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3 years ago

In addition to 1 m = 39.37 in, the following exact conversion equivalents are

Physics

1 answer:

Norma-Jean [14]3 years ago

5 0

Answer:

3.76 m/s.

Explanation:

Velocity in mile per hour (mph) = 8.4 miles per hour

Velocity in metre per second (m/s) =?

Next, we shall convert 8.4 mph to ft/h. This can be obtained as follow :

1 mph = 5280 ft/h

Therefore,

8.4 mph = 8.4 mph / 1 mph × 5280 ft/h

8.4 mph = 44352 ft/h

Next, we shall convert 44352 ft/h to in/h. This is illustrated below:

1 ft/h = 12 in/h

Therefore,

44352 ft/h = 44352 ft/h / 1 ft/h × 12 in/h

44352 ft/h = 532224 in/h

Next, we shall convert 532224 in/h to m/h. This can be obtained as follow:

39.37 in/h = 1 m/h

Therefore,

532224 in/h = 532224 in/h / 39.37 in/h × 1 m/h

532224 in/h = 13518.51664 m/h

Next, we shall convert 13518.51664 m/h to m/min. This is illustrated below:

1 m/h = 1/60 m/min

Therefore,

13518.51664 m/h = 13518.51664 m/h / 1 m/h × 1/60 m/min

13518.51664 m/h = 225.30861 m/min

Finally, we shall convert 225.30861 m/min to m/s. This is illustrated below:

1 m/min = 1/60 m/s

Therefore,

225.30861 m/min = 225.30861 m/min / 1 m/min × 1/60 m/s

225.30861 m/min = 3.76 m/s

Therefore,

8.4 miles per hour is equivalent to 3.76 m/s.

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A 1640 kg merry-go-round with a radius of 7.50 m accelerates from rest to a rate of 1.00 revolution per 8.00 s. Estimate the mer

son4ous [18]

Solution :

Given data :

Mass of the merry-go-round, m= 1640 kg

Radius of the merry-go-round, r = 7.50 m

Angular speed, $\omega = \frac{1}{8}$ rev/sec

$=\frac{2 \pi \times 7.5}{8}$ rad/sec

= 5.89 rad/sec

Therefore, force required,

$F=m.\omega^2.r$

$$$=1640 \times (5.89)^2 \times 7.5$

= 427126.9 N

Thus, the net work done for the acceleration is given by :

W = F x r

= 427126.9 x 7.5

= 3,203,451.75 J

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2 years ago

A force of 1 N will cause a mass of 1 kg to have an acceleration of 1 m/s2. Therefore, a force of 7 N applied to a mass of 7 kg

Lana71 [14]

1 m/s^2

Using F=ma,

7=7a

a=1

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3 years ago

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Explain how potential difference produces a current in a conductor.

Yuki888 [10]

Potential difference is the work done In moving a charge from one point to another in a conductor

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3 years ago

Cuanto cambia la entropía de 0.50 kg de vapor de mercurio [Lv: 2.7 x 10⁵ j/kg ] al calentarse en su punto de ebullición de 357°

lord [1]

Answer:

La entropía del vapor de mercurio cambia en 214.235 joules por Kelvin.

Explanation:

Por definición de entropía ( $S$ ), medida en joules por Kelvin, tenemos la siguiente expresión:

$dS = \frac{\delta Q}{T}$ (1)

Donde:

$Q$ - Ganancia de calor, en joules.

$T$ - Temperatura del sistema, en Kelvin.

Ampliamos (1) por la definición de calor latente:

$dS = \frac{L_{v}}{T}\cdot dm$ (1b)

Donde:

$m$ - Masa del sistema, en kilogramos.

$L_{v}$ - Calor latente de vaporización, en joules

Puesto que no existe cambio en la temperatura durante el proceso de vaporización, transformamos la expresión diferencial en expresión de diferencia, es decir:

$\Delta S = \frac{\Delta m \cdot L_{v}}{T}$

Como vemos, el cambio de la entropía asociada al cambio de fase del mercurio es directamente proporcional a la masa del sistema. Si tenemos que $m = 0.50\,kg$ , $L_{v} = 2.7\times 10^{5}\,\frac{J}{kg}$ and $T = 630.15\,K$ , entonces el cambio de entropía es: