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3 years ago

If a North pole and a South pole cross what will the results be?

1 answer:

Taya2010 [7]3 years ago

8 0

If the north and south pole crosses, radiations can enter Earth's atmosphere easily, continents will lurch in one directions causing natural calamities and species extinction.

<u>Explanation:</u>

The North and South pole forms the Earth's magnetic field which protects us from harmful radiation entering the atmosphere. If these poles get flipped, a global cataclysm may occur.
The continents will move to one direction causing massive earthquakes, rapid change of weather and climate and results in species extinction.
Over the past 150 years, the magnetic poles are casually moving but now it is moving very fast which may result in a flip of the magnetic poles.
Earth's magnetic field is what shields us from harmful space radiation which can damage cells, causing cancer, and fry electronic circuits and electrical grids. With a weaker field in place, This will disrupt the internal compass and many navigations may get interrupted

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The radius of a planet is 2400 km, and the acceleration due to gravity at its surface is 3.6 m/s2.

kiruha [24]

Answer:

$3.1\cdot10^{23}\:\mathrm{kg}$

Explanation:

We can use Newton's Universal Law of Gravitation to solve this problem:

$g_P=G\frac{m}{r^2}$ ., where $g_P$ is acceleration due to gravity at the planet's surface, $G$ is gravitational constant $6.67\cdot 10^{-11}$ , $m$ is the mass of the planet, and $r$ is the radius of the planet.

Since acceleration due to gravity is given as $m/s^2$ , our radius should be meters. Therefore, convert $2400$ kilometers to meters:

$2400\:\mathrm{km}=2,400,000\:\mathrm{m}$ .

Now plugging in our values, we get:

$3.6=6.67\cdot10^{-11}\frac{m}{(2,400,000)^2}$ ,

Solving for $m$ :

$m=\frac{2,400,000^2\cdot3.6}{6.67\cdot 10^{-11}},\\m=\fbox{$3.1\cdot10^{23}\:\mathrm{kg}$}$ .

6 0

2 years ago

When a 0.350-kg package is attached to a vertical spring and lowered slowly, the spring stretches 12.0 cm. The package is now di

AleksandrR [38]

To solve the problem it is necessary to use the concepts related to the calculation of periods by means of a spring constant.

We know that by Hooke's law

$F=kx$

Where,

k = Spring constant

x = Displacement

Re-arrange to find k,

$k= \frac{F}{x}$

$k= \frac{mg}{x}$

$k= \frac{(0.35)(9.8)}{12*10^{-2}}$

$k = 28.58N/m$

Perioricity in an elastic body is defined by

$T = 2\pi \sqrt{\frac{m}{k}}$

Where,

m = Mass

k = Spring constant

$T = 2\pi \sqrt{\frac{0.35}{28.58}}$

$T = 0.685s$

Therefore the period of the oscillations is 0.685s

4 0

2 years ago

Which statement describes the horizontal acceleration of a projectile? It is –9.8 m/s2. It is 9.8 m/s2. It is constant. It is ze

lina2011 [118]

In a projectile, the horizontal acceleration is zero. The velocity remains constant at all times. However, the <u>vertical acceleration</u> is -9.81m/s^2.

Hope this helps!

6 0

3 years ago

Read 2 more answers

Acceleration usually has the symbol a. It is a vector. What is the correct way

Ghella [55]

Explanation:

a straight line under the letter