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lbvjy [14]
3 years ago
14

when you inhale (breathe in) your ribs move upwards and outwards and your diaphragm moves down. Explain in terms of pressure and

volume how this process forces air into your lungs
Physics
1 answer:
katrin2010 [14]3 years ago
3 0
As you inhale, your ribs move up and out decreasing the internal pressure and increasing the volume of the lungs forcing air into the lungs from an area of externally high pressure in comparison to the low internal pressure. 
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In a race, a runner traveled 12 meters in
Nastasia [14]
     Since it is a race, the runner start from rest, so we can use the Velocity Hourly Equation to describle yours moviment.

S=S_{o}+V_{o}t+ \frac{at^2}{2}  \\ S=\frac{at^2}{2}
 
     Entering the unknowns, we have:

S=\frac{at^2}{2} \\ 12*2=a*4^2 \\ \boxed {a=1.5m/s^2}

Number 2

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3 years ago
1. Which direction will the box move in the diagram below? (1 point)
svetoff [14.1K]

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4 years ago
should I tell penny, potato chip and used napkin that I've made new friends and moved on. I don't want to make them sad. :(
MrRa [10]

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7 0
3 years ago
Read 2 more answers
Consider two particles A and B. The angular position of particle A, with constant angular acceleration, depends on time accordin
Doss [256]

Answer:

t - t_1 = \frac{-\omega_o + \sqrt{\omega_o^2 + 8\alpha(2\omega_o t + \alpha t^2)}}{4\alpha}

Explanation:

After time "t" the angular position of A is given as

\theta_a = \theta_o + \omega_o t + \frac{1}{2}\alpha t^2

now we know that B start motion after time t1

so its angular position is also same as that of position of A after same time "t"

so we have

\theta_b = \theta_o + \frac{\omega_o}{2} (t - t_1) + \frac{1}{2}(2\alpha) (t - t_1)^2

now since both positions are same

\theta_a = \theta_b

\omega_o t + \frac{1}{2}\alpha t^2 = \frac{\omega_o}{2}(t - t_1} + \alpha(t - t_1)^2

2\omega_o t + \alpha t^2 = \omega_o(t - t_1) + 2\alpha(t - t_1)^2

t - t_1 = \frac{-\omega_o + \sqrt{\omega_o^2 + 8\alpha(2\omega_o t + \alpha t^2)}}{4\alpha}

3 0
3 years ago
brainly a child on a swing set swings back and forth with a period of 3.3 s and an amplitude of 25°. what is the maximum speed o
beks73 [17]

Maximum speed of the child as she swings is  2.23 m/s.

<h3>Step by Step Calculation:</h3>

T=3.3 s is the oscillation's time period.

The swing's greatest angle is 25° (max).

The swing's bottom will have the following kinetic energy:

k=12mv2...........(1)

The mass in this situation is m, and the speed is v.

The potential energy change is expressed as,

∆U=mgL1-cosθmax...............(2)

Here, L is a string's length and g is the acceleration caused by gravity. L is given as,

L=gT24π2

Combine equation (1) with (2)

12mv2=mgL1-cos, maxv=2g, maxv=gT24, maxv=g2T22, maxv=9.8 m/s

22.33 m/s, 22.31 s22.31 cos25°

Therefore, the child's top speed is 2.23 m/s.

<h3>What is Oscillation ?</h3>
  • The process of any quantity or measure fluctuating repeatedly about its equilibrium value in time is known as oscillation.
  • A periodic change in a substance's value between two values or around its central value is another way to define oscillation.

To learn more about Oscillation refer to:

brainly.com/question/28312746

#SPJ1

6 0
1 year ago
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