Which sporting event is symbolic of American culture?
2 answers:
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The radius of the Ferris wheel is r = 83/2 = 91.5 m.
The angular velocity is
ω = (2π rad)/(37.3*60 s) = 2.8075 rad/s
The tangential velocity is
v = rω = 0.2569 m/s
The arc length traveled in 8.6 min (= 8.6*60 s = 516 s) is
s = (91.5 m)*(2.8075 rad/s)*(516 s) = 132.553 m
The central angle swept is
θ = 132.553/91.5 = 1.4487 rad
From the vector diagram, the change in velocity is (from the Law of Cosines)
Δv² = v²(1 - 2 cosФ)
where Ф = Π - 1.4487 = 1.6929 rad
Δv² = 0.2569²[1 - 2*(-0.1218)] = 0.0821
Δv = 0.2865 m/s
The acceleration is
a₁ = (0.2865 m/s)/(516 s) = 5.6 x 10⁻⁴ m/s²
The actual centripetal acceleration is directed toward the center of the wheel, and its value is
a = v²/r = 0.2569²/91.5 = 7.2 x 10⁻⁴ m/s²
Answer:
a = 7.2 x 10⁻⁴ m/s², the centripetal acceleration acting toward the center of the wheel.
The magnitude of a₁ is 5.6 x 10⁻⁴ m/s², but it is not directed toward the center of the wheel.
The angular velocity is
ω = (2π rad)/(37.3*60 s) = 2.8075 rad/s
The tangential velocity is
v = rω = 0.2569 m/s
The arc length traveled in 8.6 min (= 8.6*60 s = 516 s) is
s = (91.5 m)*(2.8075 rad/s)*(516 s) = 132.553 m
The central angle swept is
θ = 132.553/91.5 = 1.4487 rad
From the vector diagram, the change in velocity is (from the Law of Cosines)
Δv² = v²(1 - 2 cosФ)
where Ф = Π - 1.4487 = 1.6929 rad
Δv² = 0.2569²[1 - 2*(-0.1218)] = 0.0821
Δv = 0.2865 m/s
The acceleration is
a₁ = (0.2865 m/s)/(516 s) = 5.6 x 10⁻⁴ m/s²
The actual centripetal acceleration is directed toward the center of the wheel, and its value is
a = v²/r = 0.2569²/91.5 = 7.2 x 10⁻⁴ m/s²
Answer:
a = 7.2 x 10⁻⁴ m/s², the centripetal acceleration acting toward the center of the wheel.
The magnitude of a₁ is 5.6 x 10⁻⁴ m/s², but it is not directed toward the center of the wheel.
The magnetic force experienced by the proton is given by
where q is the proton charge, v its velocity, B the magnitude of the magnetic field and
the angle between the direction of v and B. Since the proton moves perpendicularly to the magnetic field, this angle is 90 degrees, so 
and we can ignore it in the formula.
For Netwon's second law, the force is also equal to the proton mass times its acceleration:
So we have
from which we can find the magnitude of the field:
where q is the proton charge, v its velocity, B the magnitude of the magnetic field and
For Netwon's second law, the force is also equal to the proton mass times its acceleration:
So we have
from which we can find the magnitude of the field: