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romanna [79]
3 years ago
13

La masa de un objeto en la tierra es de 100 kg.

Physics
1 answer:
andreev551 [17]3 years ago
3 0

Answer:

a) 980 N

b) 100 kg

c)  163.33 N

Explanation:

a)

The weight of the object on Earth is the product of its mass times the acceleration due to gravity, that is:

W=m\,g=(100\,\,kg)\,g=(100\,\,kg)\,(9.8\,\,m/s^2)=980\,\,N

b)

The mass of the object on the moon, is exactly the same (100 kg) since mass is an inherited value of the object.

c)

Considering that the acceleration of gravity on the Moon is about 1/6 that we have on earth (g), then the weight of the object on the Moon becomes:

W=m\,\frac{g}{6} =\frac{980}{6}\,\,N=163.33\,\,N

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Answer:

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<em>The</em><em> </em><em>second</em><em> </em><em>law</em><em> </em><em>states</em><em> </em><em>that</em><em> </em>The line joining the planet to the sun sweeps out equal areas in equal time intervals.

<em>The</em><em> </em><em>third</em><em> </em><em>law</em><em> </em><em>states</em><em> </em><em>that</em><em> </em>The squares of the period of revolution is proportional to the cubes of the mean distance between the planet and the sun

5 0
3 years ago
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Aneli [31]

Answer:

B.electromagnetic spectrum

Explanation:

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Electromagnetic spectrum

3 0
3 years ago
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
Make the following conversion. 120 mL = _____ cm³ 12.0 1200 120 1.20
omeli [17]
120 is your answer. 120 mL = 120 cm^3
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3 years ago
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The magnitude​ R, measured on the Richter​ scale, of an earthquake of intensity I is defined as Requalslog StartFraction Upper I
lapo4ka [179]

Answer:

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Explanation:

Given data:

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R = log(10^{6.8})

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6 0
3 years ago
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