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kipiarov [429]
3 years ago
7

A 174 pound Jimmy Cheek is riding on a 54 ft diameter Ferris Wheel. The normal force on Jimmy Cheek is 146 pounds when Jimmy is

at the top of the wheel. Determine the angular velocity of the Ferris Wheel.
Physics
1 answer:
Veronika [31]3 years ago
6 0

To solve this problem we will apply the concepts related to the balance of Forces, the centripetal Force and Newton's second law.

I will also attach a free body diagram that allows a better understanding of the problem.

For there to be a balance between weight and normal strength, these two must be equivalent to the centripetal Force, therefore

F_c = W-N

m\omega^2r = W-N

Here,

m = Net mass

\omega= Angular velocity

r = Radius

W = Weight

N = Normal Force

m\omega^2r = 174-146

The net mass is equivalent to

F = mg \rightarrow m = \frac{F}{g}

Then,

m = \frac{174lb}{32.17ft/s^2}

Replacing we have then,

(\frac{174lb}{32.17ft/s^2})\omega^2 (54ft) =174lb-146lb

Solving to find the angular velocity we have,

\omega = 0.309rad/s

Therefore the angular velocity is 0.309rad/s

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Answer:

0.013N

Explanation:

F=\frac{mv^{2} }{r}

m=0.04, v=4.4m/s, r=60

F=0.013 N

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Which of the following statements is false?
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A & B

Observe the path of the light ray as it passes through the lenses as shown in the attached images. Concave lenses diverge light rays while the convex lens converges the light rays.

Explanation:

Real images are formed where the rays converge, a property of images by convex lenses. Convex lenses can be used to magnify objects. If the image occurs before the focal point of the lens then the image will be upright but smaller. The images inverts and gets bigger past the focal point.

Virtual images are property of concave lenses. These images appear closer but smaller than the real object.

Learn More:

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A projectile is fired at time t= 0.0 s from point 0 at the edge of a cliff, with initial velocity components of Vox = 30 m/s and
BARSIC [14]

Answer:

At t = 15.0 s the magnitude of the velocity is 58.31 m/s

Explanation:

The given parameters are;

V₀ₓ = 30 m/s

V_{0y} = 100 m/s

The time of flight of the projectile = 25 s

For projectile motion;

Vₓ = V₀ × cos(θ₀)

The magnitude of the velocity V = √(V₀ₓ² + V_{0y}²)

We have the magnitude of the initial velocity = √(30² + 100²) = 10·√109 m/s

cos(θ₀) = V₀ₓ/V₀ = 30/(10·√109) = 3/√109

θ₀ = cos⁻¹(3/√109) = 73.3°

The components of the velocity after time t is given by the relations;

Vₓ = V₀ × cos(θ₀) = 30 m/s

V_y =  V₀ × sin(θ₀) - g×t

When V_y = 0, we have;

0 =  V₀ × sin(θ₀) - g×t

g×t  =  V₀ × sin(θ₀)  = 10·√109×0.958 = 100 m/s

t = 100/g = 100/10 = 10 s

The time to reach maximum height = 10 s

At 15.0 seconds, we have;

V_y =  V₀ × sin(θ₀) - g×t = 10·√109×0.958  - 10×15 = -50 m/s

Therefore, the projectile is returning at 50 m/s

The magnitude of the velocity =√(30² + 50²) = 10·√34 m/s = 58.31 m/s.

5 0
3 years ago
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