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2 years ago

Why we don’t feel atmospheric pressure?

Physics

2 answers:

Veseljchak [2.6K]2 years ago

8 0

Answer:

The reason we can't feel it is that the air within our bodies (in our lungs and stomachs, for example) is exerting the same pressure outwards, so there's no pressure difference and no need for us to exert any effort.

lesantik [10]2 years ago

4 0

Answer:We don't feel atmospheric pressure because the pressure in our bodies is almost the same as the atmospheric pressure outside

Explanation:

You might be interested in

How old is a bone that has 12.5% of the original amount radioactive carbon 14 remaining?

nlexa [21]

Nuclear decay formula is N(t)=N₀*2^-(t/T), where N(t) is the amount of nuclear material in some moment t, N₀ is the original amount of nuclear material, t is time and T is the half life of the material, in this case carbon 14. In our case N(t)=12.5% of N₀ or N(t)=0.125*N₀, T=5730 years and we need to solve for t:

0.125*N₀=N₀*2^-(t/T), N₀ cancels out and we get:

0.125=2^-(t/T),

ln(0.125)=ln(2^-(t/T))

ln(0.125)=-(t/T)*ln(2), we divide by ln(2),

ln(0.125)/ln(2)=-t/T, multiply by T,

{ln(0.125)/ln(2)}*T=-t, divide by (-1) and plug in T=5730 years,

{ln(0.125)/[-ln(2)]}*5730=t

t=3*5730=17190 years.

The bone is t= 17190 years old.

4 0

3 years ago

In the first stage of a two-stage Carnot engine, energy is absorbed as heat Q1 at temperature T1 = 500 K, work W1 is done, and e

Nataly [62]

Answer:

Efficiency = 52%

Explanation:

Given:

First stage

heat absorbed, Q₁ at temperature T₁ = 500 K

Heat released, Q₂ at temperature T₂ = 430 K

and the work done is W₁

Second stage

Heat released, Q₂ at temperature T₂ = 430 K

Heat released, Q₃ at temperature T₃ = 240 K

and the work done is W₂

Total work done, W = W₁ + W₂

Now,

The efficiency is given as:

$\eta=\frac{\textup{Total\ work\ done}}{\textup{Energy\ provided}}$

Work done = change in heat

thus,

W₁ = Q₁ - Q₂

W₂ = Q₂ - Q₃

Thus,

$\eta=\frac{(Q_1-Q_2)\ +\ (Q_2-Q_3)}{Q_1}}$

$\eta=1-\frac{(Q_1-Q_3)}{Q_1}}$

$\eta=1-\frac{(Q_3)}{Q_1}}$

also,

$\frac{Q_1}{T_1}=\frac{Q_2}{T_2}=\frac{Q_3}{T_3}$

$\frac{T_3}{T_1}=\frac{Q_3}{Q_1}$

thus,

$\eta=1-\frac{(T_3)}{T_1}}$

thus,

$\eta=1-\frac{(240\ K)}{500\ K}}$

$\eta=0.52$

Efficiency = 52%

8 0

3 years ago

How far away is the closest star?

sukhopar [10]

Answer: Proxima Centauri is the closet star about 40,208,000,000,000 km away.

Explanation:

8 0

1 year ago

Read 2 more answers

Can someone help me?!!!!!

german

<h2>Hello!</h2>

The answer is:

The first option, the walker traveled 360m more than the actual distance between the start and the end points.

Why?

Since each block is 180 m long, we need to calculate the vertical and the horizontal distance, in order to calculate how farther did the travel walk between the start and the end points (displacement).

So, calculating we have:

Traveler:

$Distance=NorthCoveredDistance+EastCoveredDistance$

$Distance=4*180m+3*180m=720m+540m=1260m$

Actual distance between the start and the end point (displacement):

$ActualDistance=\sqrt{NorthDistance+EastDistance}\\\\ActualDistance=\sqrt{NorthDistance^{2} +EastDistance^{2}}\\\\ActualDistance=\sqrt{(720m)^{2} +(540m)^{2}}\\\\ActualDistance=\sqrt{518400m^{2} +291600m^{2}}\\\\ActualDistance=\sqrt{810000m^{2}}=900m$

Now, to calculate how much farter did the traveler walk, we need to use the following equation:

$DistanceDifference=WalkerCoveredDistance-ActualDistance\\\\DistanceDifference=1260m-900m=360m$

Therefore, we have that distance differnce between the distance covered by the walker and the actual distance is 360m.

Hence, we have that the walker traveled 360m more than the actual distance between the start point and the end point.

Have a nice day!

3 0

3 years ago

emmainna [20.7K]

Answer:

Increasing the mass and decreasing the distance between the two objects.

Explanation:

An increase in mass will cause them to have a stronger pull or gravity. A decrease of distance will make it easier for the objects to fall into each other because they would be further into the other objects area of influence.

3 0

2 years ago