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Serggg [28]
2 years ago
6

When you travel in an elevator (which moves linearly in space), the ___________ detect when the elevator is accelerating or dece

lerating.
Physics
1 answer:
Shtirlitz [24]2 years ago
4 0

We can confirm that when you travel in an elevator, the utricle and saccule detect when the elevator is moving.

<h3>What are the utricle and saccule?</h3>

These are small organs located in the inner ear. They serve a very important function in our everyday life. These otolith organs are responsible for the detection of movement and provide us with a sense of spatial orientation. They accomplish this by stimulating very small, hair-like cells with a specialized fluid.

Therefore, we can confirm that when you travel in an elevator, the utricle and saccule detect when the elevator is accelerating or decelerating.

To learn more about the inner ear visit:

brainly.com/question/3535321?referrer=searchResults

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This question can be solved by using the equations of motion.

a) The initial speed of the arrow is was "9.81 m/s".

b) It took the arrow "1.13 s" to reach a height of 17.5 m.

a)

We will use the second equation of motion to find out the initial speed of the arrow.

h= v_it + \frac{1}{2}gt^2\\

where,

vi = initial speed = ?

h = height = 35 m

t = time interval = 2 s

g = acceleration due to gravity = 9.81 m/s²

Therefore,

35\ m = (v_i)(2\ s)+\frac{1}{2}(9.81\ m/s^2)(2\ s)^2\\\\v_i(2\ s)=19.62\ m\\\\v_i = \frac{19.62\ m}{2\ s}

<u>vi =  9.81 m/s</u>

b)

To find the time taken by the arrow to reach 17.5 m, we will use the second equation of motion again.

h= v_it + \frac{1}{2}gt^2\\

where,

g = acceleration due to gravity = 9.81 m/s²

h = height = 17.5 m

vi = initial speed = 9.81 m/s

t = time = ?

Therefore,

17.5 = (9.81)t+\frac{1}{2}(9.81)t^2\\4.905t^2+9.81t-17.5=0

solving this quadratic equation using the quadratic formula, we get:

t = -3.13 s (OR) t = 1.13 s

Since time can not have a negative value.

Therefore,

<u>t = 1.13 s</u>

Learn more about equations of motion here:

brainly.com/question/20594939?referrer=searchResults

The attached picture shows the equations of motion in the horizontal and vertical directions.

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