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Elenna [48]
3 years ago
14

Have you ever visited an amusement park and taken a ride on a parachute drop ride? These types of rides take the passengers to a

great height, and then drop them in free fall. Before they hit the ground, the ride is slowed using a Lenz’s law mechanism thus avoiding certain death. For this discussion, first locate a photo of one of these rides (either one you’ve personally experienced or one you might like to try someday), and in your initial post, upload the photo and respond to the following:
a. Explain how Lenz’s law applies to this situation.
b. Why is the Lenz’s law mechanism ideal for such a use?
c. What other mechanisms can be used to slow the descent? Compare and contrast these options with the Lenz’s law mechanism.

Finally, be sure to respond to at least two of your peers’ discussion posts.

Physics
1 answer:
Triss [41]3 years ago
3 0

Answer & Explanation:

a)

Lenz's law states that the direction of induced electric current is always such that, it opposes the change in magnetic flux.

In a drop ride, the hub on which we sit and are hung to is an electromagnet and there are many such magnets mounted on the columns of the support. what happens is these electromagnets (in support) generate a repulsive magnetic field with respect to the field generated by the hub solenoids. this results in lift generation till the top of ride. reaching the top, the bar solenoids are at their maximum repulsive force. Then the solenoids in column are set current less means electric supply is cut off. this makes you fall under the effect of gravity. by the time you are half way down, column  solenoids are turned on again. As the hub solenoid approaches every single electromagnet in supporting columns. Due to change in magnetic field (with respect to lenz's law) an opposing current induces further providing resistance to the fall, this continues until the ride comes to rest completely. This is how it works.  

c) In addition, highly compressive springs, dampers, viscous dampers, etc. could be used in its place.

but the above listed cannot provide a differential braking,

have a limited lifecycle,

will provide resistance during lift also,

require higher maintenance

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A horizontal spring with spring constant 130 N/m is compressed 17 cm and used to launch a 2.8 kg box across a frictionless, hori
olasank [31]

Explanation:

The given data is as follows.

        k = 130 N/m,       \Delta x = 17 cm = 0.17 m   (as 1 m = 100 cm)

     mass (m) = 2.8 kg

When the spring is compressed then energy stored in it is as follows.

             Energy = \frac{1}{2}kx^{2}

Now, spring energy gets converted into kinetic energy when the box is launched.

So,    \frac{1}{2}kx^{2} = \frac{1}{2}mv^{2}

   \frac{1}{2} \times 130 \times (0.17)^{2} = \frac{1}{2} \times 2.8 \times v^{2}

          v^{2} = \frac{3.757}{2.8}

                     = 1.34

                v = 1.15 m/sec

Now,

           Frictional force = \mu \times mg

                                    = 0.15 \times 2.8 \times 9.8

                                    = 4.116 N

Also,  Kinetic energy = work done by friction

           \frac{1}{2}mv^{2} = F_{f} \times d

           \frac{1}{2} \times 2.8 \times (1.15)^{2} = 4.116 \times d

             1.8515 = 4.116 \times d

                 d = 0.449 m

Thus, we can conclude that the box slides 0.449 m across the rough surface before stopping.

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A force of 5 N is applied to the end of a lever that has a length
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Answer:

The answer is 10Nm

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8 0
3 years ago
1) You slam on the brakes of your car in a panic, and skid a certain distance on a straight level road. If you had been travelin
aleksandr82 [10.1K]

Answer:

d = 4 d₀o

Explanation:

We can solve this exercise using the relationship between work and the variation of kinetic energy

         W = ΔK

In that case as the car stops v_f = 0

the work is

          W = -fr d

we substitute

          - fr d₀ = 0 - ½ m v₀²

           d₀ = ½ m v₀² / fr

now they indicate that the vehicle is coming at twice the speed

          v = 2 v₀

using the same expressions we find

           d = ½ m (2v₀)² / fr

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3 0
3 years ago
A 70.0 kg astronaut is training for accelerations that he will experience upon reentry. He is placed in a centrifuge (r = 15.0 m
Levart [38]

Answer:

1.3823 rad/s

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28.66129935 m/s²

a=2.92164g

2006.29095 N radially outward

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m = Mass of person = 70 kg

g = Acceleration due to gravity = 9.81 m/s²

Angular velocity is given by

\omega=13.2\times \dfrac{2\pi}{60}\\\Rightarrow \omega=1.3823\ rad/s

Angular velocity is 1.3823 rad/s

Linear velocity is given by

v=r\omega\\\Rightarrow v=15\times 1.3823\\\Rightarrow v=20.7345\ m/s

The linear velocity is 20.7345 m/s

Centripetal acceleration is given by

a_c=r\omega^2\\\Rightarrow a_c=15\times 1.3823^2\\\Rightarrow a_c=28.66129935\ m/s^2

The centripetal acceleration is 28.66129935 m/s²

Acceleration in terms of g

\dfrac{a}{g}=\dfrac{28.66129935}{9.81}\\\Rightarrow a=2.92164g

a=2.92164g

Centripetal force is given by

F_c=ma_c\\\Rightarrow F_c=70\times 28.66129935\\\Rightarrow F_c=2006.29095\ N

The centripetal force is 2006.29095 N radially outward

The torque will be experienced when the centrifuge is speeding up of slowing down i.e., when it is accelerating and decelerating.

3 0
3 years ago
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