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harina [27]
2 years ago
7

A shotputter throws the shot with an initial speed of 15.6 m/s at a 30.0 degree angle to the horizontal. Calculate the horizonta

l distance traveled by the shot if it leaves the athlete's hand at a height of 2.20m above the ground.
Physics
1 answer:
vfiekz [6]2 years ago
8 0
To calculate the horizontal distance traveled by the shot if it leaves the athlete's hand at a height of 2.20 above the ground we can get the root of the quad equation for time are t=-0.24 or t =1.84 taking the t = 1.84, so the equation will be:
x = 15.6cos(30) * 1.86, x = 24.79m
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What is the peak emf generated by rotating a 940-turn, 24 cm diameter coil in the Earth’s 5·10−5 T magnetic field, given th
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Answer:

The peak emf generated by the coil is 2.67 V

Explanation:

Given;

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diameter, d = 24 cm = 0.24 m

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time, t = 5 ms = 5 x 10⁻³ s

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V₀ = NABω

Where;

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V₀ = NABω and ω = 2πf = 2π/t

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A = πd²/4

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V₀ = 940 x (π x 0.24²/4) x 5 x 10⁻⁵ x (2π/0.005)

V₀ = 940 x 0.04524 x  5 x 10⁻⁵ x 1256.8

V₀ = 2.6723 V = 2.67 V

The peak emf generated by the coil is 2.67 V

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