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2 years ago

5

A weather balloon has a volume of 90.0 L when it is released from sea level 101 kPa. What is the atmospheric pressure on the bal

loon when it has grown to a size of 175.0 L?

2 answers:

kirza4 [7]2 years ago

8 0

Answer:

<em><u>The final atmospheric pressure is 5.19 · 10⁴ Pa</u></em>

Step-by-step explanation:

Assuming that the temperature of the air does not change, we can use Boyle's law, which states that for a gas kept at constant temperature, the pressure of the gas is inversely proportional to its volume. In formula,

pV = const.

where p is the gas pressure and V is the volume

The equation can also be rewritten as

p₁ V₁ = p₂ V₂

where in our problem we have:

p₁ = 1.03 · 10₅ Pa is the initial pressure (the atmospheric pressure at sea level)

V₁ = 90.0L is the initial volume

p₂ is the final pressure

V₂ = 175.0L is the final volume

Solving the equation for p2, we find the final pressure:

p₂ = p₁ v₁ divided by V₂ = (1.01 · 10⁵)(90.0) divided by 175.0 = 5.19 · 10⁴ Pa

Anna71 [15]2 years ago

4 0

Answer:

5.19 · 10⁴ Pa

Step-by-step explanation:

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2 years ago

What is the ninth term in the binomial expansion of (X-2y)^13?

vlada-n [284]

Answer:

$329472\cdot x^5\cdot y^8$

Explanation:

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$(a+b)^n=a^n+n\cdot a^{n-1}{\cdot b}+\frac{n(n-1)}{2}\cdot a^{n-2}\cdot b^2+------+b^n$

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1 year ago

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I tell you these facts about a mystery number, $c$: $\bullet$ $1.5 < c < 2$ $\bullet$ $c$ can be written as a fraction wit

makkiz [27]

Answer:

Possible answer: $\displaystyle c = \frac{16}{10} = \frac{8}{5} = 1.6$ .

Step-by-step explanation:

Rewrite the bounds of $c$ as fractions:

The simplest fraction for $1.5$ is $\displaystyle \frac{3}{2}$ . Write the upper bound $2$ as a fraction with the same denominator:

$\displaystyle 2 = 2 \times 1 = 2 \times \frac{2}{2} = \frac{4}{2}$ .

Hence the range for $c$ would be:

$\displaystyle \frac{3}{2} < c < \frac{4}{2}$ .

If the denominator of $c$ is also $2$ , then the range for its numerator (call it $p$ ) would be $3 < p < 4$ . Apparently, no whole number could fit into this interval. The reason is that the interval is open, and the difference between the bounds is less than $2$ .

To solve this problem, consider scaling up the denominator. To make sure that the numerator of the bounds are still whole numbers, multiply both the numerator and the denominator by a whole number (for example, 2.)

$\displaystyle \frac{3}{2} = \frac{2 \times 3}{2 \times 2} = \frac{6}{4}$ .

$\displaystyle \frac{4}{2} = \frac{2\times 4}{2 \times 2} = \frac{8}{4}$ .

At this point, the difference between the numerators is now $2$ . That allows a number ( $7$ in this case) to fit between the bounds. However, $\displaystyle \frac{1}{c} = \frac{4}{7}$ can't be written as finite decimals.

Try multiplying the numerator and the denominator by a different number.

$\displaystyle \frac{3}{2} = \frac{3 \times 3}{3 \times 2} = \frac{9}{6}$ .

$\displaystyle \frac{4}{2} = \frac{3\times 4}{3 \times 2} = \frac{12}{6}$ .

$\displaystyle \frac{3}{2} = \frac{4 \times 3}{4 \times 2} = \frac{12}{8}$ .

$\displaystyle \frac{4}{2} = \frac{4\times 4}{4 \times 2} = \frac{16}{8}$ .

$\displaystyle \frac{3}{2} = \frac{5 \times 3}{5 \times 2} = \frac{15}{10}$ .

$\displaystyle \frac{4}{2} = \frac{5\times 4}{5 \times 2} = \frac{20}{10}$ .

It is important to note that some expressions for $c$ can be simplified. For example, $\displaystyle \frac{16}{10} = \frac{2 \times 8}{2 \times 5} = \frac{8}{5}$ because of the common factor $2$ .

Apparently $\displaystyle c = \frac{16}{10} = \frac{8}{5}$ works. $c = 1.6$ while $\displaystyle \frac{1}{c} = \frac{5}{8} = 0.625$ .

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Answer:

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Step-by-step explanation:

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So then you would calculate 48 divided by 4.

The weight of Carla's dog is 12 pounds.

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2 years ago