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3 years ago

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A skydiver accelerates from REST at 9.8 m/s² for the first 5m of their fall. Calculate their final speed. (Give your answer to 2

significant figures)

1 answer:

Kaylis [27]3 years ago

5 0

Answer:

The final speed is 9.89 m/s.

Explanation:

Initial speed of skydiver is 0 as it was rest.

Acceleration of the skydiver is 9.8 m/s²

It accelerates for the first 5 m of their fall.

It is required to find their final speed. Let v is the final speed. It can be calculated using third equation of motion.

$v^2-u^2=2as$

Here, u = 0

a = 9.8 m/s²

s = 5 m

$v=\sqrt{2gs} \\\\v=\sqrt{2\times 9.8\times 5} \\\\v=9.89\ m/s$

So, the final speed is 9.89 m/s.

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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

yanalaym [24]

Given:

Inductance, L = 150 mH

Capacitance, C = 5.00 mF

$\Delta V_{max}$ = 240 V

frequency, f = 50Hz

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$\X_{L} = \omega L = 100\pi \times 150\times 10^{-3} = 47.12\Omega$

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b). The capacitive reactance, $X_{C}$ is given by:

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c). Impedance, Z = $\frac{\Delta V_{max}}{I_{max}} = \frac{240}{100\times 10^{-3}} = 2400\Omega$

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d). Resistance, R is given by:

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

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

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

$\phi = 1.11^{\circ}$

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