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sertanlavr [38]

3 years ago

11

The standard molar heat of vaporization for water is 40.79 kj/mol. How much energy would be required to vaporize 4.72 x 10^10 mo

lecules of water

1 answer:

stellarik [79]3 years ago

7 0

(5.81 x 10^10 molecules) / (6.022 x 10^23 molecules/mol) x (40.79 kj/mol) = 3.94 x 10^-12 kj

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mamaluj [8]

Answer:

B electrons protons and neutrons

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2 years ago

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Elena-2011 [213]

Answer:

v = 12.86 km/h

v = 3.6 m/s

Explanation:

Given,

The distance, d = 13.5 km

The time, t = 21/20 h

= 1.05 h

The velocity of a body is defined as the distance traveled by the time taken.

v = d / t

= 13.5 km / 1.05 h

= 12.86 km/h

The conversion of km/h to m/s

1 km/h = 0.28 m/s

12.86 km/h = 12.86 x 0.28 m/s

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3 years ago

Is 5m.what<br>In si system, the length of a body is<br>does it mean?

lozanna [386]

Answer: Some conversions from one system of units to another need to be exact, without increasing or decreasing the precision of the first measurement. This is sometimes called soft conversion. It does not involve changing the physical configuration of the item being measured.

Explanation:

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2 years ago

calculate the work done in kilo joules in lifting a mass of 20kg at steady velocity through a vertical height of 20m

bulgar [2K]

Answer:

$\huge\boxed{\sf Work\ done = 4 kJ}$

Explanation:

Since work done is in the form of potential energy, we will use the formula of potential energy here.

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<h3>P.E. = mgh </h3>

Where,

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Work done = 4000 joules

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$\rule[225]{225}{2}$

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1 year ago

Consult interactive solution 2.22 before beginning this problem. a car is traveling along a straight road at a velocity of +30.0

Inessa05 [86]

Let $a_1$ be the average acceleration over the first 2.46 seconds, and $a_2$ the average acceleration over the next 6.79 seconds.

At the start, the car has velocity 30.0 m/s, and at the end of the total 9.25 second interval it has velocity 15.2 m/s. Let $v$ be the velocity of the car after the first 2.46 seconds.

By definition of average acceleration, we have

$a_1=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}$

$a_2=\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}$

and we're also told that

$\dfrac{a_1}{a_2}=1.66$

(or possibly the other way around; I'll consider that case later). We can solve for $a_1$ in the ratio equation and substitute it into the first average acceleration equation, and in turn we end up with an equation independent of the accelerations:

$1.66a_2=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}$

$\implies1.66\left(\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}\right)=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}$

Now we can solve for $v$ . We find that

$v=20.8\,\dfrac{\mathrm m}{\mathrm s}$

In the case that the ratio of accelerations is actually

$\dfrac{a_2}{a_1}=1.66$

we would instead have

$\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}=1.66\left(\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}\right)$

in which case we would get a velocity of

$v=24.4\,\dfrac{\mathrm m}{\mathrm s}$

6 0

2 years ago