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dolphi86 [110]
2 years ago
6

The New York Wheel is the world's largest Ferris wheel. It's 183 meters in diameter and rotates once every 37.3 min.

Physics
1 answer:
bagirrra123 [75]2 years ago
7 0
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.

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Explanation;

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6 0
3 years ago
Read 2 more answers
Efficiency is given by A. actual output divided by effective capacity. B. design capacity divided by utilization. C. effective c
jok3333 [9.3K]

Answer:

A

Explanation:

Actual output divided by the effective capacity. It is the ratio of output to effectiveness

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3 years ago
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Ajoba and Prav drive to work. Ajoba drives 45 miles in 2.5 hours. Prav drives 74 km in 1 hour 15 min. Work out the difference be
Serggg [28]

Explanation:

We have,

Ajoba and Prav drive to work. Ajoba drives 45 miles in 2.5 hours. Prav drives 74 km in 1 hour 15 min.

1 mile = 1.6 km

45 miles = 72.42 km

74 miles = 119.0 km

1 hour 15 min means 1.25 hours

Average speed of Ajoba is :

v_1=\dfrac{72.42 }{2.5}=28.96\ km/h

Average speed of Prav,

v_2=\dfrac{119}{1.25}=95.2\ km/h

Difference in average speed of Ajoba and Prav is :

v=v_2-v_1\\\\v=v_2-v_1\\\\v=95.2-28.96\\\\v=66.24\ km/h

So, the difference in average speed of Ajoba and Prav is 66.24 km/h.

7 0
3 years ago
Determine the magnitude of the resultant force acting on a 1.5 −kg particle at the instant t=2 s, if the particle is moving alon
Phoenix [80]

Answer:

F = 63N

Explanation:

M= 1.5kg , t= 2s, r = (2t + 10)m and

Θ = (1.5t² - 6t).

magnitude of the resultant force acting on 1.5kg = ?

Force acting on the mass =

∑Fr =MAr

Fr = m(∇r² - rθ²) ..........equation (i)

∑Fθ = MAθ = M(d²θ/dr + 2dθ/dr) ......... equation (ii)

The horizontal path is defined as

r = (2t + 10)

dr/dt = 2, d²r/dt² = 0

Angle Θ is defined by

θ = (1.5t² - 6t)

dθ/dt = 3t, d²θ/dt² = 3

at t = 2

r = (2t + 10) = (2*(2) +10) = 14

but dr/dt = 2m/s and d²r/dt² = 0m/s

θ = (1.5(2)² - 6(2) ) = -6rads

dθ/dt =3(2) - 6 = 0rads

d²θ/dt = 3rad/s²

substituting equation i into equation ii,

Fr = M(d²r/dt² + rdθ/dt) = 1.5 (0-0)

∑F = m[rd²θ/dt² + 2dr/dt * dθ/dt]

∑F = 1.5(14*3+0) = 63N

F = √(Fr² +FΘ²) = √(0² + 63²) = 63N

7 0
3 years ago
3 In a television tube, an electron starting from rest experiences a force of 4.0 × 10−15 N over a distance of 50 cm. The final
MAXImum [283]

Answer:

The final speed of the electron = 2.095×10⁸ m/s

Explanation:

From newton's fundamental equation of dynamics,

F = ma ........................Equation 1

Where F = force, m = mass of the electron, a = acceleration of the electron.

making a the subject of the equation,

a = F/m.................... Equation 2

Given: F = 4.0×10⁻¹⁵ N,

Constant: m =  9.109×10⁻³¹ kg.

Substituting into equation 2

a = 4.0×10⁻¹⁵/9.109×10⁻³¹

a = 4.39×10¹⁶ m/s².

Using newton's equation of motion,

v² = u²+2as .......................... Equation 3

Where v = final velocity of the electron, u = initial velocity of the electron, a = acceleration of the electron, s = distance covered by the electron.

Given: u = 0 m/s(at rest), s = 50 cm = 0.5 m, a = 4.39×10¹⁶ m/s²

Substituting into equation 3

v² = 0² + 2(0.5)(4.39×10¹⁶)

v = √(4.39×10¹⁶)

v = 2.095×10⁸ m/s

Thus the final speed of the electron = 2.095×10⁸ m/s

7 0
2 years ago
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