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3 years ago

14

Think about what you observed in the Sim and the information the lead chemist sent you about the soda can. Which substance in th

e Sim best represents water (substance B or C)? Explain your answer in the space below.

1 answer:

diamong [38]3 years ago

8 0

Answer:

Substance C.

Explanation:

Substance C changed phase by condense and melted faster than substance B, which best represents liquid.

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Your answer for this question is the third option.

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3 years ago

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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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3 years ago

A scientist conducted research and plotted his results in the histogram shown below. Which of the

ludmilkaskok [199]

Answer:

Median

Explanation:

we commonly use mean as the measure of the center of the data.
But Median is better measure than mean.

<u>Extra</u><u> </u><u>information</u><u>:</u><u>-</u>

<u>Median</u><u> </u><u>formula</u><u>:</u><u>-</u>

$\boxed{\begin{minipage}{6cm}$\bigstar$\:\:\sf Median = l + $\sf\dfrac{\frac{n}{2}-C.f.}{f}\times h\\\\Here: \\1)\:n = \sum f =\\2)\:l=Lower\:limit\:of\:median\:class=\\3)\:C.f.=Cumulative\:frequency\:of\:class\\preceeding\:the\:median\:class=\\4)\:f= frequency\:of\:median\:class=\\5)\:h= Class\:interval = \end{minipage}}$

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3 years ago

The weight of the North American P-51 Mustang airplane is 10,100 lb and its wing platform area is 233 ft2 . Calculate the wing l

Natali5045456 [20]

Answer

given,

weight of airplane,W = 10,100 lb

Wing area,A = 233 ft²

now,

Wing Loading Calculation in English engineering

= $\dfrac{W}{A}$

= $\dfrac{10,100}{233}$

= 43.35 lb/ft²

Wing Loading in SI unit

= $43.35\times \dfrac{4.448\ N}{1\ lb}\times (\dfrac{1\ ft}{0.3048})^2$

= $2075.51\ N/m^2$

Wing loading in kgf/m²

= $\dfrac{2075.51}{9.81}$

= 211.57 kgf/m²

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4 years ago

Where were scientific advancements being made during the dark ages

Leto [7]

Answer:

The period saw major technological advances, including the adoption of gunpowder, the invention of vertical windmills, spectacles, mechanical clocks, and greatly improved water mills, building techniques (Gothic architecture, medieval castles), and agriculture in general (three-field crop rotation).

Explanation:

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4 years ago

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