquakes would normally be shown on a map called "geologic hazards"

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3 years ago

A rope is shaken and produces 2 waves each second. Calculate the time period of the rope waves.

motikmotik

Answer:

0.5 s

Explanation:

From the question given above, the following data were obtained:

Number of circle (n) = 2

Time (t) = 1 s

Period =?

Period of a wave is simply defined as the time taken to make one complete oscillation. Mathematically, it can be expressed as:

T = t / n

Whereb

T => is the period

t => is the space time

n => is the number of circle or oscillation.

With the above formula, we can obtain the period of the wave as follow:

Number of circle (n) = 2

Time (t) = 1 s

Period =?

T = t / n

T = 1 / 2

T = 0.5 s

Thus, the period of the wave is 0.5 s

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3 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

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3 years ago

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