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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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3 years ago

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An electric turntable 0.730 mm in diameter is rotating about a fixed axis with an initial angular velocity of 0.240 rev/srev/s a

Zolol [24]

Answer:

a) $\omega = 0.421\,\frac{rev}{s}$ , b) $\Delta \theta = 0.066\,rev$ , c) $v = 0.966\,\frac{mm}{s}$ , d) $a = 3.293\,\frac{mm}{s^{2}}$

Explanation:

a) The angular velocity of the turntable after $0.200\,s$ .

$\omega = \omega_{o} + \alpha\cdot \Delta t$

$\omega = 0.240\,\frac{rev}{s} + (0.906\,\frac{rev}{s^{2}} )\cdot (0.2\,s)$

$\omega = 0.421\,\frac{rev}{s}$

b) The change in angular position is:

$\Delta \theta = \omega_{o}\cdot t + \frac{1}{2} \cdot \alpha \cdot t^{2}$

$\Delta \theta = (0.240\,\frac{rev}{s} )\cdot (0.2\,s) + \frac{1}{2}\cdot (0.906\,\frac{rev}{s^{2}} )\cdot (0.2\,s)^{2}$

$\Delta \theta = 0.066\,rev$

c) The tangential speed of a point on the rim of the turn-table:

$v = r\cdot \omega$

$v = (0.365\times 10^{-3}\,m)\cdot (0.421\,\frac{rev}{s} )\cdot (\frac{2\pi\,rad}{1\,rev} )$

$v = 9.655\times 10^{-4}\,\frac{m}{s}$

$v = 0.966\,\frac{mm}{s}$

d) The tangential and normal components of the acceleration of the turn-table:

$a_{t} = (0.365\times 10^{-3}\,m)\cdot (0.906\,\frac{rev}{s^{2}})\cdot (\frac{2\pi\,rad}{1\,rev} )$

$a_{t} = 2.078\times 10^{-3}\,\frac{m}{s^{2}}$

$a_{t} = 2.078\,\frac{mm}{s}$

$a_{n} = (0.365\times 10^{-3}\,m)\cdot \left[(0.421\,\frac{rev}{s} )\cdot (\frac{2\pi\,rad}{1\,rev} )\right]^{2}$

$a_{n} = 2.554\times 10^{-3}\,\frac{m}{s^{2}}$

$a_{n} = 2.554\,\frac{mm}{s^{2}}$

The magnitude of the resultant acceleration is:

$a = \sqrt{(2.078\,\frac{mm}{s} )^{2}+(2.554\,\frac{mm}{s} )^{2}}$

$a = 3.293\,\frac{mm}{s^{2}}$

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3 years ago

At one atmosphere of pressure and 25°C, a hot air balloon has a volume of 4,000 liters. While still tied to the ground, the air

yawa3891 [41]

Based on the options given, the most likely answer to this query is C) 4577 liters.

Upon computation of the given variables the result seems to be 4577 L

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3 years ago

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The first law of thermodynamics states that ΔE= Q− W . Is this also a statement of the principle of conservation of energy? Yes,

ozzi

Answer:

Yes, the heat that flows into the system is used to change the internal energy of the gas and becomes work done by the piston.

Explanation:

First law of thermodynamics known as Law of Conservation of Energy, states that energy can neither be created nor destroyed; energy can only be transferred or changed from one form to another.

The first law of thermodynamics states that the change in internal energy of a system equals the net heat transfer into the system minus the net work done by the system. In equation form, the first law of thermodynamics.

This is the first law of thermodynamics

ΔE= Q− W

ΔE= change internal energy of the system.

Q= heat transfer into the system

And

W= work done by the system.

Rewriting the equation

ΔE= Q− W

Q=ΔE +W

Show that the heat flowing l into the system is transferred to the internal energy of the system and the work done by the piston

So the third option is correct

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