Answer:

Explanation:
Let consider a car travelling at a speed
. The ratio of final kinetic energy to initial kinetic energy:


The final speed is:

Answer:
The total surface are of the bowl is given by: 0.0532*pi m² (approximately 0.166533 m²)
Explanation:
The total surface area of the semi-spherical bowl can be decomposed in three different sections: 1) an outer semi-sphere of radius 12 cm, 2) an inner semi-sphere of radius 10 cm, and 3) the edge, which is a 2-dimensional ring with internal radius of 10 cm and external radius of 12 cm. We will compute the areas independently and then sum them all.
a) Outer semi-sphere:
A1 = 2*pi*r² = 2*pi*(12 cm)² = 288*pi cm² = 904.78 cm²
b) Inner semi-sphere:
A2 = 2*pi*(10 cm)² = 200*pi cm² = 628.32 cm²
c) Edge (Ring):
A3 = pi*(r1² - r2²) = pi*((12 cm)²-(10 cm)²) = pi*(144-100) cm² = 44*pi cm² = 138.23 cm²
Therefore, the total surface area of the bowl is given by:
A = A1 + A2 + A3 = 288*pi cm² + 200*pi cm² + 44*pi cm² = 532*pi cm² (approximately 1665.33 cm²)
Changing units to m², as required in the problem, we get:
A = 532*pi cm² * (1 m² / 10, 000 cm²) = 0.0532*pi m² (approximately 0.166533 m²)
Answer:
but where is the question ?
Explanation:
<em>hope</em><em> it</em><em> </em><em>works</em><em> out</em>
Answer:
(a) 135 kV
(b) The charge chould be moved to infinity
Explanation:
(a)
The potential at a distance of <em>r</em> from a point charge, <em>Q</em>, is given by

where 
Difference in potential between the points is
![kQ\left[-\dfrac{1}{0.2\text{ m}} -\left( -\dfrac{1}{0.1\text{ m}}\right)\right] = \dfrac{kQ}{0.2\text{ m}} = \dfrac{9\times10^9\text{ F/m}\times3\times10^{-6}\text{ C}}{0.2\text{ m}}](https://tex.z-dn.net/?f=kQ%5Cleft%5B-%5Cdfrac%7B1%7D%7B0.2%5Ctext%7B%20m%7D%7D%20-%5Cleft%28%20-%5Cdfrac%7B1%7D%7B0.1%5Ctext%7B%20m%7D%7D%5Cright%29%5Cright%5D%20%3D%20%5Cdfrac%7BkQ%7D%7B0.2%5Ctext%7B%20m%7D%7D%20%3D%20%5Cdfrac%7B9%5Ctimes10%5E9%5Ctext%7B%20F%2Fm%7D%5Ctimes3%5Ctimes10%5E%7B-6%7D%5Ctext%7B%20C%7D%7D%7B0.2%5Ctext%7B%20m%7D%7D)

(b)
If this potential difference is increased by a factor of 2, then the new pd = 135 kV × 2 = 270 kV. Let the distance of the new location be <em>x</em>.
![270\times10^3 = kQ\left[-\dfrac{1}{x}-\left(-\dfrac{1}{0.1\text{ m}}\right)\right]](https://tex.z-dn.net/?f=270%5Ctimes10%5E3%20%3D%20kQ%5Cleft%5B-%5Cdfrac%7B1%7D%7Bx%7D-%5Cleft%28-%5Cdfrac%7B1%7D%7B0.1%5Ctext%7B%20m%7D%7D%5Cright%29%5Cright%5D)



The charge chould be moved to infinity