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dolphi86 [110]
2 years ago
10

Lexy used the formula shown to calculate the force of gravity on a space shuttle. Fg = G What does 3 × 105 kg represent? the dif

ference between Earth’s mass and the space shuttle’s mass the sum of Earth’s mass and the space shuttle’s mass the mass of Earth the mass of the space shuttle
Physics
1 answer:
STALIN [3.7K]2 years ago
5 0
<h2>Answer:</h2>

<u>This term shows the </u><u>mass of the space shuttle</u>

<h2>Explanation:</h2>

We know that the mass of the Earth is 5.972 × 10^24 kg. Similarly the sum of  mass of earth and the mass of shuttle must be a greater number as compared to the number given. It simply means that the mass of earth is itself 5.972 × 10^24 kg and the value given is 3 × 105 kg so it is obvious that if was the sum then it must be greater than the mass of earth. Therefore we can say that this not the mass of earth, neither the sum of mass of earth and shuttle, but this is only the mass of space shuttle which is the last multiple choice.

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sashaice [31]
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2 years ago
A simple circuit within a laptop has a single resistor with a resistance of 0.1 Ω and requires a current of 50 mA. Select the vo
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2 years ago
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,

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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.

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densk [106]

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so clearly gain in momentum of one object is same as the loss of momentum of other object

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