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
Answer:
Earth, Water, Air, and Fire
A child climbing a ladder is transforming kinetic energy into potential energy.
Answer:
0.25683
Explanation:
t = Time taken by the truck to stop = 11 seconds
u = Initial velocity of the truck = 62 mi/h
v = Final velocity of the truck = 0
a = Acceleration of the truck
g = Acceleration due to gravity = 9.81 m/s²
Equation of motion
The minimum coefficient of static friction between the turtle and the truck bed surface needed to prevent the turtle from sliding is 0.25683
<span>Од сила е еднаква на масата * aceleration
Ние го имаме тоа сила = 5N, маса = 0.6kg a =?
5 = 0,6 * a.
Затоа a = 5 / 0,6
a = 8.33m / s²
Ве молиме да се одбележи мене како brainliest.</span>
