Which statement about gravity is written as a law? *
1 answer:
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Answer:
a. The component of the net force which make up the apparent weight are added to each other at the bottom and subtracted (the centripetal force from the weight) at the top)
b. Apparent weight at the top is approximately 519.06 N
Apparent weight at the bottom is approximately 558.94 N
Explanation:
a. The apparent weight at the top is different from the apparent weight at the bottom of a moving Ferris wheel because of the opposite direction in which the centripetal force acts at the top and the bottom, which are upwards and downwards respectively, while the weight acts downwards constantly
b. The given parameters are
The radius of the Ferris wheel, r = 7.2 m
The period for one complete revolution, t = 28 seconds
The angle covered in one revolution, θ = 2·π radian
The mass of the person riding on the Ferris wheel, the passenger = 55 kg
Therefore, we have;
The angular speed, ω = Δθ/Δt = 2·π/(28)
From which we have;
Centripetal force, = m × ω² × r
Substituting the known values, we have = 55 kg × (2·π/(28 s))² × 7.2 m ≈ 19.94 N
The centripetal force, = 19.94 N always acting outward from the center
Weight = Mass × Acceleration due to gravity
The weight of the passenger = 55 kg × 9.8 m/s² = 539 N
The weight of the passenger = 539 N always acting downwards
At the top of the Ferris wheel the the centripetal force is acting upwards and the weight is acting downwards
Therefore;
The net force, which is the apparent weight of the passenger at the top = 539 N - 19.94 N ≈ 519.06 N
Apparent weight at the top ≈ 519.06 N
At the bottom of the Ferris wheel the weight is acting downwards and the centripetal force is also acting downwards
Therefore;
The net force at the bottom, which is the apparent weight of the passenger at the bottom = 539 N + 19.94 N ≈ 558.94 N
Apparent weight at the bottom ≈ 558.94 N.
Answer:
The minimum average speed the opponent must move so that he is in position to hit the ball is approximately 5.79 m/s
Explanation:
The given parameters of the ball are;
The initial speed of the ball = 15 m/s
The direction in which the ball is launched = 50° above the horizontal
The location of the other tennis player when the ball is launched = 10 m from the ball
The time at which the other tennis player begins to run = 0.3 seconds after the ball is launched
The height at which the ball is hit back = 2.1 m above the height from which the ball is launched
The vertical position, 'y', at time, 't', of a projectile motion is given as follows;
y = (u·sinθ)·t - 1/2·g·t²
When y = 2.1 m, we have;
2.1 = (15·sin(50°))·t - 1/2·9.8·t²
∴ 4.9·t² - (15·sin(50°))·t + 2.1 = 0
Solving with the aid of a graphing calculator function, we get;
t = 0.199776187257 s or t = 2.14525782198 s
Therefore, the ball is at 2.1 m above the start point on the other side of the court at t ≈ 2.145 seconds
The horizontal distance, 'x', the ball travels at t ≈ 2.145 seconds is given as follows;
x = u × cos(50°) × t = 15 × cos(50°) × 2.145 ≈ 20.682 m
The horizontal distance the ball travels at t ≈ 2.145 seconds, x ≈ 20.682 m
Therefore, we have;
The time the other player has to reach the ball, t₂ =2.145 s - 0.3 s ≈ 1.845 s
The distance the other player has to run, d = 20.682 m - 10 m = 10.682 m
The minimum average speed the other player has to move with, = d/t₂
∴ = 10.682 m/(1.845 s) ≈ 5.78970189702 m/s ≈ 5.79 m/s
The minimum average speed the opponent must move so that he is in position to hit the ball, ≈ 5.79 m/s.