Which core has highest temperature

Which of the following statements is true about earth's magnetic poles

Dahasolnce [82]

Answer: b) they are the areas where Earth's magnetic field is weakest

Explanation:

According to classical physics, a magnetic field always has two associated magnetic poles (north and south), the same happens with magnets. This is because for <em>classical physics</em>, naturally, magnetic monopoles can not exist.

In this context, Earth is similar to a magnetic bar with a north pole and a south pole. This means, the axis that crosses the Earth from pole to pole is like a big magnet.

Now, by convention, on all magnets the north pole is where the magnetic lines of force leave the magnet and the south pole is where the magnetic lines of force enter the magnet. Then, for the case of the Earth, the north pole of the magnet is located towards the geographic south pole and the south pole of the magnet is near the geographic north pole.

Being the magnetic poles the places where the Earth's magnetic field is weakest. And it is for this reason, moreover, that the magnetic field lines enter the Earth through its magnetic south pole (which is the geographic north pole).

4 0

3 years ago

The current theory of the structure of the

IRISSAK [1]

1) The mass of the continent is $3.3\cdot 10^{21} kg$

2) The kinetic energy of the continent is 624 J

3) The speed of the jogger must be 4 m/s

Explanation:

1)

We start by finding the volume of the continent. We have:

$L = 5850 km = 5.85\cdot 10^6 m$ is the side

$t = 35 km = 3.5\cdot 10^4 m$ is the depth

So the volume is

$V=L^2 t = (5.85\cdot 10^6)^2 (3.5\cdot 10^4)=1.20\cdot 10^{18} m^3$

We also know that its density is

$d=2750 kg/m^3$

Therefore, we can find the mass by multiplying volume by density:

$m=dV=(2750)(1.20\cdot 10^{18})=3.3\cdot 10^{21} kg$

2)

The kinetic energy of the continent is given by:

$K=\frac{1}{2}mv^2$

where

$m=3.3\cdot 10^{21} kg$ is its mass

v = 3.2 cm/year is the speed

We have to convert the speed into m/s. We have:

3.2 cm = 0.032 m

$1 year = 1(365)(24)(60)(60)=3.15\cdot 10^7 s$

So, the speed is:

$v=\frac{0.032 m}{3.15 \cdot 10^7 s}=1.02\cdot 10^{-9} m/s$

So, we can now find the kinetic energy:

$K=\frac{1}{2}(1.20\cdot 10^{21})(1.02\cdot 10^{-9})^2=624 J$

3)

Here we have a jogger of mass

m = 78 kg

And the jogger has the same kinetic energy of the continent, so

K = 624 J

The kinetic energy of the jogger is given by

$K=\frac{1}{2}mv^2$

where v is the speed of the jogger.

Solving for v, we find the speed that the jogger must have:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(624)}{78}}=4 m/s$

Learn more about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

3 0

3 years ago

In a solid , the atoms are tightly locked in position an do not change posistion , true or false

attashe74 [19]

the answer is true because they're so close together that they can't move they can't slide past each other or anything

6 0

3 years ago

Energy from the Sun arrives at the top of the Earth's atmosphere with an intensity of 1.36 kW/m 2 . How long does it take for 3.

alina1380 [7]

Answer:

The time taken by $3.55\times 10^9\ J$ to arrive on an area of $4.25\ m^2$ is $6.14\times 10^5\ seconds$ .

Explanation:

Given the intensity of the energy is $1.36\ kW/m^2$

And the arriving energy is $3.55\times 10^9\ J$

Also, the area in which energy is being arriving is $4.25\ m^2$

Now, we will use relation between energy $(E)$ , intensity of energy $(p)$ , area $(A)$ and time $(T)$ .

Where energy is in Joule, intensity is in $kW/m^2$ , area is in $m^2$ and time is in seconds.

The equation is

$E=pAT\\\\T=\frac{E}{pA}\\\\T=\frac{3.55\times 10^9}{1.36\times 10^3\times 4.25}\\\\T=0.614\times 10^6\ s\\T=6.14\times 10^5\ seconds$

So, the time taken by $3.55\times 10^9\ J$ to arrive on an area of $4.25\ m^2$ is $6.14\times 10^5\ seconds$ .

4 0

3 years ago

A box of mass 12 kg is at rest on a flat floor. The coefficient of static friction between the box and floor is 0.42. What is th

Oxana [17]

The maximum static force that can be applied is equal to the normal force*the frictional force. the normal force on the box is equal to mg since the floor is flat using 9.81m/s^2 for gravity 12kg*9.81m/s^2 = 118N multiplying the normal force by the frictional force you get a 118*.42= 49.6N so overcome the force of static friction on the box a minimum of 49.6N would need to be applied.

5 0

3 years ago

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Which core has highest temperature​

Which core has highest temperature