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weqwewe [10]
3 years ago
8

A moving walkway at an airport has a speed v1 and a length l. a woman stands on the walkway as it moves from one end to the othe

r, while a man in a hurry to reach his flight walks on the walkway with a speed of v2 relative to the moving walkway. (use any variable or symbol stated above as necessary.) (a) how long does it take the woman to travel the distance l? v1​ l​ (b) how long does it take the man to travel this distance? v2​ l​
Physics
1 answer:
sleet_krkn [62]3 years ago
4 0
L = distance traveled.

The woman:
v₁ =  the traveling speed
Time of travel, t₁ = l/v₁

The man:
v₁ + v₃ = the traveling speed.
Time of travel = l/(v₁ + v₂)

Answer:
The woman's time is l/v₁.
The man's time is l/(v₁ + v₂).
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A thin rod of length 0.75 m and mass 0.42 kg is suspended
MrRissso [65]

Answer:

a)  K = 0.63 J, b)  h = 0.153 m

Explanation:

a) In this exercise we have a physical pendulum since the rod is a material object, the angular velocity is

         w² = \frac{m g d}{I}

where d is the distance from the pivot point to the center of mass and I is the moment of inertia.

The rod is a homogeneous body so its center of mass is at the geometric center of the rod.

              d = L / 2

the moment of inertia of the rod is the moment of a rod supported at one end

              I = ⅓ m L²

we substitute

            w = \sqrt{\frac{mgL}{2}  \ \frac{1}{\frac{1}{3} mL^2} }

            w = \sqrt{\frac{3}{2}  \ \frac{g}{L} }

            w = \sqrt{ \frac{3}{2} \ \frac{9.8}{0.75}  }

            w = 4.427 rad / s

an oscillatory system is described by the expression

              θ = θ₀ cos (wt + Φ)

the angular velocity is

             w = dθ /dt

             w = - θ₀ w sin (wt + Ф)

In this exercise, the kinetic energy is requested in the lowest position, in this position the energy is maximum. For this expression to be maximum, the sine function must be equal to ±1

In the exercise it is indicated that at the lowest point the angular velocity is

           w = 4.0 rad / s

the kinetic energy is

           K = ½ I w²

           K = ½ (⅓ m L²) w²

           K = 1/6 m L² w²

           K = 1/6 0.42 0.75² 4.0²

           K = 0.63 J

b) for this part let's use conservation of energy

starting point. Lowest point

             Em₀ = K = ½ I w²

final point. Highest point

             Em_f = U = m g h

energy is conserved

             Em₀ = Em_f

             ½ I w² = m g h

             ½ (⅓ m L²) w² = m g h

             h = 1/6 L² w² / g

             h = 1/6 0.75² 4.0² / 9.8

             h = 0.153 m

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2 years ago
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Kipish [7]

Answer:

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2 years ago
Among the alkali earth metals, the tendency to react with other substances
padilas [110]
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jarptica [38.1K]
The difference in electric potential energy between the two points is
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But for energy conservation, the difference in electric potential energy \Delta U between the two points is equal to the work done to move the charge between A and B:
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