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irina [24]
3 years ago
8

A solid cylinder with a radius of 73.3 cm and a mass of 2.33 kg accelerates

Physics
1 answer:
vovangra [49]3 years ago
8 0

Answer:

(a) 20.4 rad/s

(b) 3.09 Nm

Explanation:

Given:

Δθ = 6.71 rev = 42.2 rad

ω₀ = 0 rad/s

t = 4.13 s

Find: ω

Δθ = ½ (ω + ω₀) t

42.2 rad = ½ (ω + 0 rad/s) (4.13 s)

ω = 20.4 rad/s

Find: α

Δθ = ω₀ t + ½ αt²

42.2 rad = (0 rad/s) (4.13 s) + ½ α (4.13 s)²

α = 4.94 rad/s²

∑τ = Iα

τ = ½ m r² α

τ = ½ (2.33 kg) (0.733 m)² (4.94 rad/s²)

τ = 3.09 Nm

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Explanation:

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         \frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}      

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Radius of curvature of the train is as follows.  

   \rho = \frac{[1 + (\frac{dy}{dx})^{2}]^{\frac{3}{2}}}{\frac{d^{2}y}{dx^{2}}}

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Now, we will calculate the normal component of the train as follows.

            a_{n} = \frac{\nu^{2}_{B}}{\rho}

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The magnitude of acceleration of train is calculated as follows.

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3 years ago
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Answer:

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What is the downward pull on an object due to gravity?
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Answer:

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