B. Find the centripetal force needed to keep an 60.0 kg rider traveling in a circle in the ride.
1 answer:
You might be interested in
1) Potential difference: 1 V
2)
Explanation:
1)
When a charge moves in an electric field, its electric potential energy is entirely converted into kinetic energy; this change in electric potential energy is given by
where
q is the charge's magnitude
is the potential difference between the initial and final position
In this problem, we have:
is the magnitude of the charge
is the change in kinetic energy of the particle
Therefore, the potential difference (in magnitude) is
2)
Here we have to evaluate the direction of motion of the particle.
We have the following informations:
- The electric potential increases in the +x direction
- The particle is positively charged and moves from point a to b
Since the particle is positively charged, it means that it is moving from higher potential to lower potential (because a positive charge follows the direction of the electric field, so it moves away from the source of the field)
This means that the final position b of the charge is at lower potential than the initial position a; therefore, the potential difference must be negative:
The catcher can catch the ball at a height of 0.96 m from the ground.
The distance between the pitcher's mound and the catcher's box is about 60'6", which translates to 18.44 m. An average pitcher can pitch with speeds ranging from 88 mph to 97 mph, which is from 39.3 m/s to 43.4 m/s.
Assume the pitcher pitches a ball horizontally with a speed of 40 m/s. If the catcher catches the ball in a time t, then the ball travels a horizontal distance x of 18.44 m and at the same time falls through a height y.
The horizontal motion of the ball is uniform motion since no force acts on the ball ( assuming no air resistance) and hence the acceleration of the ball along the horizontal direction is zero.
Therefore,
Calculate the time t by substituting 18.44 m for x and 40 m/s for u.
The ball is acted upon by the earth's gravitational attraction and hence it accelerates downwards with an acceleration equal to the acceleration due to gravity g.
Since a horizontal projection is assumed, the ball has no component of velocity in the downward direction.
Therefore, for vertical motion, which is an accelerated motion, the distance y, the ball falls in the time t taken by it to reach the catcher's box is given by the equation,
Substitute 9.8 m/s² for g and 0.461 s for t.
The pitcher releases the ball at a height of 1.8 m from a mound which is at a height of 0.2 m. Thus, the ball is released at a height of 2.0 m from the ground. It falls through a distance of 1.04 m in the time it takes to reach the catcher.
Therefore, the height at which the catcher needs to keep his glove so as to catch the ball is given by,
The catcher needs to hold his glove at a height of <u>0,96 m from the ground.</u>