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cupoosta [38]
3 years ago
7

Why are we not crushed by the weight of the atmosphere on our shoulders?

Physics
1 answer:
antiseptic1488 [7]3 years ago
5 0

Answer:

Due to equal pressure in all the direction at a particular level in a fluid medium (Pascal's Law)

Explanation:

We are not crushed by the weight of the atmosphere because atmosphere is a fluid and we are immersed into it. So, according to the Pascal's law the the pressure a each point in a horizontal level is equal in all the direction irrespective of the orientation of a body.

Variation of pressure in term of the height of a fluid medium is given as:

P=\rho.g.h

\rho=density of fluid

g = acceleration due to gravity

h = height of the free surface of the fluid from the immersed object.

  • And atmosphere has very less variation of pressure with change in height as it is a rare medium fluid and so for a human height there is very negligible variation of pressure at the heat of a human with respect to his toe.

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Two particles, each with charge 55.3 nC, are located on the y axis at y 24.9 cm and y -24.9cm (a) Find the vector electric field
ser-zykov [4K]

Answer:

Ex = kq 2x / ∛ (x² + y²)²  and  Ex = 2008 N / C

Explanation:

a)   The electric field is a vector quantity, so we must find the field for each particle and add them vectorially, as the whole process is on the X axis,

The equation for the electric field produced by a point charge is

         E = k q / r²

With r the distance between the point charge and the positive test charge

We look for each electric field

Particle 1.  Located at y = 24.9 m, let's use Pythagoras' theorem to find the distance

          r² = x² + y²

          E1 = k q / (x² + y²)

Particle 2.   located at x = -24.9 m

          r² = x² + y²

          E2 = k q / (x² + y²)

We can see that the two fields are equal since the particles have the same charge and coordinate it and that is squared.

In the attached one we can see that the Y components of the electric fields created by each particle are always the same and it is canceled, so we only have to add the X components of the electric fields. Let's use Pythagoras' theorem to find

Let's measure the angle from axis X

     cos θ = CA / H = x / (x2 + y2) ½

     E1x = E1 cos θ

      E2x = = E1 cos θ

The resulting field

      Ey = 0

      Ex = E1x + E2x 2 E1x

      Ex = 2 k q / (x² + y²) cos θ) = 2 k q / (x² + y²) x / √(x² + x²)

      Ex = kq 2x / ∛ (x² + y²)²

b) For this part we substitute the numerical values

      Ex = 8.99 10⁹ 55.3 10⁻⁹ x / (x² + 0.249 2) ³/₂

      Ex = 497.15   x / (x² + 0.062)  ³/₂  

Point where can the value of the electric field x = 38.1 cm = 0.381 m

       Ex = 497.15 0.381 / (0.381² + 0.062)  ³/₂  

       Ex = 497.15 0.381 / (0.1452 + 0.062) 3/2 = 189.41 / 0.2072 3/2

       Ex= 189.41 /0.0943

       Ex = 2008 N / C

c)  E = 1.00 kN / C = 1000 N / C

To solve this part we must find x in the equation

       Ex = 497.15 x / (x² + 0.062)  ³/₂  

Let's use some arithmetic

       Ex / 497.15 = x / (x² + 0.062)  ³/₂  

       [Ex / 497.15] ²/₃ = [x / (x² + 0.062) 3/2] ²/₃

       ∛[Ex / 497.15]² = (∛x²) / (x² + 0.062)                 (1)

The roots of this equation are the solution to the problem,

     

For Ex = 1.00 kN / C = 1000 N / C

 

      [Ex / 497.15] 2/3 = 1000 / 497.15) 2/3 = 1,312

       1.312 = (∛x² ) / (x² + 0.062)

       1.312 (x² + 0.062) = ∛x²

       1.312 X² - ∛x² + 1.312 0.062 = 0

       1.312 X² - ∛x² + 0.0813 = 0

We need used computer

4 0
3 years ago
A football is kicked into the air from an initial height of 4 feet. The height, in feet, of the football above the ground is giv
kakasveta [241]

Answer: 0.5 seconds or 2.625 seconds

Explanation:

At t = 0, The ball is 4 ft above the ground.

The height of the football varies with time in the following way:

s(t) = -16 t² + 50 t + 4

we need to find the time in which the height would of the football would be 25 ft:

⇒25 = -16 t² + 50 t + 4

we need to solve the quadratic equation:

⇒ 16 t² - 50 t + 21 = 0

t = \frac{50 \pm \sqrt{50^2-4\times 16\times 21}}{2\times16}

⇒ t = 0.5 s or 2.625 s

Therefore, at t = 0.5 s or 2.625 s, the football would be 25 ft above the ground.

3 0
3 years ago
Read 2 more answers
In the compound CaCO3 at the end of the compound represents the number of
Afina-wow [57]

Answer:

Calculate the mass of 6.022 × 1023 molecule of Calcium carbonate (CaCO3).

Solution —

Molar mass (Molecular mass in gram) of CaCO3 = 40+12+3×16 = 100 g

No. of moles of CaCO3

= No. of molecules/Avogadro constant

= 6.022 × 1023/ 6.022 × 1023

= 1 mole

Mass of CaCO3

= No. of moles × molar mass

= 1 × 100 g = 100 g.

5 0
3 years ago
Charges that are different each other​
adell [148]

Answer:

Like charges repel each other; unlike charges attract. Thus, two negative charges repel one another, while a positive charge attracts a negative charge. The attraction or repulsion acts along the line between the two charges. The size of the force varies inversely as the square of the distance between the two charges.

4 0
3 years ago
Read 2 more answers
Zain is standing on a bridge. He tosses a ball upwards, which rises, stops, and then falls all the way to the water (30 meters b
IgorC [24]

Answer:

A. 2.41 s.

B. 24.3 m/s.

Explanation:

vf = vi + 2a*t

where, vf = final velocity

= 0 m/s

vi = final velocity

= 1.5 m/s

a = 9.8 m/s^2

A.

ti = 1.5/(9.8 * 2)

= 0.08 s

Vf^2 = Vi^2 + 2a*s

1.5^2 = 2 * 9.8 * s

S = 0.115 m

Time taken to drop into the water,

30.115 = 1.5*to + 4.9*to^2

to = 2.33 s

Total time taken = ti + to

= 2.33 + 0.08

= 2.41 s.

B.

Vo = 0 m/s

S = 30.115 m

Vf = ?

Using,

Vf^2 = Vo^2 + 2*a*s

= sqrt (2*9.8*30.115)

= 24.3 m/s.

7 0
4 years ago
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