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3 years ago

12

Electrically charged sunspot gases which escape the sun's chromosphere and enter the earth's atmosphere near the magnetic north

pole cause the _______.

1 answer:

Kisachek [45]3 years ago

6 0

Answer:

Northern Lights ( Aurora Borealis)

Explanation:

When the electricaly charged sunspot gases (they are named a solar wind) escape the sun's chromosphere and penetrates from the earth magnetic sheild which is called earth's magnetosphere then upon there interaction with atoms and molecules of our atmosphere there are little bursts of photons in the form of light which made up these northern lights.

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Calculate the Latent Heat of Vaporization. (Please see picture attached)

Hunter-Best [27]

Answer:

20 J/g

Explanation:

In this question, we are required to determine the latent heat of vaporization

To answer the question, we need to ask ourselves the questions:

What is latent heat of vaporization?

It is the amount of heat required to change a substance from its liquid state to gaseous state without change in temperature.

It is the amount of heat absorbed by a substance as it boils.

How do we calculate the latent heat of vaporization?

Latent heat is calculated by dividing the amount of heat absorbed by the mass of the substance.

In this case;

Mass of the substance = 20 g
Heat absorbed as the substance boils is 400 J (1000 J - 600 J)

Thus,

Latent heat of vaporization = Quantity of Heat ÷ Mass

= 400 Joules ÷ 20 g

= 20 J/g

Thus, the latent heat of vaporization is 20 J/g

7 0

3 years ago

A 980 kg roller coaster cart is traveling along a track at 17 m/s before it rolls down a 30 m tall hill (Point A). What will be

MrRissso [65]

The kinetic energy halfway the hill is $2.86\cdot 10^5 J$

Explanation:

If there are no friction forces acting on the cart, we can apply the law of conservation of energy: the mechanical energy of the cart (which is the sum of potential energy + kinetic energy) must be conserved. So we can write:

$U_A +K_A = U_B + K_B$

where

$U_A=mgh_A$ is the initial potential energy, at point A, with

m = 980 kg (mass of the cart)

$g=9.8 m/s^2$ (acceleration of gravity)

$h_A = 30 m$ (height at point A)

$K_A=\frac{1}{2}mv_A^2$ is the initial kinetic energy, at point A , with

$v_A=17 m/s$ (velocity at point A)

$U_B=mgh_B$ is the final potential energy, at point B, where

$h_B = 15 m$ (height at point B)

$K_B=\frac{1}{2}mv_B^2$ is the final kinetic energy, at point B, where

$v_B$ is the velocity at point B

Here we are interested in finding $K_B$ , so by re-arranging the equation and substituting we find:

$K_B = U_A+K_B-U_B = mg(h_A-h_B)+\frac{1}{2}mv_A^2=(980)(9.8)(30-15)+\frac{1}{2}(980)(17)^2=2.86\cdot 10^5 J$

Learn more about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

8 0

3 years ago

Determine how many times per second it would move back and forth across a 6.0-m-long room on the average, assuming it made very

timofeeve [1]

Answer:

The right solution is "24.39 per sec".

Explanation:

According to the question,

⇒ $v=\frac{502.1}{\sqrt{3} }$

$=289.9 \ m/s$

The time will be:

⇒ $t=\frac{d}{v}$

$=\frac{2\times 6}{289.9}$

$=\frac{12}{289.9}$

$=0.041 \ sec$

hence,

⇒ $N=\frac{1}{t}$

$=\frac{1}{0.041}$

$=24.39 \ per \ sec$

4 0

2 years ago

Calculate the force that the 4kg block exerts on the 10kg block

Kryger [21]

Acceleration=force/mass=28/(10+4)=2m/s^2

force10kg=ma=10*2
force4kg=ma=(10*2)=20
the4 kg is pushing against the 10kg block

vf=vi+at
-10=20*28/14 * t
t=30/2=15sec

i hope this can help you.

8 0

3 years ago

A spring with a rest length of 0.7 m has a spring constant of 70 N/m. It is stretched and now has a length of 2.5 m. What is the

pentagon [3]

Answer:

<em>113.4 J</em>

Explanation:

<u>Elastic Potential Energy</u>

Is the energy stored in an elastic material like a spring of constant k, in which case the energy is proportional to the square of the change of length Δx and the constant k.

$\displaystyle PE = \frac{1}{2}k(\Delta x)^2$

The spring has a natural length of 0.7 m and a spring constant of k=70 N/m. When the spring is stretched to a length of 2.5 m, the change of length is

Δx = 2.5 m - 0.7 m = 1.8 m

The energy stored in the spring is:

$\displaystyle PE = \frac{1}{2}70(1.8)^2$

PE = 113.4 J

7 0

3 years ago