Answer:
A) The north pole of a bar magnet will attract the south pole of another bar magnet.
B) Earth's geographic north pole is actually a magnetic south pole.
E) The south poles of two bar magnets will repel each other.
Explanation:
<u>According to </u><u>classical physics</u>, a magnetic field always has two associated magnetic poles (north and south), the same happens with magnets. This means that if we break a magnet in half, we will have two magnets, where each new magnet will have a new south pole, and a new north pole.
This is because <u>for classical physics, naturally, magnetic monopoles can not exist. </u>
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.
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).
The experiments will involve two billiard balls of known masses, m₁ and m₂, and velocities u₁ and u₂. The two are allowed to collide and the velocities of the balls after the collision v₁ and v₂ are recorded.
The momentum before and after the collision is then calculated as follows:
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
<h3>What is the statement of the law of conservation of momentum?</h3>
The law of the conservation of momentum states that the momentum before and after collision in a system of colliding bodies is conserved
The momentum of a body is calculated using the formula below:
Momentum = mass * velocity.
Hence, for the two billiard balls, the momentum before and after the collision is conserved.
Learn more about momentum at: brainly.com/question/1042017
#SPJ1
Answer:
The formula i use is called, Product over Sum. Which means it is figured by their multiplied resistances divided by their sum. It is applied by pairs of known resistances. Starting with 20 and 30 Ohms, 600 is divided by 50. Using a quick mental calculation, the first pair has a resistance of 12 Ohms. Then, do that with 12 Ohms and 10 Ohms. 120 Ohms divided by 22. The answer is, about 5.5 Ohms. By this interesting development, we are reminded that resistances in parallel are effectively never more than the least one.
The students decide to assemble a convenient experiment and will run one amp through them all in parallel and measure their voltage. Watching the Amperage gauge on their teacher’s power supply. As one begins to turn it up to an Amp, another is watching its voltage till an Amp is perfectly applied. But as they carefully do that, watching the Amp gauge, another screams, their 10 Ohm resistor turns black and smokes as they were only pumping out 2 or 3 tenths of an Amp. What happened? What did they need, to make this simple experiment not so embarass-king?
Buy room air freshener?
I don’t speak Spanish blah blah
Answer:
E) be two times larger.
Explanation:
As we know that the relation between the resistance and the resistivity of the wire is given as:
![R=\rho.\frac{l}{a}](https://tex.z-dn.net/?f=R%3D%5Crho.%5Cfrac%7Bl%7D%7Ba%7D)
where:
resistivity of the wire
length of wire
area of wire
resistance
Now, when the length of the wire is four times the initial length then for the resistance to remain constant:
![R=\rho.\frac{4l}{a'}](https://tex.z-dn.net/?f=R%3D%5Crho.%5Cfrac%7B4l%7D%7Ba%27%7D)
where:
area of the new wire
![\rho.\frac{l}{a} =\rho.\frac{4l}{a'}](https://tex.z-dn.net/?f=%5Crho.%5Cfrac%7Bl%7D%7Ba%7D%20%3D%5Crho.%5Cfrac%7B4l%7D%7Ba%27%7D)
![a'=4a](https://tex.z-dn.net/?f=a%27%3D4a)
we know that area of the cross section of wire is given as:
![a=\pi.r^2](https://tex.z-dn.net/?f=a%3D%5Cpi.r%5E2)
![\pi.r'^2=4\times \pi.r^2](https://tex.z-dn.net/?f=%5Cpi.r%27%5E2%3D4%5Ctimes%20%5Cpi.r%5E2)
![r'=2r](https://tex.z-dn.net/?f=r%27%3D2r)
Hence the radius must be twice of the initial radius for the resistance to be constant when length is taken four times.