Answer:
0.82 m
Explanation:
The ball is in free fall - uniform accelerated motion with constant acceleration downward, 
(acceleration of gravity). So we can use the following suvat equation to solve the problem:
where
v is the final velocity
u = 4 m/s is the initial velocity
a is the acceleration
s is the displacement
At the maximum displacement, v = 0 (the velocity becomes zero). Substituting and solving for s, we find:
The answer should be A weighlifter lifting 100lb. bell bar from his chest to a certain height.
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
Betelgeuse is one of the largest known stars and is probably at least the size of the orbits of Mars or Jupiter around the sun. That's a diameter about 700 times the size of the Sun or 600 million miles. For a star it has a rather low surface temperature (6000 F compared to the Sun's 10,000 F).
The velocity of the ball when it strikes the ground, given the data is 21.56 m/s
<h3>Data obtained from the question</h3>
From the question given above, the following data were obtained:
- Time to reach ground from maximum height (t) = 2.2 s
- Initial velocity (u) = 0 m/s
- Acceleration due to gravity (g) = 9.8 m/s²
- Final velocity (v) =?
<h3>How to determine the velocity when the ball strikes the ground</h3>
The velocity of the ball when it strikes the ground can be obtained as illustrated below:
v = u + gt
v = 0 + (9.8 × 2.2)
v = 0 + 21.56
v = 21.56 m/s
Thus, the velocity of the ball when it strikes the ground is 21.56 m/s
Learn more about motion under gravity:
brainly.com/question/22719691
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