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2 years ago

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

Physics

1 answer:

antiseptic1488 [7]2 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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A box is placed on a conveyor belt that moves with a constant speed of 1.05 m/s. The coefficient of kinetic friction between the

sveticcg [70]

Answer:

The box stops in 0.139 seconds, after moving 7.29cm (0.0729m) backwards relative to the belt.

Explanation:

As the box is initially at rest relative to the earth, it is moving backwards with a speed of 1.05m/s relative to the belt. Then, the frictional force acts on the box to make it stop relative to the belt. So, we first have to write the equations of motion of the box in each axis:

$x: f_k=ma\implies a=\frac{f_k}{m} \\\\y: N-mg=0\implies N=mg$

Since the frictional force $f_k$ is equal to $f_k=\mu_k N=\mu_k mg$ , then we have that the acceleration is:

$a=\frac{\mu_k mg}{m}=\mu_k g$

Now, from the definition of acceleration we get:

$a=\frac{v-v_0}{t}\implies t=\frac{v-v_0}{a}$

And, as the final velocity is zero because the box gets to a stop, we have:

$t=-\frac{v_0}{a}=-\frac{v_0}{\mu_k g}$

(Don't worry about the negative sign. It will disappear because the initial velocity is also negative, since we take the box initially moving backwards)

Then, plugging in the given values, we calculate the time:

$t=-\frac{(-1.05m/s)}{0.770(9.81m/s^{2})}=0.139s$

In words, the time the box takes to stop sliding relative to the belt is 0.139s.

The displacement of the box in this time, is given by the kinematics formula:

$v^{2}=v_0^{2}+2ax\implies x=-\frac{v_0^{2}}{2\mu_kg}$

Finally, we calculate the displacement:

$x=-\frac{(1.05m/s)^{2} }{2(0.770)(9.81m/s^2)}=-0.0729m=-7.29cm$

This means that the box moves 7.29cm backwards relative to the belt.

4 0

2 years ago

A loud sound is produced in the downtown section of a city. Which of the following is least likely to occur with the sound wave

JulsSmile [24]

Answer:

A. The sound wave will reflect off Buildings and automobiles.

Explanation:

This is because the sound waves would more likely propagate through diffraction through buildings and transmission through the air. It is also more likely to be absorbed by buildings than for multiple reflections to occur off buildings and automobiles. In the process of reflection, these materials would absorb the sound energy thereby reducing its ability to reflect.

6 0

2 years ago

In Physics, work depends on two factors. What are those two<br> factors?

Aloiza [94]

Answer:

Force and displacement.

Explanation:

Work done is positive when we push table and it move in the direction of applied force.

5 0

2 years ago

WOULUJUTUL RECIPECUIUS.

3241004551 [841]

The force between the two objects is 19.73 nN.

<u>Explanation: </u>

Any force acting between two objects tends to be directly proportional to the product of their masses and inversely proportional to the square of the distance between the two objects. And this kind of attraction force between two objects is termed as gravitational force.

So if we consider $M_{1}$ and $M_{2}$ as the masses of both objects and let d be the distance of separation of two objects. Then the force between the two objects can be determined as below:

$\text {Gravitational force}=\frac{G \times M_{1} \times M_{2}}{d^{2}}$

As gravitational constant $G=6.67 \times 10^{-11} \mathrm{m}^{3} \mathrm{kg}^{-1} \mathrm{s}^{-2}$ , $M_{1}$ = 20 kg and $M_{2}$ = 100 kg, while d = 2.6 m, then