Which statement best describes a scientific law?

If the surface air pressure is 1000 mb and the pressure at the top of the atmosphere (75 km) is 0 mb, at what altitude would I f

Lana71 [14]

Answer: 5.5km

Explanation:

Atmospheric pressure will be 500 mb (that is half of the total 1000mb air pressure).

Pressure decreases with increasing altitude. This is because at At higher altitudes, there are fewer air molecules above a the known or given surface than a similar surface at lower levels.

Pressure decreasing with higher altitudes also means that air pressure decreases rapidly at lowerevels but more slowly at higher levels.

It is also known that more than half of the atmospheric molecules are located below 5.5 km(that is atmospheric pressure decreases within the lowest 5.5 km to about fifty(50) percent( that is 500 millibar).

8 0

3 years ago

What is one of the forms of a gene called?

vaieri [72.5K]

Answer:

Explanation:

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4 0

3 years ago

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The current that charges a capacitor transfers energy that is stored in the capacitor’s electric field. Consider a 2.0 μF capaci

lapo4ka [179]

Answer:

the capacitor voltage is V = 20 V

Explanation:

Given,

Capacitance of the capacitor = 2.0 μF

energy stored = 200 W

time (t) =2.0 μs

Capacitor voltage = ?

$\dfrac{dE}{dt}=200 W$

$E =200\times 2 \times 10^{-6}$

$E =400 \times 10^{-6}\ J$

we know,

$E = \dfrac{1}{2}CV^2$

$V =\sqrt{\dfrac{2E}{C}}$

$V =\sqrt{\dfrac{2\times 400 \times 10^{-6}}{2\times 10^{-6}}}$

$V =\sqrt{400}$

V = 20 V

so , the capacitor voltage is V = 20 V

7 0

3 years ago

A spring gun is made by compressing a spring in a tube and then latching the spring at the compressed position. A 4.97-g pellet

dimaraw [331]

Answer:

$v = 2.8898 \frac{m}{s}$

Explanation:

This is a problem easily solve using energy conservation. As there are no non-conservative forces, we know that the energy is conserved.

When the spring is compressed downward, the spring has elastic potential energy. When the spring is relaxed, there is no elastic potential energy, but the pellet will have gained gravitational potential energy and kinetic energy. Lets see what are the terms for each of this.

<h3>Elastic potential energy</h3>

We know that a spring following Hooke's Law has a elastic potential energy:

$E_{ep} = \frac{1}{2} k (\Delta x)^2$

where $\Delta x$ is the displacement from the relaxed length and k is the spring's constant.

To obtain the spring's constant, we know that Hooke's law states that the force made by the spring is :

$\vec{F} = - k \Delta \vec{x}$

as we need 9.12 N to compress 4.60 cm, this means:

$k = \frac{9.12 \ N}{4.6 \ 10^{-2} \ m}$

$k = 198.26 \ \frac{ N}{m}$

So, the elastic energy of the compressed spring is:

$E_{ep} = \frac{1}{2} 198.26 \ \frac{ N}{m} (4.6 \ 10^{-2} \ m)^2$

$E_{ep} = 0.209759 \ Joules$

And when the spring is relaxed, the elastic potential energy will be zero.

<h3>Gravitational potential energy</h3>

To see how much gravitational potential energy will the pellet win, we can use

$\Delta E_{gp} = m g \Delta h$

where m is the mass of the pellet, g is the acceleration due to gravity and \Delta h is the difference in height.

Taking all this together, the gravitational potential energy when the spring is relaxed will be:

$\Delta E_{gp} = 4.97 \ 10^{-3} kg \ 9.8 \frac{m}{s^2} 4.6 \ 10^{-2} m$

$\Delta E_{gp} = 0.00224 \ Joules$

<h3>Kinetic Energy</h3>

We know that the kinetic energy for a mass m moving at speed v is:

$E_k = \frac{1}{2} m v^2$

so, for the pellet will be

$E_k = \frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2$

<h3>All together</h3>

By conservation of energy, we know:

$E_{ep} = \Delta E_{gp} + E_k$

$0.209759 \ Joules = 0.00224 \ Joules + \frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2$

So

$\frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2 = 0.209759 \ Joules - 0.00224 \ Joules$

$\frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2 = 0.207519 \ Joules$

$v = \sqrt{ \frac{ 0.207519 \ Joules}{ \frac{1}{2} \ 4.97 \ 10^{-3} kg } }$

$v = 2.8898 \frac{m}{s}$

7 0

3 years ago

What is elasticity in polymer generally related to?

Andrej [43]

The elasticity of a polymer is primarily due to the structure of the molecule and the cross-linking between strands. Hydrogen bonding is a contributor to the shape of the molecule, but not a major player in terms of elasticity. We would have to answer "false".
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6 0

3 years ago

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