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3 years ago

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What is the relationship between air pressure and air temperature?

1 answer:

LenKa [72]3 years ago

8 0

The relationship between the two is that air temperature changes the air pressure. For example, as the air warms up the molecules in the air become more active and they use up more individual space even though there is the same<span> number of molecules. This causes an </span>increase<span> in the air pressure.</span>

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You lift a stuffed toy up above the ground (over your head).

BigorU [14]

Answer:

Gravitational energy

Explanation:

when over head , when it hits the floor thermal energy

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3 years ago

Engineers and science fiction writers have proposed designing space stations in the shape of a rotating wheel or ring, which wou

Gennadij [26K]

Answer:

a. The station is rotating at $1.496 \frac{rev}{min}$

b. the rotation needed is $2.8502 \frac{rev}{min}$

Explanation:

We know that the centripetal acceleration is

$a_{c}= \omega ^2 r$

where $\omega$ is the rotational speed and r is the radius. As the centripetal acceleration is feel like an centrifugal acceleration in the rotating frame of reference (be careful, as the rotating frame of reference is <u>NOT INERTIAL,</u> the centrifugal force is a fictitious force, the real force is the centripetal).

<h3>a. </h3>

The rotational speed is :

$2.7 \frac{m}{s^2} = \omega ^2 * 110 \ m$

$\omega ^2 = \frac{2.7 \frac{m}{s^2}} {110 \ m}$

$\omega = \sqrt{ 0.02454 \frac{rad^2}{s^2} }$

$\omega = 0.1567 \frac{rad}{s}$

Knowing that there are $2\pi \ rad$ in a revolution and 60 seconds in a minute.

$\omega = 0.1567 \frac{rad}{s} \frac{1 \ rev}{2\pi \ rad} \frac{60 \ s}{1 \ min}$

$\omega = 1.496 \frac{rev}{min}$

<h3>b. </h3>

The rotational speed needed is :

$9.8 \frac{m}{s^2} = \omega ^2 * 110 \ m$

$\omega ^2 = \frac{9.8 \frac{m}{s^2}} {110 \ m}$

$\omega = \sqrt{ 0.08909 \frac{rad^2}{s^2} }$

$\omega = 0.2985 \frac{rad}{s}$

Knowing that there are $2\pi \ rad$ in a revolution and 60 seconds in a minute.

$\omega = 0.2985 \frac{rev}{min} \frac{1 \ rev}{2\pi \ rad} \frac{60 \ s}{1 \ min}$

$\omega = 2.8502 \frac{rev}{min}$

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3 years ago

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Lithium (chemical symbol Li) is located in Group 1, Period 2. Which is lithium most likely to be? O A. A soft, shiny, highly rea

algol13

A. A soft, shiny, highly reactive metal

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3 years ago

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The electric potential difference across the membrane of a body cell is 0.070 V (higher on the outside than on the inside). The

12345 [234]

Answer:

The electric field is $8.75 \times 10^{6}~v~m^{-1}$ and the ditection is from outer to inner side of the membrane.

Explanation:

We know the electric field ( $\vec{E}$ ) is given by $\vec{E} = - \nabla V$ , 'V' being the potential.

In 1-D, it can be written as

$E=\dfrac{V}{d}$

where 'd' is the separation of space in between the potential difference is created.

Given, $V = 0.070~V~$ and the thickness of the cell membrane is $d = 8.0 \times 10^{-9}~m$ .

Therefore the created electric field through the cell membrane is

$E = \dfrac{0.07~V}{8 \times 10^{-9}~m} = 8.75 \times 10^{6}~m~s^{-1}$

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In mathematics, a percentage is a number or ratio expressed as a fraction of 100.

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