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AleksAgata [21]

3 years ago

12

A train started from rest and moved with constant acceleration. At one time it was traveling 27 m/s, and 150 m farther on it was

traveling 54 m/s. Calculate (a) the acceleration, (b) the time required to travel the 150 m mentioned, (c) the time required to attain the speed of 27 m/s, and (d) the distance moved from rest to the time the train had a speed of 27 m/s.

1 answer:

AlekseyPX3 years ago

7 0

Explanation:

(a) Given:

Δx = 150 m

v₀ = 27 m/s

v = 54 m/s

Find: a

v² = v₀² + 2aΔx

(54 m/s)² = (27 m/s)² + 2a (150 m)

a = 7.29 m/s²

(b) Given:

Δx = 150 m

v₀ = 0 m/s

a = 7.29 m/s²

Find: t

Δx = v₀ t + ½ at²

150 m = (0 m/s) t + ½ (7.29 m/s²) t²

t = 6.42 s

(c) Given:

v₀ = 0 m/s

v = 27 m/s

a = 7.29 m/s²

Find: t

v = at + v₀

27 m/s = (7.29 m/s²) t + 0 m/s

t = 3.70 s

(d) Given:

v₀ = 0 m/s

v = 27 m/s

a = 7.29 m/s²

Find: Δx

v² = v₀² + 2aΔx

(27 m/s)² = (0 m/s)² + 2 (7.29 m/s²) Δx

Δx = 50 m

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A series AC circuit contains a resistor, an inductor of 150 mH, a capacitor of 5.00 mF, and a generator with DVmax 5 240 V opera

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Given:

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Capacitance, C = 5.00 mF

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frequency, f = 50Hz

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e). Phase angle between current and the generator voltage is given by:

$tan\phi = \frac{X_{L} - X_{C}}{R}$

$\phi =tan^{-1}( \frac{X_{L} - X_{C}}{R})$

$\phi =tan^{-1}( \frac{47.12 - 0.636}{2399.5}) = tan^{-1}{0.0.01937}$

$\phi = 1.11^{\circ}$

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