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Serggg [28]
1 year 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]1 year 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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Which declares two related integer vectors named personname and personage each with 50 elements?
Serjik [45]

Both vectors should be declared to have the same number of elements.

vector<int> personName(50); vector<int> personAge(50)

Vector is a quantity that has both magnitude and direction. It is typically represented by an arrow whose direction is the same as that of the quantity and whose length is proportional to the quantity's magnitude. Although a vector has magnitude and direction, it does not have position like a point.

Learn more about Vectors here:

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4 0
1 year ago
A primary succession does not include any invasive species <br><br> True or false
yan [13]
To be referenced, it would be true
7 0
3 years ago
 why are waves faster in liquids and gases than in air
Alika [10]

Answer:

Sound travels faster in liquids than in gases because molecules are packed more closely together. This means that when the water molecules begin to vibrate, they quickly begin to collide with each other forming a rapidly moving compression wave. Sound travels over four times faster than in air

Explanation:

4 0
2 years ago
Read 2 more answers
A uniform meterstick of mass 0.20 kg is pivoted at the 40 cm mark. where should one hang a mass of 0.50 kg to balance the stick?
Tcecarenko [31]
The weight of the meterstick is:
W=mg=0.20 kg \cdot 9.81 m/s^2 = 1.97 N
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance 
d_1 = 0.50 m - 0.40 m=0.10 m
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
M_w = W d_1 = (1.97 N)(0.10 m)=0.20 Nm

To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
(mg) d_2 = 0.20 Nm
from which we find the value of d2:
d_2 =  \frac{0.20 Nm}{mg}= \frac{0.20 Nm}{(0.5 kg)(9.81 m/s^2)}=0.04 m

So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
4 0
3 years ago
In an old-fashioned amusement park ride, passengers stand inside a 3.0-m-tall, 5.0-m-diameter hollow steel cylinder with their b
lesya692 [45]

Answer:2.55 rad/s

Explanation:

Given

Diameter of ride=5 m

radius(r)=2.5 m

Static friction coefficient range=0.60-1

Here Frictional force will balance weight

And limiting  frictional force is provided by Centripetal force

f=\mu N=\mu m\omega ^2\cdot r

weight of object=mg

Equating two

f=mg

\mu m\omega ^2\cdot r=mg

\omega ^2=\frac{g}{\mu r}

\omega =\sqrt{\frac{g}{\mu r}}

\omega =2.55 rad/sec

8 0
3 years ago
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