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Sergeu [11.5K]
3 years ago
5

Why did you spin counterclockwise rather than clockwise

Physics
2 answers:
Ilya [14]3 years ago
7 0
Because I am left-handed. When combined with the Coriolis force in a leap year, it causes me to do everything backwards.
rodikova [14]3 years ago
6 0
<span>Because of the orbit of the earth and the sun and the moon. </span>
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What is the quantity of work done when a crane lifts a 100-n block from 2 m above the ground to 6 m above the ground?.
mylen [45]

Answer: by definition work = Force * distanceForce = 100Ndistance = 6 - 2 = 4work = 100 * 4 = 400 J or CStill stuck? Get 1-on-1 help from an expert tutor now.

Explanation:

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2 years ago
A rock at rest falls off a tall cliff and hits the valley below after 3.5s. What is the rocks velocity as it hits the ground
pogonyaev
T = 3.5 secs

Velocity (v) = g * t = 10 m/s^2 * 3.5 sec = 35 m/s
7 0
3 years ago
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A box has a 20 N force applied to it to move it 5 m. What is the work done on the box? 4 J 4 N 25 J 100 J
Sunny_sXe [5.5K]
100 J

Explanation:
multiply the force by the distance
20 N x 5 meters = 100 J
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3 years ago
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HELPPpPpPpPpPp!!!! ASAP !!!
DENIUS [597]

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50 N

Explanation:

8 0
2 years ago
Two polarizing sheets have their transmission axes crossed so that no light is transmitted. A third sheet is inserted so that it
jek_recluse [69]

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

              θ₂ = 90- θ’ = θ

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

        I= I₀ (cos²θ - cos⁴θ)

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

          \frac{dI}{d \theta}= Io (2 cos θ - 4 cos³θ) = 0

whereby

            cos θ - 2 cos³ θ = 0

            cos θ ( 1 - 2 cos² θ) = 0  

The zeros of this function are in

           θ = 90º

           1-2cos²θ =0       cos θ = 0.25  θ =  75.5º

Let's analyze this two results for the angle of 90º the intnesidd is zero with respect to the first polarizer, so it is not an acceptable solution.

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

5 0
3 years ago
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