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3 years ago

9

What is the most appropriate SI unit to express the speed of a cyclist in a 10-km race? km/s cm/h km/h mm/s

1 answer:

Murljashka [212]3 years ago

5 0

You sure wouldn't want something like cm/s or (yikes cm/hr). You want a reasonable number for sports usually between 0 and 100

Km / hour would be a good choice.

The next town to where I live is 25 km away. On a good day, I can make it there in about 3/4 of an hour.

Speed = 25 km / 0.75 hour = 33.3 km/hour. That's actually a little fast most of the time. But you should understand what I mean.

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Answer:

$4.81*10^{29}J$

Explanation:

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2 years ago

A block with mass m =6.4 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.28 m.

Zanzabum

Answer

given,

mass of block (m)= 6.4 Kg

spring is stretched to distance, x = 0.28 m

initial velocity = 5.1 m/s

a) computing weight of spring

k x = m g

$k = \dfrac{mg}{x}$

$k = \dfrac{6.4 \times 9.8}{0.28}$

k = 224 N/m

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$\omega = \sqrt{\dfrac{k}{m}}= \sqrt{\dfrac{224}{6.4}} = 5.92 \ rad/s$

$f = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

$f = \dfrac{1}{2\pi}\sqrt{\dfrac{224}{6.4}}$

$f =0.94\ Hz$

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$v_b = -5.1 \times cos (5.92 \times 0.42)$

$v_b = 4.04\ m/s$

d) $a_{max} = v \omega$

$a_{max} = 4.04 \times 5.92$

$a_{max} =23.94\ m/s^2$

e) $Y =- A sin (\omega t)$

$A = \dfrac{v}{\omega}$

$A = \dfrac{4.04}{5.92}$

A = 0.682 m

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$Y =- 0.42$

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= $6.4 \times 5.92^2\times 0.42$

F = 94.20 N

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3 years ago

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Juli2301 [7.4K]

Answer:

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Explanation:

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2 years ago

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fenix001 [56]

Answer:

See below

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