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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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During Cardiovascular Excercise, what do the heart and lungs do?

skad [1K]

They circulate blood and oxygen throughout your body

6 0

3 years ago

An electron is accelrated by a unifor electric field (1000v/m) pointing vertically upward. Use energy methods to get the magnitu

ExtremeBDS [4]

Explanation:

In the given situation two forces are working. These are:

1) Electric force (acting in the downward direction) = qE

2) weight (acting in the downward direction) = mg

Therefore, work done by all the forces = change in kinetic energy

Hence, $qE \times S + mg \times S = 0.5 \times mv^{2}$

$1.6 \times 10^{-19} \times 1000 + 9.1 \times 10^{-31} \times 9.8 \times (\frac{0.10}{100}) = 0.5 \times 9.1 \times 10^{-31} \times v^{2}$

It is known that the weight of electron is far less compared to electric force. Therefore, we can neglect the weight and the above equation will be as follows.

$(1.6 \times 10^{-19} \times 1000) \times (\frac{0.10}{100}) = 0.5 \times 9.1 \times 10^{-31} \times v^{2
}$

v = $sqrt{\frac{1.6 \times 10^{-19}}{(0.5 \times 9.1 \times 10^{-31})}$

= 592999 m/s

Since, the electron is travelling downwards it means that it looses the potential energy.

8 0

3 years ago

Merry-go-rounds are a common ride in park playgrounds. The ride is a horizontal disk that rotates about a vertical axis at their

Vera_Pavlovna [14]

Answer:

A = 2.36m/s

B = 3.71m/s²

C = 29.61m/s2

Explanation:

First, we convert the diameter of the ride from ft to m

10ft = 3m

Speed of the rider is the

v = circumference of the circle divided by time of rotation

v = [2π(D/2)]/T

v = [2π(3/2)]/4

v = 3π/4

v = 2.36m/s

Radial acceleration can also be found as a = v²/r

Where v = speed of the rider

r = radius of the ride

a = 2.36²/1.5

a = 3.71m/s²

If the time of revolution is halved, then radial acceleration is

A = 4π²R/T²

A = (4 * π² * 3)/2²

A = 118.44/4

A = 29.61m/s²

7 0

3 years ago

A ball is dropped from an upper floor, some unknown distance above your apartment. As you look out of your window, which is 1.50

laila [671]

Answer:

The ball is dropped at a height of 9.71 m above the top of the window.

Explanation:

<u>Given:</u>

Height of the window=1.5 m
Time taken by ball to cover the window height=0.15

Now using equation of motion in one dimension we have

$s=ut+\dfrac{at^2}{2}$

Let u be the velocity of the ball when it reaches the top of the window

then

$1.5=0.15u+\dfrac{9.8\times0.15^2}{2}\\u=3.96\ \rm m/s\\$

Now u is the final velocity of the ball with respect to the top of the building

so let t be the time taken for it to reach the top of the window with this velocity

$3.96=gt\\t=0.4\ \rm s\\$

Let h be the height above the top of the window

$h=\dfrac{gt^2}{2}\\\\h=\dfrac{9.8\times 0.4^2}{2}\\h=9.71\ \rm m$

3 0

3 years ago

A box is moving along the x-axis and its position varies in time according to the expression:

Colt1911 [192]

Answer:

38.4 m/s

Explanation:

a) at t = 3.2s. $x = 6 * 3.2^2 = 61.44 m$

b) at t = 3.2 + Δt. $x = 6*(3.2 + \Delta t)^2$

c) As Δt approaches 0. We can find the velocity by the ratio of Δx/Δt

$v = \frac{\Delta x}{\Delta t} = \frac{x_2 - x_1}{\Delta t}$

$v = \frac{6*(3.2 + \Delta t)^2 - 61.44}{\Delta t}$

$v = \frac{6(3.2^2 + 6.4\Delta t + \Delta t^2) - 61.44}{\Delta t}$

$v = \frac{61.44 + 38.4\Delta t + \Delta t^2 - 61.44}{\Delta t}$

$v = \frac{\Delta t(38.4 + \Delta t)}{\Delta t}$

$v = 38.4 + \Delta t$

As Δt approaches 0, v = 38.4 + 0 = 38.4 m/s

3 0

3 years ago