All categories

3 years ago

How does energy move through the atmosphere hydrosphere and biosphere

1 answer:

5 0

When the Sun’s energy moves through space, it reaches Earth’s atmosphere and finally the surface. This radiant solar energy warms the atmosphere and becomes heat energy. This heat energy is transferred throughout the planet’s systems in three ways: by radiation, conduction, and convection

You might be interested in

The kinetic energy of a proton and that of an a-particle are 4 eV and 1 eV,

madam [21]

Answer:

(b) 1:1

Explanation:

1. Formulae:

E = hf

E = h.v/λ

λ = h/(mv)

E = (1/2)mv²

Where:

E = kinetic energy of the particle
λ = de-Broglie wavelength
m = mass of the particle
v = speed of the particle
h = Planck constant

2. Reasoning

An alha particle contains 2 neutrons and 2 protons, thus its mass number is 4.

A proton has mass number 1.

Thus, the relative masses of an alpha particle and a proton are:

$\dfrac{m_\alpha}{m_p}=4$

For the kinetic energies you find:

$\dfrac{E_\alpha}{E_p}=\dfrac{m_\alpha \times v_\alpha^2}{m_p\times v_p^2}$

$\dfrac{1eV}{4eV}=\dfrac{4\times v_\alpha^2}{1\times v_p^2}\\\\\\\dfrac{v_p^2}{v_\alpha^2}=16\\\\\\\dfrac{v_p}{v_\alpha}=4$

Thus:

$\dfrac{m_\alpha}{m_p}=4=\dfrac{v_p}{v__\alpha}$

$m_\alpha v_\alpha=m_pv_p$

From de-Broglie equation, λ = h/(mv)

$\dfrac{\lambda_p}{\lambda_\alpha}=\dfrac{m_\lambda v_\lambda}{m_pv_p}=\dfrac{1}{1}=1:1$

5 0

3 years ago

Hydrogen bonds are too weak to bind atoms together to form molecules, but they do hold different parts of a single large molecul

pogonyaev

Hydrogen bonds are too weak to bind atoms together to form molecules, but they do hold different parts of a single large molecule in a specific three-dimensional shape. The given statement is true.

<h3>What are hydrogen bonds?</h3>

A hydrogen bond is an electrostatic force of attraction among a hydrogen atom tightly attached to a more electronegative "donor" atom or group and another electronegative atom bearing a lone pair of electrons, known as the hydrogen bond acceptor.

Hydrogen bonds are too flimsy to connect atoms to form molecules, but they do hold various portions of a single large molecule together in a specific three-dimensional shape.

Thus, the given statement is true.

For more details regarding hydrogen bonding, visit:

brainly.com/question/10904296

#SPJ1

5 0

1 year ago

1. The horizontal and vertical components of a projectile's velocity are

Anni [7]

The horizontal and vertical components of a projectile's velocity are independent of each other.

Answer: Option C

Explanation:

The path of a projectile is determined by two components of motion. They are termed as horizontal and the vertical components. Since both components velocity are perpendicular to each other, so it can stated that they are independent of each other.

Even it can seen that when the horizontal components of velocity is constant, then there will be change in the vertical components of velocity leading to free fall projectile path.

And in the absence of gravity, there will be change in the horizontal components of velocity with zero vertical component of velocity. Thus, the horizontal and the vertical components of a projectile’s velocity are seemed to be independent of each other.

5 0

3 years ago

A: hydrogen and hydrogen b: copper and copper c: copper and oxygen Rank the above bonds in terms of increasing bond strength. Ch

pashok25 [27]

Answer: Copper and oxygen

Explanation:

Copper and oxygen shares the ionic bond. As we know that ionic bond is the most strongest bond. Here is the order:

Hydrogen bond< Metallic bond< Ionic bond.

That means order in terms of increasing bond strength is :

Hydrogen and hydrogen< Copper and copper< Copper and oxygen.

5 0

3 years ago

Two equal charges with magnitude Q and Q experience a force of 12.3442 when held at a distance r. What is the force between two

andre [41]

Answer:

197.5072.

Explanation:

According to the Coulomb's law, the magnitude of the electrostatic force of interaction between two charges $\rm q_1$ and $\rm q_2$ which are separated by the distance $\rm d$ is given by

$\rm F = \dfrac{kq_1q_2}{d^2}.$

where, k is the Coulomb's constant.

For the case, when,

$\rm q_1 = Q.$
$\rm q_2 = Q.$
$\rm d=r.$
$\rm F=12.3442.$

Then, using Coulomb's law,

$\rm 12.3442 = \dfrac{kQQ}{r^2}=\dfrac{kQ^2}{r^2}\ \ \ \ .......\ (1).$

For the case, when,

$\rm q_1 = 2Q.$
$\rm q_2 = 2Q.$
$\rm d=\dfrac r2.$

Then, using Coulomb's law, the new electric force between the charges is given by,

$\rm F' = \dfrac{k(2Q)(2Q)}{\left (\dfrac r2\right )^2}\\=\dfrac{k\ 4Q^2}{\dfrac{r^2}{4}}\\=4\times 4 \times \dfrac{kQ^2}{r^2}\\=16\ \dfrac{kQ^2}{r^2}\\=16\times 12.3442\ \ \ \ \ \ \ \ (Using\ (1))\\=197.5072.$

8 0

3 years ago