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3 years ago

15

a 64 l gas tank is filled completely is 15 degree celsius how much gasoline overflows into can get up to 35 degree celsius while

the car is parked in direct sunlight

1 answer:

o-na [289]3 years ago

3 0

Answer:

Volume of gasoline overflow(v)= 40/9 L (I.e. 4.44 L)

Explanation:

Use <u>v1</u><u>/</u><u>T1</u><u>=</u><u>v2</u><u>/</u><u>T2</u>

.....overflow(V)=v2-v1

<u>Note</u><u>;</u> <em>Take</em><em> </em><em>temperature</em><em> </em><em>in</em><em> </em><em>absolute</em><em> </em><em>scale</em><em> </em><em>or</em><em> </em><em>kelvin</em><em> </em><em>scale</em><em> </em>

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Based on the graph, which data point is most likely to have experimental

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Answer:

B. 59 kg

Explanation:

From the graph you notice that a linear relation in indicated by the line joining the points such that the points on the line represent the data that show a correct relationship in the experiment.

This means that the point outside the line has an error .

This point is the value 59 kg that does not align with other values which are included in the graph.

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Liquids have what kind of volume and shape

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During a solar eclipse, the Moon, Earth, and Sun all lie on the same line, with the Moon between the Earth and the Sun.

lord [1]

Answer:

(a) $F_{sm} = 4.327\times 10^{20}\ N$

(b) $F_{em} = 1.983\times 10^{20}\ N$

(c) $F_{se} = 3.521\times 10^{20}\ N$

Solution:

As per the question:

Mass of Earth, $M_{e} = 5.972\times 10^{24}\ kg$

Mass of Moon, $M_{m} = 7.34\times 10^{22}\ kg$

Mass of Sun, $M_{s} = 1.989\times 10^{30}\ kg$

Distance between the earth and the moon, $R_{em} = 3.84\times 10^{8}\ m$

Distance between the earth and the sun, $R_{es} = 1.5\times 10^{11}\ m$

Distance between the sun and the moon, $R_{sm} = 1.5\times 10^{11}\ m$

Now,

We know that the gravitational force between two bodies of mass m and m' separated by a distance 'r' is given y:

$F_{G} = \frac{Gmm'_{2}}{r^{2}}$ (1)

Now,

(a) The force exerted by the Sun on the Moon is given by eqn (1):

$F_{sm} = \frac{GM_{s}M_{m}}{R_{sm}^{2}}$

$F_{sm} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 7.34\times 10^{22}}{(1.5\times 10^{11})^{2}}$

$F_{sm} = 4.327\times 10^{20}\ N$

(b) The force exerted by the Earth on the Moon is given by eqn (1):

$F_{em} = \frac{GM_{s}M_{m}}{R_{em}^{2}}$

$F_{em} = \frac{6.67\times 10^{- 11}\times 5.972\times 10^{24}\times 7.34\times 10^{22}}{(3.84\times 10^{8})^{2}}$

$F_{em} = 1.983\times 10^{20}\ N$

(c) The force exerted by the Sun on the Earth is given by eqn (1):

$F_{se} = \frac{GM_{s}M_{m}}{R_{es}^{2}}$

$F_{se} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 5.972\times 10^{24}}{((1.5\times 10^{11}))^{2}}$

$F_{se} = 3.521\times 10^{20}\ N$

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