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dolphi86 [110]
3 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]3 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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<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>H</em><em>=</em><em>h</em><em>e</em><em>i</em><em>g</em><em>h</em><em>t</em>

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<em>M</em><em>g</em><em>h</em>

<em>(</em><em>M</em><em>=</em><em>6</em><em>, </em><em>g</em><em>=</em><em>10</em><em>,</em><em>h</em><em>=</em><em>?</em><em>) </em>

6×10×h

=60joules

7 0
2 years ago
What’s the equalibrium rule?
ale4655 [162]
The vector sum of forces acting on a non-accelerating object equals zero.
equation form: ΣF = 0
4 0
3 years ago
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A gold sphere of radius R=100 μm and density 19g/cm^3 falls through water. Given the viscosity of water is about 10^-3​ Pa s and
icang [17]

The terminal velocity of gold sphere is 39.2 cm/s

<h3>What is terminal velocity?</h3>

Terminal velocity is the maximum velocity attainable for an object as it falls through a fluid.

<h3>How to calculate the terminal velocity of the gold sphere?</h3>

The terminal velocity of the gold sphere is given by v = 2gr²(ρ - σ)/9η where

  • g = acceleration due to gravity = 9.8 m/s²,
  • r = radius of sphere = 100 μm = 100 × 10⁻⁶ m = 10⁻⁴ m = 10⁻² cm,
  • ρ = density of sphere = 19 g/cm³,
  • σ = density of water = 1.0 g/cm³ and
  • η = viscosity of water = 10⁻³ Pa-s

So, susbtituting the values of the variables into the equation, we have that

v = 2gr²(ρ - σ)/9η

v = 2 × 9.8m/s²× (10⁻² cm)²(19 g/cm³ - 1.0 g/cm³)/(9 × 10⁻³ Pa-s)

v = 2 × 9.8 m/s² × 10⁻⁴ cm² × (18 g/cm³)/(9 × 10⁻³ Pa-s)

v = 2 × 980 cm/s² × 10⁻⁴ cm² × 2 g/cm³/(1 × 10⁻³ Pa-s)

v = 3920 g/s² × 10⁻⁴/(1 × 10⁻³ Pa-s)

v = 392 cm/s × 10³ × 10⁻⁴

v = 392 × 10⁻¹ cm/s

v = 39.2 cm/s

So, the terminal velocity is 39.2 cm/s

Learn more about terminal velocity of sphere here:

brainly.com/question/21684177

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4 0
1 year ago
two negative charges that are both -3.0 C push each other apart with a force of 19.2 N how far apart are the charges
Hoochie [10]
The electrostatic force between two charges q1 and q2 is given by
F=k_e  \frac{q_1 q_2}{r^2}
where k_e =8.99 \cdot 10^9 N m^2 C^{-2} is the Coulomb's constant and r is the distance between the two charges.

If we use F=19.2 N and q1=q2=-3.0 C, we can find the value of r, the  distance between the two charges by re-arranging the previous formula:
r= \sqrt{k_e \frac{q_1 q_2}{F} }= \sqrt{ 8.99 \cdot 10^9 N m^2 C^{-2} \frac{(-3.0C)^2}{19.2 N} } =6.49 \cdot 10^4 m=64.9 km
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3 years ago
A 25.0-kg child is standing at the edge of a horizontal merry-go-round with a radius of 2.40 m and a moment of inertia of 356 kg
motikmotik

Answer:

\omega_{f}=1.634\ rad/s  

Explanation:

given,  

diameter of merry - go - round = 2.40 m  

moment of inertia = I = 356 kg∙m²

speed of the merry- go-round = 1.80 rad/s

mass of child = 25 kg  

initial angular momentum of the system  

L_i = I\omega_i  

L_i =356\times 1.80  

L_i =640.8\ kg.m^2/s  

final angular momentum of the system  

L_f = (I_{disk}+mR^2)\omega_{f}  

L_f = (356 + 25\times 1.2^2)\omega_{f}  

L_f= (392)\omega_{f}  

from conservation of angular momentum  

L_i = L_f  

640.8= (392)\omega_{f}  

\omega_{f}=1.634\ rad/s  

8 0
3 years ago
Read 2 more answers
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