For scientist Could it be possible to run at the speed of sound with some future tool
2 answers:
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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:
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:
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:
- 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:
- (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:
- Prior to solve (5) we need to convert ω from rev/min to rad/sec, as follows:
- Replacing by the givens in (5), we can solve for μsmín, as follows:
1) The grapefruit is in free fall, so it moves by uniformly accelerated motion, with constant acceleration 
. Calling h its height at t=0, the height at time t is given by
We are told thatn when
the grapefruit hits the ground, so h(0.75 s)=0. If we substitute these data into the equation, we can find the initial height h of the grapefruit:
2) The speed of the grapefruit at time t is given by
where
is the initial speed of the grapefruit. Substituting t=0.75 s, we find the speed when the grapefruit hits the ground:
We are told thatn when
2) The speed of the grapefruit at time t is given by
where
(a) The vertical motion is accelerated by gravity. The horizontal component is constant (neglecting air resistance, if this is not a projectile motion, the horizontal component would also be accelerated)
(b) Vertical:
Horizontal:
(c) Use the kinematic equation for distance. Calculate only the vertical component (horizontal is irrelevant):
The ball will be in the air for about 3.9 s.
(d) The range is the horizontal distance traveled. We know the ball is in the air for 3.9s and it moves with a horizontal velocity of 23 m/s. So:
The range is 89.7 meters.