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damaskus [11]
3 years ago
5

a firework rock is burning at the rate of 14g/s all the products of combustion are ejected in one direction at the speed of 210m

/s what is the greatest height the rockect can have if it is moving vertically upward Give your answer in newtons
Physics
1 answer:
Naddika [18.5K]3 years ago
5 0

SO MY NAME IS ROHITH AND I LIKE MM

Explanation:

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R  =  5.91 m

now we have

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\omega^2 * 5.91 = 34.1

\omega^2 = 5.77

\omega = 2.4 rad/s

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An airplane is flying through a thundercloud at a height of 2 000 m. (this is a very dangerous thing to do because of updrafts,
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It is given that an<span> airplane is flying through a thundercloud at a height of 2000 m.
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Read 2 more answers
The energy from 0.015 moles of octane was used to heat 250 grams of water. The temperature of the water rose from 293.0 K to 371
arsen [322]

Answer : The correct option is, (B) -5448 kJ/mol

Explanation :

First we have to calculate the heat required by water.

q=m\times c\times (T_2-T_1)

where,

q = heat required by water = ?

m = mass of water = 250 g

c = specific heat capacity of water = 4.18J/g.K

T_1 = initial temperature of water = 293.0 K

T_2 = final temperature of water = 371.2 K

Now put all the given values in the above formula, we get:

q=250g\times 4.18J/g.K\times (371.2-293.0)K

q=81719J

Now we have to calculate the enthalpy of combustion of octane.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy of combustion of octane = ?

q = heat released = -81719 J

n = moles of octane = 0.015 moles

Now put all the given values in the above formula, we get:

\Delta H=\frac{-81719J}{0.015mole}

\Delta H=-5447933.333J/mol=-5447.9kJ/mol\approx -5448kJ/mol

Therefore, the enthalpy of combustion of octane is -5448 kJ/mol.

5 0
3 years ago
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