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skad [1K]
2 years ago
12

In Michael Johnson's world-record 400 m sprint, he ran the first 100 m in 11.20 s; then he reached the 200 m mark after a total

time of 21.32 s had elapsed, reached the 300 m mark after 31.76 s,and finished in 43.18 s.
During what 100 m segment was his speed the highest?

a) Between the start and the 100 m mark.

b) Between the 100 m mark and the 200 m mark.

c) Between the 200 m mark and the 300 m mark.

d)Between the 300 m mark and the finish.
Physics
1 answer:
GrogVix [38]2 years ago
6 0

Answer:

b)

Explanation:

Assuming that we are talking about average speed during any segment, we can apply the definition of average speed, as follows:

v(avg) = Δx / Δt = (xf-x₀) / (tfi-t₀)

Using this definition for the 4 segments, we have:

1) v(0-100m) = 100 m / 11.20 sec = 8.93 m/s

2) v(100m-200m) = 100 m / (21.32 s - 11.2 s) = 100 m / 10.12 s = 9.88 m/s

3) v(200m -300m) = 100 m / (31.76 s- 21.32s) = 100 m / 10.44 s = 9.58 m/s

4) v(300m-400m) = 100 m / (43.18 s - 31.76 s) = 100 m / 11.42 s = 8.76 m/s

As we can see, the highest speed was reached between the 100m mark and the 200m mark, so the statement b) is the one that results to be true.

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A long, straight wire carrying a current of 3.45 A moves with a constant speed v to the right. A 5-turn circular coil of diamete
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Answer:

I = 69.3  μA

Explanation:

Current through the straight wire, I = 3.45 A

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Diameter of the coil, D = 1.25 cm

Resistance of the coil, R = 3.25 \mu ohms

Distance of the wire from the center of the coil, d = 20 cm = 0.2 m

The magnetic field, B₁, when the wire is at a distance, d, from the center of the coil.

B_{1} = \frac{\mu_{0}I }{2\pi d}

B_{1} = \frac{4\pi* 10^{-7}  *3.45 }{2\pi *0.2}\\B_{1} =0.00000345 T

Magnetic field B₂ when the wire is at a distance, 2d from the center of the coil

B_{2} = \frac{\mu_{0}I }{2\pi(2d)) } \\B_{2} = \frac{\mu_{0}I }{4\pi d } \\

B_{2} = \frac{4\pi* 10^{-7}  *3.45 }{2\pi *2*0.2}\\B_{2} = 0.000001725 T

Change in the magnetic field, ΔB = B₂ - B₁ = 0.00001725 - 0.0000345

ΔB = -0.000001725

Induced current, I = \frac{E}{R}

E = -N (Δ∅)/Δt

Δ∅ = A ΔB

Area, A = πr²

diameter, d = 0.0125 m

Radius, r = 0.00625 m

A = π* 0.00625²

A = 0.0001227 m²

Δ∅ =  -0.000001725 * 0.0001227

Δ∅ = -211.6575 * 10⁻¹²

E = -N (Δ∅)/Δt

E = -5\frac{-211.6575 * 10^{-12} }{4.70} \\E = 225.17 * 10^{-12} V

Resistance, R = 3.25 μ ohms = 3.25 * 10⁻⁶ ohms

I = E/R

I = \frac{225.17 * 10^{-12} }{3.25 * 10^{-6} }

I = 0.0000693 A

I = 69 .3 * 10⁻⁶A

I = 69.3  μA

3 0
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Answer:

252.45 hours or 908820 seconds

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Time=\dfrac{Distance}{Speed}\\\Rightarrow Time=\dfrac{2\pi 60268}{1500}\\\Rightarrow Time=252.45\ h=252.45\times 60\times 60=908820\ s

It will take 252.45 hours or 908820 seconds for the equatorial flow to encircle the planet.

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