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The correct answer is 7.24
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Answer:
θ = 14.27°
Explanation:
The only force acting on the puck is the gravitational force. Since the track is banked with an angle θ, we have to separate the components of the weight.
For the sake of simplicity, I will denote the perpendicular direction to the truck as the y-direction, and the direction along the radius as the x-direction.
So, the free-body diagram of the puck is as follows:
1- x-component of the weight of the puck: mgsinθ
2- y-component of the weight of the puck: mgcosθ
3- Normal force in the y-direction perpendicular to the track.
Since there is no motion on the y-direction, normal force is equal to the y-component of the weight of the puck.
The x-component of the weight of the puck is equal to the centripetal force according to Newton's Second Law:
Substituting the variables given in the question, the angle of the track can be found:
Answer:
5.53 m
Explanation:
= initial speed of shuffleboard disk =
= final speed of shuffleboard disk =
= Coefficient of kinetic friction between the disk and concrete court = 0.31
acceleration due to friction is given as
= distance traveled by disk before it stops
using the kinematics equation that fits the above the data, we have
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Answer:
g = 15.5 m/s²
Explanation:
In order to find the acceleration due to gravity near the surface of this planet can be calculated by using 2nd equation of motion. The 2nd equation of motion is given as:
h = Vi t + (0.5)gt²
where,
h = height covered by the wrench = 5 m
Vi = Initial Velocity = 0 m/s
t = Time Taken to hit the ground = 0.804 s
g = acceleration due to gravity near the surface of the planet = ?
Therefore,
5 m = (0 m/s)(0.804 s) + (0.5)(g)(0.804 s)²
g = (5 m)/(0.3232 s²)
<u>g = 15.5 m/s²</u>