Let
be the height of the building and thus the initial height of the ball. The ball's altitude at time
is given by

where
is the acceleration due to gravity.
The ball reaches the ground when
after
. Solve for
:


so the building is about 16 m tall (keeping track of significant digits).
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
Answer:
8 N North.
Explanation:
Given that,
One force has a magnitude of 10 N directed north, and the other force has a magnitude of 2 N directed south.
We need to find the magnitude of net force acting on the object.
Let North is positive and South is negative.
Net force,
F = 10 N +(-2 N)
= 8 N
So, the magnitude of net force on the object is 8 N and it is in North direction (as it is positive). Hence, the correct option is (d) "8N north".
Answer: 9.9%
Explanation: efficiency = (work output /work input) × 100
Note that, 1 kilocalorie = 4184 joules, hence 22kcal = 22× 4184 = 92048 joules.
Work output = 9200 j and work input = 92048 j
Efficiency = (9200/92048) × 100 = 0.099 × 100 = 9.9%
Answer:
The angle of banked curve that makes the reliance on friction unnecessary is

Explanation:
In order the car to stay on the curve without friction, the net force in the direction of radius should be equal or smaller than the centripetal force. Otherwise the car could slide off the curve.
The only force in the direction of radius is the sine component of the weight of the car

The cosine component is equivalent to the normal force, which we will not be using since friction is unnecessary.
Newton’s Second Law states that

Also, the car is making a circular motion:

Combining the equations:

Finally the angle is
