A merry go round exerts a force of 1000 N on a rider on the
1 answer:
Answer:
The radius of the circle made by the person on the merry go round is 74.55 meters
Explanation:
The given parameters are;
The force the merry go round exerts on the rider = 1000 N
The time it takes the merry go round to make one complete revolution = 15 seconds
The weight of the person = 750 N
The radius of the circle made by the person on the merry go round = r
We have;
Where;
m = The mass of the person
v = The velocity of the person
= The centrifugal force acting on the person = 1,000 N
r = The radius of the circle made by the person on the merry go round
ω = Angular velocity = 2·π/15 rad/s
We have;
The mass of the person = The weight/(The acceleration due to gravity, g)
∴ The mass of the person = 750/9.81 ≈ 76.45 kg
By substituting the calculated and known values into the equation for the centripetal force, we have;
= m × ω² × r
1000 = 76.45 × (2·π/15)² × r
r = 1000/(76.45 × (2·π/15)²) = 74.55 m
The radius of the circle made by the person on the merry go round = r = 74.55 m.
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Answer:
(A) As it moves farther and farther from Q, its speed will keep increasing.
Explanation:
When a positive charge Q is fixed on a horizontal frictionless tabletop and a second charge q is released near to it then according to the Coulombs law the force acting on it decreases with the square of the distance between them.
Mathematically:
where:
r = distance between the charges
permittivity of free space
By the Newtons' second law of motion if the we know that the acceleration is directly proportional to the force applied. So as the distance between the charges increases the its acceleration also decreases therefore now the charge feels less acceleration but still continues to accelerate with a fading magnitude.