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zlopas [31]
2 years ago
5

Nathan is walking to the store and sees a snake slithering across the sidewalk. He jumps over it with an initial vertical veloci

ty of 10 ft/s. How long does it take for Nathan to land back on the ground
Physics
1 answer:
Travka [436]2 years ago
8 0

Answer:

0.62\:\mathrm{s}

Explanation:

Since the universal SI unit for velocity is meters/second, let's convert ft/s to m/s:

10\:\mathrm{ft/s}=3.048\:\mathrm{m/s}

We can use the following kinematics equation to solve this question:

v_f=v_i+at

What we know:

  • The initial velocity, v_i, is 3.048\:\mathrm{m/s}
  • (physics concept) The final velocity must be equal in magnitude but opposite in direction to the initial velocity (v_f=-3.048\:\mathrm{m/s})
  • Acceleration, a, is acceleration due to gravity at about 9.8\:\mathrm{m/s}

Solving for t:

-3.048=3.048+(-9.8t),\\-6.096=-9.8t,\\t=\frac{-6.096}{-9.8}\approx \boxed{0.62\:\mathrm{s}}

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A merry-go-round with a a radius of R = 1.63 m and moment of inertia I = 196 kg-m2 is spinning with an initial angular speed of
kondor19780726 [428]

Answer:

1) L = 299.88 kg-m²/s

2) L = 613.2 kg-m²/s

3) L = 499.758 kg-m²/s

4) ω₁ = 0.769 rad/s

5) Fc = 70.3686 N

6) v = 1.2535 m/s

7) ω₀ = 1.53 rad/s

Explanation:

Given

R = 1.63 m

I₀ = 196 kg-m²

ω₀ = 1.53 rad/s

m = 73 kg

v = 4.2 m/s

1) What is the magnitude of the initial angular momentum of the merry-go-round?

We use the equation

L = I₀*ω₀ = 196 kg-m²*1.53 rad/s = 299.88 kg-m²/s

2) What is the magnitude of the angular momentum of the person 2 meters before she jumps on the merry-go-round?

We use the equation

L = m*v*Rp = 73 kg*4.2 m/s*2.00 m = 613.2 kg-m²/s

3) What is the magnitude of the angular momentum of the person just before she jumps on to the merry-go-round?

We use the equation

L = m*v*R = 73 kg*4.2 m/s*1.63 m = 499.758 kg-m²/s

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L in = L fin

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where

I₁ = I₀ + m*R²

⇒  I₀*ω₀ = (I₀ + m*R²)*ω₁

Now, we can get ω₁

⇒  ω₁ = I₀*ω₀ / (I₀ + m*R²)

⇒  ω₁ = 196 kg-m²*1.53 rad/s / (196 kg-m² + 73 kg*(1.63 m)²)

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5) Once the merry-go-round travels at this new angular speed, with what force does the person need to hold on?

We have to get the centripetal force as follows

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⇒  Fc = 73 kg*(0.769 rad/s)²*1.63 m = 70.3686 N

6) Once the person gets half way around, they decide to simply let go of the merry-go-round to exit the ride.

What is the linear velocity of the person right as they leave the merry-go-round?

we can use the equation

v = ω₁*R = 0.769 rad/s*1.63 m = 1.2535 m/s

7) What is the angular speed of the merry-go-round after the person lets go?

ω₀ = 1.53 rad/s

It comes back to its initial angular speed

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