Chapter 10, Problem 16CT

(a)

To determine

To fine the distance.

Answer to Problem 16CT

  $T=16$ degree Celsius

Explanation of Solution

Given information: Speed $s=20\sqrt{T+273}$ , $s=340meterper\sec ond$ and Temperature = T.

Formula Used: Simplify

Calculation:

Speed $s=20\sqrt{T+273}$ , $s=340meterper\sec ond$ and Temperature = T.

As, Speed $s=340meterper\sec ond$

  $340=20\sqrt{T+273}$

Now,

  $\frac{340}{20}=\sqrt{T+273}$

So,

  $\sqrt{T+273}=17$

Now, squaring both sides

  ${\left(\sqrt{T+273}\right)}^2={\left(17\right)}^2$

So,

  $T+273=289$

Now,

Temperature $T=289-273$

Hence,

Temperature $T=16$ degree Celsius

(b)

To determine

To fine the time.

Answer to Problem 16CT

  $Time=5.5$ Seconds

Explanation of Solution

Given information: Speed $s=20\sqrt{T+273}$ , Distance $D=1760m$ and Temperature $t=-{17}^{\circ }C$ .

Formula Used: $Speeds=\frac{Dis\tan ce}{Time}$

Calculation:

Speed $s=20\sqrt{T+273}$ , Distance $D=1760m$ and Temperature $t=-{17}^{\circ }C$ .

As, $Speeds=\frac{Dis\tan ce}{Time}$

So,

  $Time=\frac{Dis\tan ce}{Speeds}$

Now,

  $Time=\frac{1760}{20\sqrt{T+273}}$ Seconds

As, Temperature $t=-{17}^{\circ }C$ .

  $Time=\frac{1760}{20\sqrt{-17+273}}$ Seconds

So,

  $Time=\frac{1760}{20\sqrt{256}}$ Seconds

Now, square root of 256 is 16

  $Time=\frac{1760}{20\times 16}$ Seconds

So,

  $Time=\frac{1760}{320}$ Seconds

Hence,

  $Time=5.5$ Seconds