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3 years ago

How and why dose air pressure change with altitude in the atmosphere

1 answer:

8 0

<span>Air pressure changes with altitude because of issues related to gravity molecules have more weight the closer they are to the Earth and more of them move to lower elevations as a result this causes increased pressure because there are more molecules in number and proximity</span>

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A 3.9 kg ball traveling towards a soccer player at a velocity of -3.5 m/s rebounds off the soccer player's foot at a velocity of

Tresset [83]

Answer: 2.92 s

Explanation:

Given

Mass of ball is $m=3.9\ kg$

The initial velocity of the ball is $u=-3.5\ m/s$

Velocity after the rebound is $v=15.9\ m/s$

Force during the contact is $F=25.9\ N$

We know, change in momentum is Impulse

$\Rightarrow F\cdot \Delta t=m(\Delta v)$

$\Rightarrow 25.9\cdot \Delta t=3.9(15.9-(-3.5))\\\\\Rightarrow \Delta t=\dfrac{3.9\times 19.4}{25.9}=2.92\ s$

Thus, the force is applied for 2.92 s

4 0

3 years ago

1. State the law of conservation of energy and what it means for you as a human considering how energy works.

Arturiano [62]

Answer: 1. The law of consevation of energy sates that energy can neither be created nor destroyed. It can only be transformed or transfered from one form to another. The law of conservation of energy is found everywhere for example, Water falls from the sky, converting potential energy to kinetic energy.

2. Different forms of energy are related because energy cannot be created or destroyed. they can all be transformed into from one form to another.

Explanation:

5 0

3 years ago

When a light bulb is connected to a 4.5 V battery, a current of 0.16 A passes through the bulb filament. What is the resistance

larisa [96]

Answer:

R = 28.125 ohms

Explanation:

Given that,

The voltage of a bulb, V = 4.5 V

Current, I = 0.16 A

We need to find the resistance of the filament. Using Ohm's law,

V = IR

Where

R is the resistance of the filament

So,

$R=\dfrac{V}{I}\\\\R=\dfrac{4.5}{0.16}\\\\R=28.125\ \Omega$

So, the resistance of the filament is equal to 28.125 ohms.

5 0

3 years ago

A very long, uniformly charged cylinder has radius R and linear charge density λ. Find the cylinder's electric field strength ou

mixer [17]

The cylinder's electric field magnitude, at a distance <em>r</em> from the axis of the cylinder (greater than the cylinder's radius), is equal to $E= \frac{\lambda}{2\pi \epsilon_0 \cdot r}$

<h3>Further explanation</h3>

Matter is the building block of everything that we encounter in our lives. Matter is made of atoms, which are in turn made of tiny particles which are called electrons, protons, and neutrons. The ammount of these 3 elements, and their topological configuration in the atoms, is what determines what a certain element is (like Carbon, Hydrogen, Iron, etc).

In some cases, some elements may lose or gain some electrons. Regarded that this missing or extra electrons are not very high in number, the material doesn't lose any of its properties, however it will always try to get its number of electrons back to normal. This is when we say that an element has a <em>charge</em>, which is a measure of how much electrons a body needs to get back to normal. A body has positive charge if it lacks electrons, and has negative charge if it has extra electrons.

This charge causes the material to have an Electric field, which is a measure of how much does it attract or repel electrons. In the case of our problem, we need to compute exactly that, the Electric field. In our problem, we have an infinitely long cylinder with a linear charge density $\lambda$ , this means that all parts of the cylinder have the same charge, and due to symmetry, the electric field is constant on the angular and longitudinal directions of the cylinder.

This makes easy to apply Gauss' Law, since for a Gaussian curve in the shape of a concentric cylinder (with a higher radius than that of our charged cylinder) we can write:

$\Phi = \frac{\lambda \cdot L}{\epsilon_0}$

Where $\Phi$ is called the Electric flux. Since the electric field is constant for a given distance <em>r</em> from the axis of the cylinder we can write that:

$\Phi = E \cdot 2\pi r \cdot L$

Joining both our expressions we can get that:

$E= \frac{\lambda}{2\pi \epsilon_0 \cdot r}$

<h3 /><h3>Learn more</h3>

Description on Electric fields: brainly.com/question/8971780
Relation between electric fields and magnetism: brainly.com/question/2838625
How can we use electric charges: brainly.com/question/10427437

<h3>Keywords</h3>

Electrons, protons, electric field, cylinder, electric flux

5 0

3 years ago

Read 2 more answers

What is the most likely outcome of increasing the number of slits per unit distance on a diffraction grating?1) lines become nar

Nina [5.8K]

It's important to know that diffraction gratings can be identified by the number of lines they have per centimeter. Often, more lines per centimeter is more useful because the images separation is greater when this happens. That is, the distance between lines increases.

<h2>Therefore, the answer is 2.</h2>

8 0

2 years ago