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3 years ago

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A car covers 72 kilometers in the first hour of its journey. In the next hour, it covers 90 kilometers. What is the amount of wo

rk done by the car? The total mass of the car, including its passengers, is 2.5 × 103 kilograms.

1 answer:

Nina [5.8K]3 years ago

4 0

The work done is $2.8125 \times 10^{5} \mathrm{J}$

Work Done = Change in Kinetic Energy (ΔKE)

<u>Explanation</u>

In first 1 hour it travels 72 km

So, Velocity = $\frac{\text { distance }}{\text { time }}=\frac{72}{1} k m / h=72 \mathrm{km} / \mathrm{h}=\frac{72000}{3600} \mathrm{m} / \mathrm{s}=20 \mathrm{m} / \mathrm{s}$

or, Initial Velocity (u) = 20 m/s

Similarly for the next hour it covers 90 km

So, Velocity = $\frac{\text { distance }}{\text { time }}=\frac{90}{1} k m / h=90 \mathrm{km} / \mathrm{h}=\frac{90000}{3600} \mathrm{m} / \mathrm{s}=25 \mathrm{m} / \mathrm{s}$

or, Final Velocity (v) = 20 m/s

Work done = Change in Kinetic Energy (ΔKE)

Work done = ΔKE = $\frac{1}{2} m v^{2}-\frac{1}{2} m u^{2}$

ΔKE = $\frac{1}{2} m\left(v^{2}-u^{2}\right)=\frac{1}{2} \times\left(2.5 \times 10^{3}\right) \times\left(25^{2}-20^{2}\right)$

= $\frac{2500 \times(625-400)}{2}=\frac{2500 \times 225}{2}=\frac{562500}{2}$ = 281250 joule

= $2.8125 \times 10^{5} \mathrm{J}$

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To answer this question, let's analyze the problem. Let's use conservation of energy

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we substitute

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let's use trigonometry

sin θ = h / L

we substitute

a = g sin θ \ \frac{h}{L} \ \frac{1}{1+ \frac{I}{m r^2 } }

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a₂ = g sin θ $\frac{1}{1 + \frac{1}{2} \frac{mr^2}{mr^2} }$ = g sin θ $\frac{1}{1+ \frac{1}{2} }$

a₂ = g sin θ ⅔

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a₃ = g sin θ $\frac{1}{1 + \frac{2}{3} }$

a₃ = g sin θ $\frac{3}{5}$

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a₄ = g sin θ $\frac{1 }{1 + \frac{2}{5} }$

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Answer:

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Explanation:

The maximum speed occurs when all the potential energy of the protons has been converted to kinetic energy.

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Potential Energy at maximum speed = 0 J

0 + (3.204 × 10⁻¹⁹) = mv² + 0

mv² = (3.204 × 10⁻¹⁹)

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v = √[(3.204 × 10⁻¹⁹) ÷ m]

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Force of repulsion acting on each proton is given through Coulomb's law as

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And the force acting on each proton is obtainable using Newton's law that

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