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ycow [4]
2 years ago
10

(100 POINTS AND BRAINLIEST)

Physics
1 answer:
Lostsunrise [7]2 years ago
4 0

{\huge{\fcolorbox{yellow}{red}{\orange{\boxed{\boxed{\boxed{\boxed{\underbrace{\overbrace{\mathfrak{\pink{\fcolorbox{green}{blue}{Answer}}}}}}}}}}}}}

0.72 liters

Explanation:

\sf Volume \: before \: heated(V_1) = 0.5l \\  \\  \sf Temperture \: before \: heated(T_1) = 20{ \degree} C = 20 + 273.15 = 293.15K \\  \\  \sf Volume \: after \: heated(V_2) =  ? \\  \\  \sf Temperature \: after \: heated(T_2) = 150{ \degree}C = 150 + 273.15 = 423.15K

\textsf{By using Charles's law}  \\  \\  \tt{ \frac{V_1}{T_1}  =  \frac{V_2}{T_2} }

\sf Now \: sustituting \: the \: values

\sf  \longmapsto \frac{0.5}{293.15}  =  \frac{V_2}{423.15}  \\  \\  \\  \sf \longmapsto  \frac{0.5 \times 423.15}{293.15}  = V_2 \\  \\  \\  \sf  \longmapsto  \frac{211.575}{293.15}  = V_2 \\  \\  \\  \boxed {\longmapsto 0.72l = V_2}

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<u>6.87 ft/s</u> is the rate at which the top of ladder slides down.

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Differentiating the above equation with respect to time, 't'. This gives,

\frac{d}{dt}(h^2+b^2)=\frac{d}{dt}(400)\\\\\frac{d}{dt}(h^2)+\frac{d}{dt}(b^2)=0\\\\2h\frac{dh}{dt}+2b\frac{db}{dt}=0\\\\h\frac{dh}{dt}+b\frac{db}{dt}=0--------2

In the above equation the term \frac{dh}{dt} is the rate at which top of ladder slides down and \frac{db}{dt} is the rate at which bottom of ladder slides away.

Now, as per question, h=8\ ft, \frac{db}{dt}=3\ ft/s

Plug in h=8 in equation (1) and solve for b. This gives,

8^2+b^2=400\\64+b^2=400\\b^2=400-64\\b^2=336\\b=\sqrt{336}=18.33\ ft

Now, plug in all the given values in equation (2) and solve for \frac{dh}{dt}

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Therefore, the rate at which the top of ladder slide down is 6.87 ft/s. The negative sign implies that the height is reducing with time which is true because it is sliding down.

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