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2 years ago

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car was moving in a straight road of length 320 km it covered 240 km with an average velocity 75 km/hr then it ran out of fuel a

nd stopped for 0.6 hr until it was refueled and completed its journey with a velocity 100 km/hr so the average velocity during the whole journey

1 answer:

Stella [2.4K]2 years ago

6 0

The average velocity of the car for the whole journey is 69.57 km/h.

The given parameters:

<em>Length of the road, L = 320 km</em>
<em>Distance covered = 240 km at 75 km/h</em>
<em>time spent refueling, t₂ = 0.6 hr</em>
<em>Final velocity, = 100 km/hr</em>

The time spent by the before refueling is calculated as follows;

$t = \frac{d}{v} \\\\t_1 = \frac{240}{75} \\\\t_1 = 3.2 \ hours$

The time spent by the car for the remaining journey;

$t_3 = \frac{320 - 240}{100} \\\\t_3 = 0.8 \ hr$

The total time of the journey is calculated as follows;

$t = t_1 + t_2 + t_3\\\\t = 3.2 \ hr \ + \ 0.6 \ hr \ + \ 0.8 \ hr\\\\t = 4.6 \ hours$

The average velocity of the car for the whole journey is calculated as follows;

$v = \frac{total \ distance }{total \ time} \\\\v = \frac{320}{4.6} \\\\v = 69.57 \ km/h$

Learn more about average velocity here: brainly.com/question/6504879

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kati45 [8]

The ball has height <em>y</em> and velocity <em>v</em> at time <em>t</em> according to

<em>y</em> = <em>v</em>₀ <em>t</em> - 1/2 <em>g</em> <em>t</em> ²

and

<em>v</em> = <em>v</em>₀ - <em>g t</em>

where <em>v</em>₀ is its initial speed and <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity.

The ball is falling with a velocity of 7.94 m/s when it's 2.72 m below the release point, which at time <em>t </em>such that

-2.72 m = <em>v</em>₀ <em>t</em> - 1/2 <em>g</em> <em>t</em> ²

-7.94 m/s = <em>v</em>₀ - <em>g t</em>

Solve for <em>t</em> in the second equation:

<em>t </em>= (<em>v</em>₀ + 7.94 m/s)/<em>g</em>

Substitute this into the first equation and solve for <em>v</em>₀ :

-2.72 m = <em>v</em>₀ (<em>v</em>₀ + 7.94 m/s) /<em>g</em> - 1/2 <em>g</em> ((<em>v</em>₀ + 7.94 m/s)/<em>g</em>)²

-2.72 m = <em>v</em>₀²/<em>g</em> + (7.94 m/s) <em>v</em>₀/<em>g</em> - 1/2 (<em>v</em>₀ + 7.94 m/s)²/<em>g</em>

2 (-2.72 m) <em>g</em> = 2<em>v</em>₀² + 2 (7.94 m/s) <em>v</em>₀ - (<em>v</em>₀ + 7.94 m/s)²

2 (-2.72 m) (9.80 m/s²) = 2<em>v</em>₀² + (15.9 m/s) <em>v</em>₀ - (<em>v</em>₀² + (15.9 m/s) <em>v</em>₀ + 63.0 m²/s²)

-53.3 m²/s² = <em>v</em>₀² - 63.0 m²/s²

<em>v</em>₀² = 9.73 m²/s²

<em>v</em>₀ = 3.12 m/s

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A wind turbine works by slowing the air that passes its blades and converting much of the extracted kinetic energy to electric e

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Answer:

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Explanation:

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Kinetic energy is given by

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The kinetic energy of the wind is 6624000 Joules

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Explanation:

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2 years ago

A sinusoidal voltage is given by the expression ????(????)=20cos(5π×103 ????+60°) V. Determine its (a) frequency in hertz, (b) p

MA_775_DIABLO [31]

<em>There are some placeholders in the expression, but they can be safely assumed</em>

Answer:

(a) $f=1617.9\ Hz$

(b) $T=0.618\ ms$

(c) $A=20 \ Volts$

(d) $\varphi=60^o$

Explanation:

<u>Sinusoidal Waves </u>

An oscillating wave can be expressed as a sinusoidal function as follows

$V(t)&=A\cdot \sin(2\pi ft+\varphi )$

Where

$A=Amplitude$

$f=frequency$

$\varphi=Phase\ angle$

The voltage of the question is the sinusoid expression

$V(t)=20cos(5\pi\times 103t+60^o)$

(a) By comparing with the general formula we have

$f=5\pi\times 103=1617.9\ Hz$

$\boxed{f=1617.9\ Hz}$

(b) The period is the reciprocal of the frequency:

$\displaystyle T=\frac{1}{f}$

$\displaystyle T=\frac{1}{1617.9\ Hz}=0.000618\ sec$

Converting to milliseconds

$\boxed{T=0.618\ ms}$

(c) The amplitude is

$\boxed{A=20 \ Volts}$

(d) Phase angle:

$\boxed{\varphi=60^o}$

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