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Ilia_Sergeevich [38]
3 years ago
14

A marathon is a race that commemorates the run made by a Greek soldier, Pheidippides, that took place in August 490 BC. The sold

ier ran 26.2 mi. How many kilometers did he run?
Chemistry
2 answers:
Vlad1618 [11]3 years ago
7 0
The correct answer is 42. All marathon races in Europe are 42km long, which is 26.2 miles for those who use miles.
timofeeve [1]3 years ago
5 0

the correct answer to this problem is <u>41.9 </u> to be exact

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Suppose you have a spherical balloon filled with air at room temperature and 1.0 atm pressure; its radius is 17 cm. You take the
Sladkaya [172]
<span>Answer: 17.8 cm
</span>

<span>Explanation:
</span>

<span>1) Since temperature is constant, you use Boyle's law:
</span>

<span>PV = constant => P₁V₁ = P₂V₂


</span><span>=> V₁/V₂ = P₂/P₁</span>
<span>
2) Since the ballon is spherical:


</span><span>V = (4/3)π(r)³</span>
<span>
Therefore, V₁/V₂ = (r₁)³ / (r₂)³
</span>

<span>3) Replacing in the equation V₁/V₂ = P₂/P₁:


</span><span><span>(r₁)³ / (r₂)³ </span>= P₂/P₁</span>
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And you can solve for r₂: (r₂)³ = (P₁/P₂) x (r₁)³


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r₂ = 17.8 cm</span>

4 0
3 years ago
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What animal is most like to harmed when a beaver builds a dam​
siniylev [52]

Answer:

Small fish or insects.

Explanation:

3 0
3 years ago
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A functional group of atoms, such as OH, is called a
Licemer1 [7]
<span>Hydroxy group..

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7 0
3 years ago
In an electrically heated boiler, water is boiled at 140°C by a 90 cm long, 8 mm diameter horizontal heating element immersed in
RideAnS [48]

Explanation:

The given data is as follows.

Volume of water = 0.25 m^{3}

Density of water = 1000 kg/m^{3}

Therefore,  mass of water = Density × Volume

                       = 1000 kg/m^{3} \times 0.25 m^{3}

                       = 250 kg  

Initial Temperature of water (T_{1}) = 20^{o}C

Final temperature of water = 140^{o}C

Heat of vaporization of water (dH_{v}) at 140^{o}C  is 2133 kJ/kg

Specific heat capacity of water = 4.184 kJ/kg/K

As 25% of water got evaporated at its boiling point (140^{o}C) in 60 min.

Therefore, amount of water evaporated = 0.25 × 250 (kg) = 62.5 kg

Heat required to evaporate = Amount of water evapotaed × Heat of vaporization

                           = 62.5 (kg) × 2133 (kJ/kg)

                           = 133.3 \times 10^{3} kJ

All this heat was supplied in 60 min = 60(min)  × 60(sec/min) = 3600 sec

Therefore, heat supplied per unit time = Heat required/time = \frac{133.3 \times 10^{3}kJ}{3600 s} = 37 kJ/s or kW

The power rating of electric heating element is 37 kW.

Hence, heat required to raise the temperature from 20^{o}C to 140^{o}C of 250 kg of water = Mass of water × specific heat capacity × (140 - 20)

                      = 250 (kg) × 40184 (kJ/kg/K) × (140 - 20) (K)

                     = 125520 kJ  

Time required = Heat required / Power rating

                       = \frac{125520}{37}

                       = 3392 sec

Time required to raise the temperature from 20^{o}C to 140^{o}C of 0.25 m^{3} water is calculated as follows.

                    \frac{3392 sec}{60 sec/min}

                     = 56 min

Thus, we can conclude that the time required to raise the temperature is 56 min.

4 0
3 years ago
A pacemaker:
inna [77]
Answer: D. Sends small electrical impulses to the heart, causing it to beat more

Explanation: A pacemaker is implanted in the chest and helps to treat abnormal heart rhythms especially those causing your heart to skip beats or beat too slow.
8 0
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