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Lyrx [107]
4 years ago
12

To explore the bottom of a 25-m-deep lake, your friend Tom proposes to get a long garden hose, put one end on land and the other

in his mouth for breathing underwater, and descend into the depths. Susan, who overhears the conversation, reacts with horror and warns Tom that he will not be able to inhale when he is at the lake bottom.
Why is Susan so worried?
The pressure at a depth 01 25 m is 3 x 10^5 Pa. This is almost three times greater than atmospheric pressure. But he air pressure in the hose will be only slightly higher than atmospheric pressure, because the density of air is so low. So Tom will have great difficulty breathing in the low-pressure air when the large pressure of the water is pressing in on his chest.
The pressure at a depth 01 25 m is 3.5 x 10^5 Pa. This is almost 3.5 times greater than atmospheric pressure. But he air pressure in the hose will be only slightly higher than atmospheric pressure, because the density of air is so low. So Tom will have great difficulty breathing in the low-pressure air when the large pressure of the water is pressing in on his chest.
The pressure at a depth 01 25 m is 2 x 10^5 Pa. This is almost two times greater than atmospheric pressure. But he air pressure in the hose will be only slightly higher than atmospheric pressure, because the density of air is so low. So Tom will have great difficulty breathing in the low-pressure air when the large pressure of the water is pressing in on his chest.
The pressure at a depth 01 25 m is 2.5 x 10^5 Pa. This is almost 2.5 times greater than atmospheric pressure. But he air pressure in the hose will be only slightly higher than atmospheric pressure, because the density of air is so low. So Tom will have great difficulty breathing in the low-pressure air when the large pressure of the water is pressing in on his chest.
Physics
1 answer:
Serhud [2]4 years ago
6 0

Answer:

The pressure at a depth 01 25 m is 2.5 x 10^5 Pa. This is almost 2.5 times greater than atmospheric pressure. But he air pressure in the hose will be only slightly higher than atmospheric pressure, because the density of air is so low. So Tom will have great difficulty breathing in the low-pressure air when the large pressure of the water is pressing in on his chest.

Explanation:

The depth is h = 25 m

The density of water p = 1000 kg/m^3

Acceleration due to gravity g = 9.81 m/s^3.

Pressure due to a depth is gotten from

P = pgh

P = 1000 x 9.81 x 25 =245250 Pa

==> 2.45 x 10^5 Pa

Approximately 2.5 x 10^5 Pa

Atmospheric pressure = 1.01325 x 10^5 Pa

Dividing the pressure at the bottom of the pond by the atmospheric pressure gives a value of about 2.5,which means that the pressure is 2.5 times the atmospheric pressure.

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A warehouse worker is pushing a 90.0-kg crate with a horizontal force of 282 N at a speed of v = 0.850 m/s across the warehouse
Elanso [62]

Answer:

v_{f} = 0.51 \frac{m}{s}

Explanation:

We apply Newton's second law at the crate :

∑F = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Data:

m=90kg :  crate mass

F= 282 N

μk =0.351 :coefficient of kinetic friction

g = 9.8 m/s² : acceleration due to gravity

Crate weight  (W)

W= m*g

W= 90kg*9.8 m/s²

W= 882 N

Friction force : Ff

Ff= μk*N Formula (2)   

μk: coefficient of kinetic friction

N : Normal force (N)  

Problem development

We apply the formula (1)

∑Fy = m*ay    , ay=0

N-W = 0

N = W

N = 882 N

We replace the  data in the formula (2)

Ff= μk*N  = 0.351* 882 N

Ff=  309.58 N

We apply the formula (1) in x direction:

∑Fx = m*ax    , ax=0

282 N - 309.58 N = 90*a  

a=  (282 N - 309.58 N ) / (90)

a= - 0.306 m/s²

Kinematics of the crate

Because the crate moves with uniformly accelerated movement we apply the following formula :

vf²=v₀²+2*a*d Formula (3)

Where:  

d:displacement in meters (m)  

v₀: initial speed in m/s  

vf: final speed in m/s  

a: acceleration in m/s²

Data

v₀ = 0.850 m/s

d = 0.75 m

a= - 0.306 m/s²

We replace the  data in the formula (3)

vf²=(0.850)²+(2)( - 0.306 )(0.75 )

v_{f} = \sqrt{(0.850)^{2} +(2)( - 0.306 )(0.75 )}

v_{f} = 0.51 \frac{m}{s}

8 0
3 years ago
A satellite is to be launched into an orbit of radius,r Show that v = 2gr, where V is the
Snezhnost [94]

Explanation:

We start by using the conservation law of energy:

\Delta{K} + \Delta{U} = 0

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Simplifying the above equation, we get

v^2 = 2G\dfrac{M}{r}

We can rewrite this as

v^2 = 2\left(G\dfrac{M}{r^2}\right)r

Note that the expression inside the parenthesis is simply the acceleration due to gravity g so we can write

v^2 = 2gr

where v is the launch velocity.

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