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3 years ago

Which of these statements best describes an effect of changes in Earth's magnetic field?

Physics

2 answers:

Alona [7]3 years ago

8 0

Answer:

The magnetic north and south poles of Earth are interchanged.

Explanation:

Earth acts as a huge bar magnet due to the presence of iron and nickel in its core. The magnetic field surrounding Earth is known as magnetosphere which constantly protects from solar wind. Auroras are visible in the sky when charged particles from the sun interact with the magnetic field of the earth especially near the poles where strength of the magnetic field is maximum.

Over the years, the magnetic field might weaken as it is continuously repelling the charged particles from the Sun. As a consequence, the poles of this magnetic field might reverse.

Thus, changes in Earth's magnetic field can reverse the poles.

Sveta_85 [38]3 years ago

6 0

The only statement that makes sense as to which one describes an effect of changes in Earth's magnetic field is the first one - Southern lights are seen at high altitude places like Alaska.
Normally, Southern lights are seen around the South Pole, which means that it is quite odd for them to be seen in Alaska, which is near the North Pole.

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3 years ago

A train travels 120 km In 2 hours and 30 minutes what’s its average speed

dangina [55]

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3 years ago

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Different _______ of light through two separate mediums causes the bending of wave fronts associated with light rays.

eimsori [14]

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3 years ago

(a) What is the current involved when a truck battery sets in motion 720 C of charge in 4.00 s while starting an engine? (b) How

OLEGan [10]

Answer:

(a) Current flowing through truck battery is 180 A

(b) Time taken in calculator is 333.33 s

Explanation:

(a) Given:

The charge on the truck battery,q = 720 C

Time, t = 4.00 s

Consider I be the current flowing through truck battery.

The relation between current, charge and time is:

I = q/t

Substitute the suitable values in the above equation.

$I=\frac{720}{4}$

I = 180 A

(b) Given:

The charge on the calculator,q = 7.00 C

The current flowing through calculator, I = 0.3 mA = 0.3 x 10⁻³ A

Consider t be the time.

The relation between current, charge and time is:

t = q/I

Substitute the suitable values in the above equation.

$t=\frac{1}{0.3\times10^{-3} }$

I = 333.33 s

8 0

3 years ago

(b) A ball is thrown from a point 1.50 m above the ground. The initial velocity is 19.5 m/s at

Nataly_w [17]

The answers are:

(i) 6.35m

(ii) 20.2 m/s

It seems like you already have the answer, but let me show you how to get it:

You have two givens:

Vi = 19.5m/s

Θ = 30°

dy = 1.50m (This is not the maximum height, just to be clear)

When working with these types of equations, you just need to know your kinematics equations. For projectiles launched at an angle, you will need to first break down the initial velocity (Vi) into its horizontal (x) and vertical (y) components.

*Now remember this, if you are solving for something in the horizontal movement always use only x-components. When solving for vertical movements, always use y-components.

Let's move on to breaking down the initial velocity into both y and x components.

Viy = SinΘVi = (Sin30°)(19.5m/s) = 9.75 m/s

Vix = CosΘVi = (Cos30°)(19.5m/s) = 16.89 m/s

Okay, so we have that down now. The next step is to decide which kinematics equation you will use. Because you have no time, you need to use the kinematic equation that is not time dependent.

(i) Maximum height above the ground

Remember that the object was thrown 1.50m above the ground. So we save that for later. First we need to solve for the maximum height above the horizontal, or the point where it was thrown.

The kinematics equation you will use is:

$Vf^{2} = Vi^{2}+2ad$

Where:

Vf = final velocity

Vi = inital velocity

a = 9.8m/s²

d = displacement

We will derive our displacement from this equation. And you will come up with this:

$d = \dfrac{Vf^{2}-Vi^{2}}{2a}$

Again, remember that we are looking for a vertical component or y-component because we are looking for HEIGHT. So we use this plugging in vertical values only.

Vf at maximum height is always 0m/s because at maximum height, objects stop. Also because gravity is a downwards force you will use -9.8m/s².

Vfy = 0 m/s a = -9.8m/s² Viy = 9.75m/s

$dy = \dfrac{Vfy^{2}-Viy^{2}}{2a}$

$dy = \dfrac{0^{2}-(9.75m/s)^{2}}{2(-9.8m/s^{2})}$

$dy = \dfrac{-95.0625m^{2}/s^{2}}{-19.6m/s^{2}}$

$dy = 4.85m$

So from the point it was thrown, it reached a height of 4.85m. Now we add that to the height it was thrown to get the MAXIMUM HEIGHT ABOVE THE GROUND.

4.85m + 1.50m = 6.35m

(ii) Speed before it strikes the ground. (Vf=resultant velocity)

Okay, so here we need to consider a couple of things. To get the VF we need to first figure out the final velocities of both the x and y components. We are combining them to get the resultant velocity.

Vfx = horizontal velocity = Initial horizontal velocity (Vix). This is because gravity is not acting upon the horizontal movement so it remains constant.

Vfy = ?

VF =?

We need to solve this, again, using the same formula, but this time, you need to consider we are moving downwards now. So this time, instead of Vfy being 0 m/s, Viy is now 0 m/s. This is because it started moving from rest.

$Vfy^{2} = Viy^{2}+2ad$