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viva [34]
3 years ago
5

A 100-g toy car moves along a curved frictionless track. At first, the car runs along a flat horizontal segment with an initial

velocity of 2.77 m/s. The car then runs up the frictionless slope, gaining 0.184 m in altitude before leveling out to another horizontal segment at the higher level. What is the final velocity of the car if we neglect air resistance?
Physics
1 answer:
aleksandrvk [35]3 years ago
7 0

Answer:

The final velocity of the car is 2.02 m/s

Explanation:

Hi there!

The kinetic energy of the car as it runs along the first flat horizontal segment can be calculated using the following equation:

KE = 1/2 · m · v²

Where:

KE =  kinetic energy

m = mass

v = velocity

Then, the initial kinetic energy will be:

KE = 1/2 · 0.100 kg · (2.77 m/s)²

KE = 0.384 J

When the car gains altitude, it gains potential energy. The amount of gained potential energy will be equal to the loss of kinetic energy. So let´s calculate the potential energy of the car as it reaches the top:

PE = m · g · h

Where:

PE = potential energy.

m = mass

g = acceleration due to gravity.

h = height.

PE = 0.100 kg · 9.8 m/s² · 0.184 m

PE = 0.180 J

Then, the final kinetic energy will be (0.384 J - 0.180 J) 0.204 J

Using the equation of kinetice energy, we can obtain the velocity of the car:

KE =  1/2 · m · v²

0.204 J = 1/2 · 0.100 kg · v²

2 · 0.204 J  / 0.100 kg = v²

v = 2.02 m/s

The final velocity of the car is 2.02 m/s

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0.46Ω

<h2>Explanation:</h2>

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<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}

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

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V = 12.0V

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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>

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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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