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kow [346]
3 years ago
10

Two elevators begin descending from the same height. Elevator A has descended 4 feet after one second, 9 feet after two seconds,

14 feet after three seconds, and so on. Elevator B has descended 3.5 feet after one second, 6.5 feet after two seconds, 9.5 feet after three seconds, and so on. How many feet would each elevator descend in 10 seconds?
Physics
2 answers:
solniwko [45]3 years ago
8 0

Answer:

a) y = 400 ft  b) y = 350 ft

Explanation:

We can solve that problem with kinematic relationships

    y = v₀ t + ½ a t²

Let's apply this equation to the elevator 1

     y₁ = v₀ t₁ + ½ to t₁²

time t= 1 s

    y₁ = v₀ 1 + ½ a 1²

    y₁ = v₀ + ½ a

For t = 2s

    y₂ = v₀ 2 + ½ a 2²

    y₂ = 2 v₀ + 2 a

Let's write the equations and solve the system

    4 = v₀ + ½ a

    9 = 2 v₀ + 2 a

Let's multiply the first by -2

    -8 = -2v₀ -a

     9 = 2v₀ + 2 a

Let's add

      1 = a

We replace in the first

      4 = v₀ + ½ 1

      v₀ = 4- 1/2

      v₀ = 3.5

The equation for the first elevator is

      y = 3.5 t + ½  t²

For t = 10 s

      y = 3.5 10 + ½ 10²  

      y = 400 ft

We repeat the process for the second elevator

t = 1s

     y₁ = v₀ 1 + ½ a 1²

     3.5 = v₀ + ½ a

t = 2 s

     y₂ = v₀ 2 + ½ a 2²

     6.5 = 2 v₀ +2 a

multiply by -2

    -7 = -2 v₀ - a

     6.5 = 2 v₀ + 2 a

Let's add

    -0.5 = a

I replace in the first equation

    3.5 = v₀ + ½ (-0.5)

    v₀ = 3.5 + 0.25

    v₀ = 3.75

The equation is

    y = 3.75 t -0.25 t²

For t = 10s

    y = 3.75 10 - 0.25 10²

   y = 350 ft

alexandr1967 [171]3 years ago
5 0

Answer:

 

A: 49 ft; B: 30.5 ft

Explanation:

correct answer

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If the separation distance is doubled, then the electric field decreases by a factor of 4.

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We know that the electric field strength is known to depend on the magnitude of the charge and the distance of separation. We know that the electric field refers to the region in which the influence of a charge is felt. Recall that a charge is a specie that is positively or negatively charged. The charge on a specie must always be shown by its sign.

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Answer:

a)    x = v₀² sin 2θ / g

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c)    x = 16.7 m

Explanation:

This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity

        sin θ = v_{oy} / vo

        cos θ = v₀ₓ / vo

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         v₀ₓ = 13.5 cos 32 = 11.45 m / s

a) In the x axis there is no acceleration so the velocity is constant

         v₀ₓ = x / t

          x = v₀ₓ t

the time the ball is in the air is twice the time to reach the maximum height, where the vertical speed is zero

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          t = v_{o} sin θ / g

         

we substitute

       x = v₀ cos θ (2 v_{o} sin θ / g)

       x = v₀² /g      2 cos θ sin θ

       x = v₀² sin 2θ / g

at the point where the receiver receives the ball is at the same height, so this coincides with the range of the projectile launch,

b) The acceleration to which the ball is subjected is equal in the rise and fall, therefore it takes the same time for both parties, let's find the rise time

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