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padilas [110]
3 years ago
7

What part of the diagram is labeled by the number 3?

Physics
1 answer:
lutik1710 [3]3 years ago
5 0

Answer:

The part of the diagram thta is labeled number 3 is D. bronchi.

Explanation:

Bronchi are the airways that lead from the trachea to the lungs and they branch off into progressively smaller parts.

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Scientists use a substance called Iodine-131 to treat cancer. As Iodine-131 undergoes radioactive decay and emits beta particles
olganol [36]

Answer:

I think it's strong I'm not to sure I'm sorry if it's wrong

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an audio CD has a diameter of 120 mm and spins at up to 540 rpm. When a CD is spinning at its maximum rate, how much time is req
Andru [333]

Answer:

 t = 0.1111 s

Explanation:

Let's reduce the magnitudes to the SI system

    d = 120 mm (1m / 1000 mm)

    d= 0.120 m

    w = 540 rpm (2pi rad / 1 rev) (1 min / 60s)  

     w= 56.55 rad / s

When at maximum speed we can use angular kinematic relationships to find the time for a sperm revolution with zero angular acceleration

     W = θ / t

     t = θ / w

     t = 2π / 56.55

     t = 0.1111 s

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2 years ago
Calculate the horizontal force that must be applied to a 1,300 kg vehicle to give it an acceleration of 2.6 m/s² on a level road
ANEK [815]
The answer is 3,380 N
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2 years ago
People living near wind farms sometimes object to them because of visual pollution and what other
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Wind turbines are a source of clean renewable energy, but some people who live nearby describe the shadow flicker, the audible sounds and the subaudible sound pressure levels as "annoying." They claim this nuisance negatively impacts their quality of life

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2 years ago
A rock is suspended by a light string. When the rock is in air, the tension in the string is 51.9 N . When the rock is totally i
Luden [163]

Answer:

\rho _{liquid}=1995.07kg/m^{3}

Explanation:

When the rock is immersed in unknown liquid the forces that act on it are shown as under

1) Tension T by the string

2) Weight W of the rock

3) Force of buoyancy due to displaced liquid B

For equilibrium we have T_{3}+B = W_{rock}

T_{3}+\rho _{Liquid}V_{rock}g=W_{rock}.....(\alpha)

When the rock is suspended in air for equilibrium we have

T_{1}=W_{rock}....(\beta)

When the rock is suspended in water for equilibrium we have

T_{2} + \rho _{water}V_{rock}g=W_{rock}.....(\gamma)

Using the given values of tension and solving α,β,γ simultaneously for \rho _{Liquid} we get

W_{rock}=51.9N\\31.6+1000\times V_{rock}\times g=51.9N\\\\11.4+\rho _{liquid}V_{rock}g=51.9N\\\\

Solving for density of liquid we get

\rho _{liquid}=\frac{51.9-11.4}{51.9-31.6}\times 1000

\rho _{liquid}=1995.07kg/m^{3}

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2 years ago
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