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4vir4ik [10]
2 years ago
14

Evangelista Torricelli was the first person to realize that we live at the bottom of an ocean of air. He correctly surmised that

the pressure of our atmosphere is attributable to the weight of the air. The density of air at 0°C at the Earth's surface is 1.29 kg/m3. The density decreases with increasing altitude (as the atmosphere thins). On the other hand, if we assume that the density is constant at 1.29 kg/m3 up to some altitude h, and zero above that altitude, then h would represent the depth of the ocean of air.
(a) Use this model to determine the value of h that gives a pressure of 1.00 atm at the surface of the Earth.
(b) Would the peak of Mt. Everest rise above the surface of such an atmosphere?
Yes or no?
Physics
1 answer:
Radda [10]2 years ago
8 0

Answer:

a)    h = 8.02 10³ m  b) yes

Explanation:

a) The pressure in a fluid is given by

      P = ρ g h

The pressure in this case is the atmospheric pressure, 1.013 105 Pa, let's clear the height (h)

      h = P / ρ g

      h = 1.013 10⁵ / (1.29 9.8)

      h = 8.02 10³ m

b) The height of Mount Everest is 8848 m

It is above this height, according to this model there would be no air to breathe

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A projectile is fired over level ground with an initial velocity that has a vertical component of 20 m/s and a horizontal compon
Anettt [7]
First of all, let's write the equation of motions on both horizontal (x) and vertical (y) axis. It's a uniform motion on the x-axis, with constant speed v_x=30 m/s, and an accelerated motion on the y-axis, with initial speed v_y=20 m/s and acceleration g=9.81 m/s^2:
S_x(t)=v_xt
S_y(t)=v_y t- \frac{1}{2} gt^2
where the negative sign in front of g means the acceleration points towards negative direction of y-axis (downward).

To find the distance from the landing point, we should find first the time at which the projectile hits the ground. This can be found by requiring
S_y(t)=0
Therefore:
v_y t -  \frac{1}{2}gt^2=0
which has two solutions:
t=0 is the time of the beginning of the motion,
t= \frac{2 v_y}{g} = \frac{2\cdot 20 m/s}{9.81 m/s^2}=4.08 s is the time at which the projectile hits the ground.

Now, we can find the distance covered on the horizontal axis during this time, and this is the distance from launching to landing point:
S_x(4.08 s)=v_x t=(30 m/s)(4.08 s)=122.4 m
4 0
3 years ago
Dropping a stone from a height is work or not?​
kvasek [131]

Answer:

Dropping a stone from a height is work  

Explanation:

Work is the action of a force through a distance.

w = fd

The Earth's gravitational attractive force is pulling the stone to its centre, so the Earth is doing work on the stone.

7 0
3 years ago
In your own words, what was the purpose of the field study?
DanielleElmas [232]

Answer:

Hi there! I have only the procedure and the scientific names of the creatures. Hope this helps!

Procedure

I studied the physical features of ten creatures and classified them using the key. I completed each section below and recorded the creature's scientific name in the data section.

The Creatures Scientific Names:

1  

Fuzzus tallywag

2  

Fuzzus pointilus

3  

Silkus duosquirmus

4  

Fuzzus chompilus

5  

Silkus stretchilus

6  

Silkus tallyhas

7  

Fuzzus feelzalot

8  

Silkus monosquirmus

9  

Fuzzus squarilus

10  

Silkus monowrestle

Have a thrilling Thursday!

~Lola

6 0
3 years ago
How does potential difference behave in a parallel circuit
Illusion [34]
The same voltage will appear across all resistors in parallel. 
6 0
3 years ago
Read 2 more answers
What is the frequency of sound waves whose wavelength is 0.25 m on a day when the air temperature is 25 degree Celsius? Group of
Dvinal [7]

Answer:1384 Hz

Explanation:

Given

wavelength(\lambda )=0.25 m

Temperature T=25^{\circ}

at T=25^{\circ} velocity of sound is 346 m/s

and we know

velocity=frequency\times \lambda

v=f\times \lambda

346=f\times 0.25

f=1384 Hz

5 0
3 years ago
Read 2 more answers
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