Answer:
When it comes to serving, the court is divided into six zones. Right back is zone one, right front is zone two, middle front is zone three, left front is zone four, left back is zone five and middle back is zone six.
Explanation:
4 for sure because you dont want anything spilling on you or others that is harmful.
Answer:
μsmín = 0.1
Explanation:
- There are three external forces acting on the riders, two in the vertical direction that oppose each other, the force due to gravity (which we call weight) and the friction force.
- This friction force has a maximum value, that can be written as follows:
![F_{frmax} = \mu_{s} *F_{n} (1)](https://tex.z-dn.net/?f=F_%7Bfrmax%7D%20%3D%20%5Cmu_%7Bs%7D%20%2AF_%7Bn%7D%20%281%29)
where μs is the coefficient of static friction, and Fn is the normal force,
perpendicular to the wall and aiming to the center of rotation.
- This force is the only force acting in the horizontal direction, but, at the same time, is the force that keeps the riders rotating, which is the centripetal force.
- This force has the following general expression:
![F_{c} = m* \omega^{2} * r (2)](https://tex.z-dn.net/?f=F_%7Bc%7D%20%3D%20%20m%2A%20%5Comega%5E%7B2%7D%20%2A%20r%20%282%29)
where ω is the angular velocity of the riders, and r the distance to the
center of rotation (the radius of the circle), and m the mass of the
riders.
Since Fc is actually Fn, we can replace the right side of (2) in (1), as
follows:
![F_{frmax} = m* \mu_{s} * \omega^{2} * r (3)](https://tex.z-dn.net/?f=F_%7Bfrmax%7D%20%3D%20m%2A%20%5Cmu_%7Bs%7D%20%2A%20%5Comega%5E%7B2%7D%20%2A%20r%20%283%29)
- When the riders are on the verge of sliding down, this force must be equal to the weight Fg, so we can write the following equation:
![m* g = m* \mu_{smin} * \omega^{2} * r (4)](https://tex.z-dn.net/?f=m%2A%20g%20%3D%20m%2A%20%5Cmu_%7Bsmin%7D%20%2A%20%5Comega%5E%7B2%7D%20%2A%20r%20%284%29)
- (The coefficient of static friction is the minimum possible, due to any value less than it would cause the riders to slide down)
- Cancelling the masses on both sides of (4), we get:
![g = \mu_{smin} * \omega^{2} * r (5)](https://tex.z-dn.net/?f=g%20%3D%20%5Cmu_%7Bsmin%7D%20%2A%20%5Comega%5E%7B2%7D%20%2A%20r%20%285%29)
- Prior to solve (5) we need to convert ω from rev/min to rad/sec, as follows:
![60 rev/min * \frac{2*\pi rad}{1 rev} *\frac{1min}{60 sec} =6.28 rad/sec (6)](https://tex.z-dn.net/?f=60%20rev%2Fmin%20%2A%20%5Cfrac%7B2%2A%5Cpi%20rad%7D%7B1%20rev%7D%20%2A%5Cfrac%7B1min%7D%7B60%20sec%7D%20%3D6.28%20rad%2Fsec%20%286%29)
- Replacing by the givens in (5), we can solve for μsmín, as follows:
![\mu_{smin} = \frac{g}{\omega^{2} *r} = \frac{9.8m/s2}{(6.28rad/sec)^{2} *2.5 m} =0.1 (7)](https://tex.z-dn.net/?f=%5Cmu_%7Bsmin%7D%20%3D%20%5Cfrac%7Bg%7D%7B%5Comega%5E%7B2%7D%20%2Ar%7D%20%20%3D%20%5Cfrac%7B9.8m%2Fs2%7D%7B%286.28rad%2Fsec%29%5E%7B2%7D%20%2A2.5%20m%7D%20%3D0.1%20%287%29)
Answer:
42 grams
Explanation:
you multiply the value by 1000 to get your answer
Even tho one is stronger then the other... they are both alike because they are still nuclear forces.