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elena55 [62]
3 years ago
10

The elevators in the Landmark Tower in Yokohama, Japan are among the fastest in the world. After starting from rest, they reach

a maximum speed of 120 meters per minute. It takes the elevator 4.0 seconds to reach maximum speed. What is its acceleration in m/s/s (meters per second per second)?
Physics
1 answer:
MariettaO [177]3 years ago
3 0

Answer:

The acceleration is 30 m/s²

Explanation:

Given that,

Maximum speed = 120 m/s

Time = 4.0 sec

We need to calculate the acceleration

Using equation of motion

v=u+at

Where, v = speed

u = initial speed

a = acceleration

t = time

Put the value into the formula

120=0+a\times4.0

a=\dfrac{120}{4.0}

a=30\ m/s^2

Hence, The acceleration is 30 m/s²

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The equation for the speed of a satellite in a circular orbit around the earth depends on mass. Which mass?
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<h3><u>Question: </u></h3>

The equation for the speed of a satellite in a circular orbit around the Earth depends on mass. Which mass?

a. The mass of the sun

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The equation for the speed of a satellite orbiting in a circular path around the earth depends upon the mass of Earth.

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Two polarizing sheets have their transmission axes crossed so that no light is transmitted. A third sheet is inserted so that it
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Answer:

a)    I= I₀ (cos²θ - cos⁴θ)    b) 75.5º

Explanation:

a) For this exercise we must use Malus's law

         I = I₀ cos² θ

where tea is the angle between the two polarizers.

We apply this expression to our case

* Polarizer 1 suppose that it is vertical and polarizer 2 (intermediate) is at an angle θ with respect to the vertical

         I₁ = I₀ cos² θ

* We analyze for the polarity 2 and the last polarizer 3 which indicate that it must be at 90º from the first one, therefore it must be horizontal.

The angle of polarizers 2 and 3 is θ' measured from the horizontal, if we measure with respect to the vertical

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fiate that in the exercise we must take a reference system and measure everything with respect to this system.

          I = I₁ cos² θ'

       

we substitute

         I = (I₀ cos² tea) cos² (θ - 90)

        cos (θ -90) = cos θ cos 90 + sin θ sin 90 = sin θ

         I = Io cos² θ sin² θ

        1= cos²θ+ sin²θ

       sin²θ = 1 - cos²θ

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b) to find when the intensity is maximum,

we can use that we have an extreme point when the drift is zero

          \frac{dI}{d \theta} = 0

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The zeros of this function are in

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Consequently, the angle that allows the maximum intensity to pass is 75.5º

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