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2 answers:
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Answer:
a) μ = 0.475 , b) μ = 0.433
Explanation:
a) For this exercise of Newton's second law, we create a reference system with the x-axis parallel to the plane and the y-axis perpendicular to it
X axis
Wₓ - fr = m a
the friction force has the expression
fr = μ N
y Axis
N - = 0
let's use trigonometry for the components the weight
sin 27 = Wₓ / W
Wₓ = W sin 27
cos 27 = W_{y} / W
W_{y} = W cos 27
N = W cos 27
W sin 27 - μ W cos 27 = m a
mg sin 27 - μ mg cos 27 = m a
μ = (g sin 27 - a) / (g cos 27)
very = tan 27 - a / g sec 27
μ = 0.510 - 0.0344
μ = 0.475
b) now the block starts with an initial speed of 3m / s. In Newton's second law velocity does not appear, so this term does not affect the result, the change in slope does affect the result
μ = tan 25 - 0.3 / 9.8 sec 25
μ = 0.466 -0.03378
μ = 0.433
Answer:
(a) 0.261 rad/s
(b) 1007.72 m
Explanation:
The angular velocity of the Ferris wheel is 150.0 revolutions per hour.
(a) To calculate the angular velocity of the wheel in units of radians per second, you take into account the following equivalence:
1 hour = 3600 seconds
1 revolution = 2π radians
You use the previous conversion factors:
In units of radians per seconds the wheel turns at 0.261 rad/s
(b) To find the arc length described by the wheel, you first calculate the angle described by the wheel in the time t, by using the following formula:
(1)
ω: angular velocity = 0.261 rad/s
t: time = 4.95 min
You first convert the time to units of seconds
Next, you replace the values of the parameters in the equation (1):
Next, you use the following formula for the arc length:
(2)
r: radius of the wheel = 13.0 m
You replace the values of the parameters in the equation (2):
The arc length described by the wheel is 1007.72m