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vredina [299]
3 years ago
11

On a 150 m straight sprint, a cyclist accelerates from rest for 4.5 s at 3.8 m/s2. How long will it take her to complete the 150

m track, assuming she maintains her speed for the remaining part of the track?
Physics
1 answer:
balu736 [363]3 years ago
5 0

Answer:

It takes 6.52 s to complete the 150 m track.

Explanation:

Let's find the position of the cyclist during the acceleration.

x_{1}=v_{0}t+0.5at^{2}

x_{1}=0.5(3.8)(4.5)^{2}

x_{1}=38.48\: m

Now, the final speed at this distance will be:

v_{f}=v_{0}+at

v_{f}=3.8*4.5

v_{f}=17.1\: m/s

Now, we know that x(tot) = 150 m, the remaining part will be:

x_{2}=150-38.48=111.5\: m

Therefore, the time is:

t=\frac{x_{2}}{V_{f}}

t=\frac{111.5}{17.1}

t=6.52\: s

I hope it helps you!

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Answer:

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A)

<u>From the Coulomb's Law the force between the charges is given as:</u>

F=\frac{1}{4\pi.\epsilon_0} .\frac{q_1.q_2}{r^2}

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B)

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You want to produce three 2.00-mm-diametercylindrical wires, each with a resistance of 1.00 Ω at room temperature. One wire is g
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Answer:

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(c) L =  114.28 m

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Explanation:

The resistance of the wire is given as:

R = ρL/A

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R = Resistance

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A = cross-sectional area

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For Gold Wire:

ρ = 2.44 x 10⁻⁸ Ω.m

A = πd²/4 = π(2 x 10⁻³ m)²/4 = 3.14 x 10⁻⁶ m²

R = 1 Ω

Therefore,

1 Ω = (2.44 x 10⁻⁸ Ω.m)L/(3.14 x 10⁻⁶ m²)

L = (1 Ω)(3.14 x 10⁻⁶ m²)/(2.44 x 10⁻⁸ Ω.m)

<u>L =  128.75 m</u>

<u></u>

(b)

For Copper Wire:

ρ = 1.72 x 10⁻⁸ Ω.m

A = πd²/4 = π(2 x 10⁻³ m)²/4 = 3.14 x 10⁻⁶ m²

R = 1 Ω

Therefore,

1 Ω = (1.72 x 10⁻⁸ Ω.m)L/(3.14 x 10⁻⁶ m²)

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R = 1 Ω

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