The force (attractive if the charges are dissimilar, else repulsive) is along a line that connects the two particles.
The cyclist's final velocity is 10 m/s.
From the question,
We are to determine the cyclist's final velocity.
<h3>Linear motion</h3>
From one of the equations of motion for linear motion, we have
![v=u+at](https://tex.z-dn.net/?f=v%3Du%2Bat)
Where
v is the final velocity
u is the initial velocity
a is the acceleration
and t is the time
From the given information,
The cyclist starts at rest, this means the initial velocity is 0 m/s
That is,
u = 0 m/s
Also
a = 0.5 m/s²
and t = 20 s
Putting the parameters into the equation, we get
![v = 0 + 0.5 \times 20](https://tex.z-dn.net/?f=v%20%3D%200%20%2B%200.5%20%5Ctimes%2020)
![v = 0 + 10](https://tex.z-dn.net/?f=v%20%3D%200%20%2B%2010)
![v = 10 \ m/s](https://tex.z-dn.net/?f=v%20%3D%2010%20%5C%20m%2Fs)
Hence, the cyclist's final velocity is 10 m/s.
Learn more on linear motion here: brainly.com/question/19365526
Answer:
The value ![t = 1.995 \ s](https://tex.z-dn.net/?f=t%20%3D%201.995%20%5C%20%20s%20)
Explanation:
From the question we are told that
The acceleration due to gravity is a = 2g
The maximum height is h= 39 m
Generally from the kinematic equation
![s = ut + \frac{1}{2} at^2](https://tex.z-dn.net/?f=s%20%3D%20%20ut%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20at%5E2)
Here u is the velocity of the ball at maximum height before it start falling and the value is 0 m/s
So
![39 = 0* t + \frac{1}{2} (2g)t^2](https://tex.z-dn.net/?f=39%20%3D%20%200%2A%20t%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20%282g%29t%5E2)
![39 = (9.8)t^2](https://tex.z-dn.net/?f=39%20%3D%20%289.8%29t%5E2)
![t = \sqrt{3.9795918}](https://tex.z-dn.net/?f=t%20%3D%20%20%5Csqrt%7B3.9795918%7D)
![t = 1.995 \ s](https://tex.z-dn.net/?f=t%20%3D%201.995%20%5C%20%20s%20)
Answer:
Option (B)
Explanation:
To measure the distance between two objects we use the units of length.
To measure small distances we use, mm or cm or m. But if we measure large distances between two objects we use large units such that kilometer or miles.
To measure the very large distances between two objects we use large units such as Astronomical unit or Parsec.
The Astronomical unit is defied as the distance between earth and Sun.
Thus, the distance between earth and the sun is defined as 1 Astronomical unit.
1 Astronomical unit = 1 A.U. = 1.5 x 10^8 km = 1.5 x 10^11 m