Answer:
we cannot view through an opague object since it does not alow light through but only cast shadows
Explanation:
Answer:
(a)
(b)
Explanation:
Part (a)
The total length of copper cord L=86.3 m
The cross sectional area A=1.71×10⁻⁶m²
The resistivity of copper p=1.72×10⁻⁸Ω
Thus the resistance of extension cord is
Part (b)
The resistance of trimmer Rt=17.9 ohms
When voltage of 120V is applied then the current I is passing through series circuit is
Thus the voltage across the trimmer is:
solution:
E\delta =\frac{R}{\epsilon0}(1-\frac{A}{\sqrt{4R^{2}}-ac}
=\frac{R}{\epsilon0}(1-\frac{1}{\sqrt{4r^{2}/^{_a{2}}+1}})
=\frac{R}{\epsilon0}(1-\frac{1}{\sqrt{4x^2+1}})
x=\frac{r}{a}
infinite case,
Ei=\frac{r}{\epsilon0}
\therefore e\delta =ei(1-\frac{1}{\sqrt{4x^{2}+1}})
we have to find x when,
ei-e\delta =1% ,y=ei=1/100 ei
or,ei-ei+\frac{ei}{\sqrt{4x^2+1}} = 1/100ei
\frac{1}{\sqrt{4x^2+1}}=\frac{1}{100}
4x^2+1 =10^4
x=\frac{\sqrt{\frac{10^4-1}{4}}}=49.99\approx 50
\therefore \frac{r}{a}\approx 50
The frequency of the wave is the frequency of that sine function ... 2,000π ... but that's the <u>angular </u>frequency, in radians per second.
To convert angular frequency to Hz, we have to divide by 2π.
So the frequency of this wave is <em>1,000 Hz. </em>
Answer:
Explanation:
Given
volume
Suppose base is square with side L
height of crate is h
Volume
Cost of top and bottom area
Cost of Side area
Total Cost
Total Cost
Differentiate C w.r.t Length
Dimensions are