Chapter 9.5, Problem 27AYU

To determine

To find:

a. $I$ and $A$, and interpret the meaning of each.

Answer to Problem 27AYU

Solution:

a. The intensity of the sun rays is approximately $0.022w/c{m}^2$, the area of the solar panel is $500c{m}^2$.

Explanation of Solution

Given:

$I=\langle -0.02,-0.01\rangle$ and $A=\langle 300,400\rangle$. The total number of watts collected by the panel is given by $W=I\cdot A$.

Calculation:

a. If $\langle a,b\rangle$ denotes the position vector of a vector $v$, then the vector $v$ is expressed as,

$\langle a,b\rangle =a\langle 1,0\rangle +b\langle 0,1\rangle =ai+bj$

Thus the vector $v$ is $\langle -0.02,-0.01\rangle =a\langle -0.02,0\rangle +b\langle 0,-0.01\rangle =-0.02i-0.01j$,

$I=-0.02i-0.01j$

Now the area of the vector $A$ is,

$\langle 300,400\rangle =a\langle 300,0\rangle +b\langle 0,400\rangle =300i+400j$

Therefore the vector $A$ is $A=300i+400j$,

$\Vert I\Vert =\sqrt{{\left(-0.02\right)}^2+{\left(-0.01\right)}^2}=\sqrt{0.0005}=0.022$

Therefore the intensity of the sun rays is approximately $0.022w/c{m}^2$,

$\Vert A\Vert =\sqrt{{\left(300\right)}^2+{\left(400\right)}^2}=500$

The area of the solar panel is $500c{m}^2$.

To determine

To find:

b. Compute $W$ and interpret its meaning.

Answer to Problem 27AYU

Solution:

b. $w=10$ watts of energy is collected.

Explanation of Solution

Given:

$I=\langle -0.02,-0.01\rangle$ and $A=\langle 300,400\rangle$. The total number of watts collected by the panel is given by $W=I\cdot A$.

Calculation:

b. $W=I\cdot A=-0.02\left(300\right)-0.01\left(400\right)=10$

$w=10$ watts of energy is collected.

To determine

To find:

c. If the solar panel is to collect the maximum number of watts, what must be true about $I$ and $A$?

Answer to Problem 27AYU

Solution:

c. The vectors $I$ and $A$ should be parallel with solar panel facing the sun.

Explanation of Solution

Given:

$I=\langle -0.02,-0.01\rangle$ and $A=\langle 300,400\rangle$. The total number of watts collected by the panel is given by $W=I\cdot A$.

Calculation:

c. $\cos \theta =\frac{I\cdot A}{\Vert I\Vert \Vert A\Vert }=\frac{10}{0.02\left(500\right)}=1$

$\theta ={\cos }^{-1}\left(1\right)=0$

Therefore the vectors $I$ and $A$ should be parallel with solar panel facing the sun.