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3 years ago

In avarage,How many times do a child breathe in a

2 answers:

6 0

There are 60 minutes in an hour, therefore:

Breaths per hour = 12 X 60, 15 X 60
= 720 to 900 breaths per hour

There are 24 hours in a day, therefore:

Breaths per day = 720 X 24, 900 X 24
= 17,280 to 21,600 breaths per day.
This differs between adults and kids, so:
A child breathes around 28,800 times a day.
An Adult breathes around 23040 times a day.

Hope this helps!

ollegr [7]3 years ago

5 0

On an approximate scale, A child breaths 20 times a minute as compared to only 12 to 16 in resting phase of an Adult.

So, In 60 minutes (1 hour), They breathe = 20 * 60 = 1200
In 24 hours (1 day), They breathe = 1200 * 24 = 28,800

In short, Your Answer would be: 28,800

Hope this helps!

You might be interested in

A rocket travels in the x-direction at speed 0.70c with respect to the earth. An experimenter on the rocket observes a collision

marishachu [46]

Answer:

A) The space time coordinate x of the collision in Earth's reference frame is

$x \approx 103,46x10^{9}m$ .

B) The space time coordinate t of the collision in Earth's reference frame is

$t=377,29s$

Explanation:

We are told a rocket travels in the x-direction at speed v=0,70 c (c=299792458 m/s is the exact value of the speed of light) with respect to the Earth. A collision between two comets is observed from the rocket and it is determined that the space time coordinates of the collision are (x',t') = (3.4 x 10¹⁰ m, 190 s).

An event indicates something that occurs at a given location in space and time, in this case the event is the collision between the two comets. We know the space time coordinates of the collision seen from the reference frame of the rocket and we want to find out the space time coordinates in Earth's reference frame.

Lorentz transformation

The Lorentz transformation relates things between two reference frames when one of them is moving with constant velocity with respect to the other. In this case the two reference frames are the Earth and the rocket that is moving with speed v=0,70 c in the x axis.

The Lorentz transformation is

$x'=\frac{x-vt}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$y'=y$

$z'=z$

$t'=\frac{t-\frac{v}{c^{2}}x}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

prime coordinates are the ones from the rocket reference frame and unprimed variables are from the Earth's reference frame. Since we want position x and time t in the Earth's frame we need the inverse Lorentz transformation. This can be obtained by replacing v by -v and swapping primed an unprimed variables in the first set of equations

$x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$y=y'$

$z=z'$

$t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

First we calculate the expression in the denominator

$\frac{v^{2}}{c^{2}}=\frac{(0,70)^{2}c^{2}}{c^{2}} =(0,70)^{2}$

$\sqrt{1-\frac{v^{2}}{c^{2}}} =0,714$

then we calculate t

$t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$t=\frac{190s+\frac{0,70c}{c^{2}}.3,4x10^{10}m}{0,714}$

$t=\frac{190s+\frac{0,70c .3,4x10^{10}m}{299792458\frac{m}{s}}}{0,714}$

$t=\frac{190s+79,388s}{0,714}$

finally we get that

$t=377,29s$

then we calculate x

$x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$x=\frac{3,4x10^{10}m+0,70c.190s}{0,714}}$

$x=\frac{3,4x10^{10}m+0,70.299792458\frac{m}{s}.190s}{0,714}}$

$x=\frac{3,4x10^{10}m+39872396914m}{0,714}}$

$x=\frac{73872396914m}{0,714}}$

$x=103462740775,91m$

finally we get that

$x \approx 103,46x10^{9} m$

5 0

3 years ago

Listed below are the measured radiation absorption rates (in W/kg) corresponding to 11 cell phones. Use the given data to const

Fantom [35]

Answer:

The 5-number summary is

1. Median = 0.93 W/kg

2. Lower quartile = 0.69 W/kg

3. Upper quartile = 1.16 W/kg

4. Minimum value = 0.54 W/kg

5. Maximum value = 1.42 W/kg

Explanation:

We are given the measured radiation absorption rates (in W/kg) corresponding to 11 cell phones.

1.16 0.85 0.69 0.75 0.95 0.93 1.18 1.17 1.42 0.54 0.57

What is 5-number summary?

A 5-number summary refers to a box plot that basically shows 5 statistical characteristics of a data set.

These statistical characteristics are:

1. Median

2. Lower quartile

3. Upper quartile

4. Minimum value

5. Maximum value

1. Median:

Arrange the data in ascending order

0.54 0.57 0.69 0.75 0.85 0.93 0.95 1.16 1.17 1.18 1.42

(n+1)/2 gives the median value of the data set.

(11 + 1)/2 = 6th position

Therefore, 0.93 W/kg is the median of the data set.

2. Lower quartile:

Divide the data set into two equal halfs (include median in both if n = odd)

Lower half = 0.54 0.57 0.69 0.75 0.85 0.93

Upper half = 0.93 0.95 1.16 1.17 1.18 1.42

The lower quartile is the median of the lower half of the data set.

Lower half = 0.54 0.57 0.69 0.75 0.85 0.93

The median is 6/2 = 3rd position

Therefore, the lower quartile of the data set is 0.69 W/kg

3. Upper quartile:

Divide the data set into two equal halfs (include median in both if n = odd)

Lower half = 0.54 0.57 0.69 0.75 0.85 0.93

Upper half = 0.93 0.95 1.16 1.17 1.18 1.42

The upper quartile is the median of the lower half of the data set.

Upper half = 0.93 0.95 1.16 1.17 1.18 1.42

The median is 6/2 = 3rd position

Therefore, the upper quartile of the data set is 1.16 W/kg

4. Minimum value:

The minimum value is the least value in the data set.

0.54 0.57 0.69 0.75 0.85 0.93 0.95 1.16 1.17 1.18 1.42

Therefore, the minimum value of the data set is 0.54 W/kg

5. Maximum value

The maximum value is the least value in the data set.

0.54 0.57 0.69 0.75 0.85 0.93 0.95 1.16 1.17 1.18 1.42

Therefore, the maximum value of the data set is 1.42 W/kg

The box plot is illustrated in the attached diagram.

6 0

3 years ago

The next four questions refer to the situation below.

Anna11 [10]

Answer:

t_{out} = $\frac{v_s - v_r}{v_s+v_r}$ t_{in}, t_{out} = $\frac{D}{v_s +v_r}$

Explanation:

This in a relative velocity exercise in one dimension,

let's start with the swimmer going downstream

its speed is

$v_{sg 1} = v_{sr} + v_{rg}$

The subscripts are s for the swimmer, r for the river and g for the Earth

with the velocity constant we can use the relations of uniform motion

$v_{sg1}$ = D / $t_{out}$

D = v_{sg1} t_{out}

now let's analyze when the swimmer turns around and returns to the starting point

$v_{sg 2} = v_{sr} - v_{rg}$

$v_{sg 2}$ = D / $t_{in}$

D = v_{sg 2} t_{in}

with the distance is the same we can equalize

$v_{sg1} t_{out} = v_{sg2} t_{in}$

t_{out} = t_{in}

t_{out} = $\frac{v_s - v_r}{v_s+v_r}$ t_{in}

This must be the answer since the return time is known. If you want to delete this time

t_{in}= D / $v_{sg2}$

we substitute

t_{out} = \frac{v_s - v_r}{v_s+v_r} ()

t_{out} = $\frac{D}{v_s +v_r}$

7 0

2 years ago

Before Freddy lands on the skateboard it has a certain momentum. After landing, the skateboards momentum

Nadya [2.5K]

Answer:

remains the same

Explanation:

Momentum refers to the quantity of motion of a body. When any body of mass moves, it possess momentum. Numerically,

Momentum = mass x velocity

i.e. momentum is the product of the mass x velocity

Momentum of a body is always conserved.

In the context, the skateboard has certain momentum before Freddy lands on it. After Freddy lands, the momentum of skateboard remains the same, there is no change in the momentum.

This is because, here the momentum is conserved. After Freddy lands on the skateboard, the total mass on the skateboard increases and so the velocity decreases making the momentum same before the landing.

3 0

2 years ago

Which statement describes the relationship between bond strength and the melting and boiling points of a substance? A. As the fo

icang [17]

Answer:

Explanation:

4 0

3 years ago