The data in the table above show how far

Part 1: Use complete sentences to explain why solar winds occur. Part 2: Give two examples in which solar winds impact Earth.

Troyanec [42]

Part 1

When the solar atmosphere accumulates a lot of magnetic energy to a point that cannot accumulate more, all that magnetic energy is suddenly released, and with it, a lot of radiation. So much, that in fact it covers all of the electromagnetic spectrum; from radio waves to gamma rays. That burst of radiation is called a solar flare. In a single solar flare the amount of radiation released is millions of times greater than all the nuclear bombs in the face if the earth exploding together. Lucky for us, most of the high-energy radiation dissipates before reaching the Earth, and the radiation that do reach us, is deflected by the Earth’s magnetic field.

Part 2

1. Not all the radiation of solar flares that reach the Earth is deflected by its magnetic field; some of them reach us and charges the upper atmosphere with ionized particles. Those particles react with the gases in the atmosphere and produce a light; that light is what we call Auroras borealis or southern nights; One the most beautiful natural spectacles in earth, who thought Auroras begin their lives as deadly solar flares.

2. Solar flares contain a lot of high-energy radiation that is extremely dangerous for our electronic devices; when they reach the Earth, they can damage sensible electronics like satellites. A very powerful solar flare could even damage all the electronic devices on the surface of the Earth.

4 0

3 years ago

PLS ANSWER FAST WILL GIVE BRAINLY!!!! ANSWER ONLY IF YOU KNOW!!!!

timurjin [86]

Answer:

The nucleus has an overall positive charge as it contains the protons. Every atom has no overall charge (neutral). This is because they contain equal numbers of positive protons and negative electrons. These opposite charges cancel each other out making the atom neutral.

5 0

3 years ago

Derive the equation of motion of the block of mass m1 in terms of its displacement x. The friction between the block and the sur

Alenkasestr [34]

Answer:

the equivalent mass : $m_e = m_1+m_2+\frac{I}{R^2}$

the equation of the motion of the block of mass $m_1$ in terms of its displacement is = $(m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)$

Explanation:

Let use m₁ to represent the mass of the block and m₂ to represent the mass of the cylinder

The radius of the cylinder be = R

The distance between the center of the pulley to center of the block to be = x

Also, the angles of inclinations of the cylinder and the block with respect to the ground to be $\phi$ and $\beta$ respectively.

The velocity of the block to be = v

The equivalent mass of the system = $m_e$

In the terms of the equivalent mass, the kinetic energy of the system can be written as:

$K.E = \frac{1}{2} m_ev^2$ --------------- equation (1)

The angular velocity of the cylinder = $\omega$ : &

The inertia of the cylinder about its center to be = I

The angular velocity of the cylinder can be written as:

$v = \omega R$

$\omega =\frac{v}{R}$

The kinetic energy of the system in terms of individual mass can be written as:

$K.E = \frac{1}{2}m_1v^2+\frac{1}{2} m_2v^2+\frac{1}{2}I\omega^2$

By replacing $\omega$ with $\frac{v}{R}$ ; we have:

$K.E = \frac{1}{2}m_1v^2+\frac{1}{2} m_2v^2+\frac{1}{2}I(\frac{v}{R})^2$

$K.E = \frac{1}{2}(m_1+ m_2+ \frac{I}{R} )v^2$ ------------------ equation (2)

Equating both equation (1) and (2); we have:

$m_e = m_1+m_2+\frac{I}{R^2}$

Therefore, the equivalent mass : $m_e = m_1+m_2+\frac{I}{R^2}$ which is read as;

The equivalent mass is equal to the mass of the block plus the mass of the cylinder plus the inertia by the square of the radius.

The expression for the force acting on equivalent mass due to the block is as follows:

$f_{block }=m_1gsin \beta$

Also; The expression for the force acting on equivalent mass due to the cylinder is as follows:

$f_{cylinder} = m_2gsin \phi$

Equating the above both equations; we have the equation of motion of the equivalent system to be

$m_e \bar x = f_{cylinder}-f_{block}$

which can be written as follows from the previous derivations

$(m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)$

Finally; the equation of the motion of the block of mass $m_1$ in terms of its displacement is = $(m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)$

8 0

3 years ago

Water ice has a density of 0.91 g/cm2, so it will float in liquid water. Imagine you have a cube of ice, 10 cm on a side. a. Wha

Reptile [31]

Answer:

(i) W = 8.918 N

(ii) $V = 9.1 \times 10^{-4} m^3$

(iii) d = 9.1 cm

Explanation:

Part a)

As we know that weight of cube is given as

$W = mg$

$W = \rho V g$

here we know that

$\rho = 0.91 g/cm^3$

$Volume = L^3$

$Volume = 10^3 = 1000 cm^3$

now the mass of the ice cube is given as

$m = 0.91 \times 1000 = 910 g$

now weight is given as

$W = 0.910 \times 9.8 = 8.918 N$

Part b)

Weight of the liquid displaced must be equal to weight of the ice cube

Because as we know that force of buoyancy = weight of the of the liquid displaced

$W_{displaced} = 8.918 N$

So here volume displaced is given as

$\rho_{water}Vg = 8.918$

$1000(V)9.8 = 8.918$

$V = 9.1 \times 10^{-4} m^3$

Part c)

Let the cube is submerged by distance "d" inside water

So here displaced water weight is given as

$W = \rho_{water} (L^2 d) g$

$8.918 = 1000(0.10^2 \times d) 9.8$

$d = 0.091 m$

so it is submerged by d = 9.1 cm inside water

4 0

3 years ago

What value do we use to describe acceleration due to gravity

andrey2020 [161]

Answer:

9.8 m/s2

Explanation:

In the first equation above, g is referred to as the acceleration of gravity. Its value is 9.8 m/s2 on Earth. That is to say, the acceleration of gravity on the surface of the earth at sea level is 9.8 m/s2.

Got it from the internet, hope it helps though ^^

7 0

3 years ago