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3 years ago

When a perfume bottle is opened, some liquid changes to gas and the fragrances spreads around the room.

1 answer:

8 0

The liquid changes to gas by the process of 'vaporization' or 'evaporation.
The gas spreads around the room by the process of 'diffusion'.

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What does this meannn plsssssss

VARVARA [1.3K]

Answer:

the answer is B

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Explanation:

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3 years ago

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An electric clothes dryer draws 20 A of current from a 240-V wall outlet. How much power, in watts, does it use?

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The answer is 4800 watts.

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4 years ago

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The armature windings of a dc motor have a resistance of 3.0 Î©. The motor is connected to a 120V line, and when the motor reach

svlad2 [7]

Answer:

0.6 A

Explanation:

As the motor gears towards full speed, the voltage of the circuit tends to become the difference that exists between the line voltage and the that of the back emf. Remembering Ohm's law, we then apply it to get the final, lower current that is based on the reduced voltage Ef. We use the provided resistance in the question, that is 3 ohms.

Ef = (120 V) - (117 V) = 3 V

If = Ef/R = (3 V) / (5.0 Ω) = 0.6 A

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3 years ago

NaOH is__<br> an acid <br> a base <br> a salt

masya89 [10]

Answer:

Sodium Hydroxide

Explanation:

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4 years ago

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In class we calculated the range of a projectile launched on flat ground. Consider instead, a projectile is launched down-slope

zysi [14]

Answer:

With an initial speed of 10m/s at an angle 30° below the horizontal, and a height of 8m, the projectile travels 7.49m horizontally before it lands.

Explanation:

Since the horizontal motion is independent from the vertical motion, we can consider them separated. The horizonal motion has a constant speed, because there is no external forces in the horizontal axis. On the other hand, the vertical motion actually is affected by the gravitational force, so the projectile will be accelerated down with a magnitude $g$ .

If we have the initial velocity $v_o$ and its angle $\theta$ , we can obtain the vertical component of the velocity $v_{oy}$ using trigonometry:

$v_{oy}=v_osin\theta$

Therefore, if we know the height at which the projectile was launched, we can obtain the final velocity using the formula:

$v_{fy}^{2} =v_{oy}^{2}+2gy\\\\ v_{fy}=\sqrt{v_{oy}^{2}+2gy }$

Next, we compute the time the projectile lasts to reach the ground using the definition of acceleration:

$g=\frac{v_{fy}-v_{oy}}{\Delta t} \\\\\Delta t= \frac{v_{fy}-v_{oy}}{g}=\frac{\sqrt{v_{oy}^{2}+2gy} -v_{oy}}{g}$

Finally, from the equation of horizontal motion with constant speed, we have that:

$x=v_{ox}\Delta t= v_{ox}\frac{\sqrt{v_{oy}^{2}+2gy} -v_{oy}}{g}$

For example, if the projectile is launched at an angle 30° below the horizontal with an initial speed of 10m/s and a height 8m, we compute:

$v_{ox}=10\frac{m}{s} cos30=8.66\frac{m}{s}\\v_{oy}=10\frac{m}{s} sin30=5\frac{m}{s}\\\\x=8.66\frac{m}{s} \frac{\sqrt{(5\frac{m}{s}) ^{2}+2(9.8\frac{m}{s^{2}})8m}-5\frac{m}{s} }{9.8\frac{m}{s^{2} } } =7.49m$

In words, the projectile travels 7.49m horizontally before it lands.

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4 years ago