Chapter 9.5, Problem 27AYU

To determine

To find:

a 1 and $A$ , and interpret the meaning of each.

Answer to Problem 27AYU

Solution:

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

Explanation of Solution

Given:

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

Calculation:

a. If $(a,b)$ denotes the position vector of a vector $v$ , then the vector $v$ is expressed as,

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

Thus the vector $v$ is $(-0.02,-0.01)=a(-0.02,0)+b(0,-0.01)=-0.02t-0.01j$

  $I=-0.02i-0.01j$

Now the area of the vector $A$ is,

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

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

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

Therefore the intensity of the sun rays is approximately $0.022\text{w}/{\text{cm}}^2$ ,

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

The area of the solar panel is $500{\text{cm}}^2$ .

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 $A=\langle 300,400).$ The total number of watts collected by the panel is given by $W=I^{-}A$

Calculation:

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

  $w=10$ 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 $A=(300,400).$ The total number of watts collected by the panel is given by $W=I^{-}$ A.

Calculation:

c. $\cos \theta =\frac{t\cdot A}{\Vert r\Vert u\Vert }=\frac{10}{0.02(500)}=1$

  $\theta ={\cos }^{-1}(1)=0$

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