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2 years ago

What is the tube that carries air inside a lung?

Physics

2 answers:

VladimirAG [237]2 years ago

6 0

the tube that carries air inside the lungs is called bronchus

Elis [28]2 years ago

4 0

The trachea is a tube that carries air inside the lungs.

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1) A car is moving to the right with the constant speed of 1.2 m/s. If the car starts

Marina CMI [18]

(100, 108)
Due to
1.2x90=108

100, 108

7 0

2 years ago

Normal atmospheric pressure is 1.013 105 Pa. The approach of a storm causes the height of a mercury barometer to drop by 27.1 mm

Burka [1]

Answer:

The atmospheric pressure is $0.97622\times10^{5}\ Pa$ .

Explanation:

Given that,

Atmospheric pressure $P_{atm}= 1.013\times10^{5}\ Pa$

drop height h'= 27.1 mm

Density of mercury $\rho= 13.59 g/cm^3$

We need to calculate the height

Using formula of pressure

$p = \rho g h$

Put the value into the formula

$1.013\times10^{5}=13.59\times10^{3}\times9.8\times h$

$h =\dfrac{1.013\times10^{5}}{13.59\times10^{3}\times9.8}$

$h=0.76\ m$

We need to calculate the new height

$h''=h - h'$

$h''=0.76-27.1\times10^{-3}$

$h''=0.76-0.027$

$h''=0.733\ m$

We need to calculate the atmospheric pressure

Using formula of atmospheric pressure

$P=\rho g h$

Put the value into the formula

$P= 13.59\times10^{3}\times9.8\times0.733$

$P=0.97622\times10^{5}\ Pa$

Hence, The atmospheric pressure is $0.97622\times10^{5}\ Pa$ .

7 0

2 years ago

Investigators are most likely to use the case history method when they study

user100 [1]

Investigators are most likely to use the case history method when they study <span>a rare behavior or an unusual person.
They do this to obtain some sort of basis that they could use as a pointer to make their decision regarding the similar case (after figuring out the difference in situation between each period)</span>

4 0

2 years ago

A simple harmonic oscillator consists of a block (m = 0.50 kg) attached to a spring (k = 128 N/m). The block is pulled a certain

Zinaida [17]

Answer:

0.5 m

14.00595

8 m/s, 0.0625 s

5.71314 m/s

Explanation:

k = Spring constant = 128 N/m

A = Amplitude

E = Energy in spring = 16 J

Energy in spring is given by

$E=\dfrac{1}{2}kA^2\\\Rightarrow A=\sqrt{\dfrac{2E}{k}}\\\Rightarrow A=\sqrt{\dfrac{2\times 16}{128}}\\\Rightarrow A=0.5\ m$

The amplitude is 0.5 m

Time period is given by

$T=2\pi\sqrt{\dfrac{m}{k}}\\\Rightarrow T=2\pi\sqrt{\dfrac{0.5}{128}}\\\Rightarrow T=0.39269\ s$

Number of oscillations is given by

$N=\dfrac{5.5}{0.39269}\\\Rightarrow N=14.00595$

The number of oscillations is 14.00595

For maximum speed

$\dfrac{1}{2}mv^2=16\\\Rightarrow v=\sqrt{\dfrac{16\times 2}{0.5}}\\\Rightarrow v=8\ m/s$

The maximum speed is 8 m/s

For a distance of 0.5 m which is the amplitude

$Time=\dfrac{Distance}{Speed}\\\Rightarrow Time=\dfrac{0.5}{8}\\\Rightarrow Time=0.0625\ s$

The time taken would be 0.0625 s

The maximum kinetic energy is equal to the mechanical energy

$\dfrac{1}{2}mv^2+\dfrac{1}{2}kx^2=16$

At x = 0.35 m

$v=\sqrt{\dfrac{16-\dfrac{1}{2}kx^2}{\dfrac{1}{2}m}}\\\Rightarrow v=\sqrt{\dfrac{16-\dfrac{1}{2}128\times 0.35^2}{\dfrac{1}{2}0.5}}\\\Rightarrow v=5.71314\ m/s$

The speed of the block is 5.71314 m/s

4 0

2 years ago

Read 2 more answers

A pendulum with a 1.0-kg weight is set in motion from a position 0.04 m above the lowest point on the path of the weight. What i

Solnce55 [7]

Answer:

The kinetic energy of the pendulum at the lowest point is 0.393 joules.

Explanation:

Under the assumption that effects from non-conservative forces can be neglected, the maximum kinetic energy of the pendulum (lowest point) ( $K_{2}$ ), measured in joules, is equivalent to the maximum gravitational potential energy (highest point) ( $U_{g,1}$ ), measured in joules, by th Principle of Energy Conservation:

$U_{g,1} = K_{2}$ (1)

By the definition of potential gravitational energy and under the assumption that the height of the lowest point is zero, we conclude that the kinetic energy of the pendulum is:

$K_{2} = m\cdot g\cdot y_{2}$ (1b)