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2 years ago

Why don't you feel the pressure exerted by the atmosphere?

Physics

2 answers:

mariarad [96]2 years ago

8 0

Answer:

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.

Explanation:

Galina-37 [17]2 years ago

6 0

Because in our body blood also exerts pressure that's why we don't feel atmospheric pressure.

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At what common temperature will a block of wood and a block of metal both feel neither hot nor cold to the touch ?

Anastaziya [24]

When you touch an object and heat flows OUT of it, INTO your finger, you say the object feels hot.

When you touch an object and heat flows INTO it, OUT of your finger, you say the object feels cold.

If the object has the same temperature as your finger ... around the mid-90s ... then no heat flows in or out of your finger when you touch the object, and the object doesn't feel hot or cold.

6 0

3 years ago

5. Dry ice is an example of _________, which is the process of a solid turning directly into a gas. (1 point) precipitation melt

ki77a [65]

5. Dry ice is an example of _________, which is the process of a solid turning directly into a gas. (1 point)

sublimation

6. The ____ is a unit of force. (1 point)

newton

7. Which of the following is the boiling point of water? (1 point)

100°C

8. Which of the following describes the molecular structure of water at 40°C? (1 point)

water molecules are close together and moving freely around each other

6 0

3 years ago

A circular loop of wire with a diameter of 0.626 m is rotated in a uniform electric field to a position where the electric flux

Mila [183]

Answer:

C) 2.44 × 106 N/C

Explanation:

The electric flux through a circular loop of wire is given by

$\Phi = EA cos \theta$

where

E is the electric field

A is the cross-sectional area

$\theta$ is the angle between the direction of the electric field and the normal to A

The flux is maximum when $\theta=0^{\circ}$ , so we are in this situation and therefore $cos \theta =1$ , so we can write

$\Phi = EA$

Here we have:

$\Phi = 7.50\cdot 10^5 N/m^2 C$ is the flux

d = 0.626 m is the diameter of the coil, so the radius is

r = 0.313 m

and so the area is

$A=\pi r^2 = \pi (0.313 m)^2=0.308 m^2$

And so, we can find the magnitude of the electric field:

$E=\frac{\Phi}{A}=\frac{7.50\cdot 10^5 Nm^2/C}{0.308 m^2}=2.44\cdot 10^6 N/C$

3 0

3 years ago

The answer and how to do it?? Thanks

denis-greek [22]

Answer:

14 m/s²

Explanation:

Start with Newton's 2nd law: Fnet=ma, with F being force, m being mass, and a being acceleration. The applied forces on the left and right side of the block are equivalent, so they cancel out and are negligible. That way, you only have to worry about the y direction. Don't forget the force that gravity has the object. It appears to me that the object is falling, so there would be an additional force from going down from weight of the object. Weight is gravity (can be rounded to 10) x mass. Substitute 4N+weight in for Fnet and 1kg in for m.

(4N + 10 x 1kg)=(1kg)a

14/1=14, so the acceleration is 14 m/s²

4 0

2 years ago

A potential difference of 71 mV is developed across the ends of a 12.0-cm-long wire as it moves through a 0.27 T uniform magneti

Dafna1 [17]

Answer:

Explanation: please see attached file I attached the answer to your question.

7 0

2 years ago