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katrin [286]
2 years ago
6

Describe an Activity to show magnetic poles doesn't seperated​

Physics
1 answer:
mojhsa [17]2 years ago
7 0

Answer:

We know that pole is the point where the strength of the magnet is maximum. So more and more iron particles will be attracted at poles of a magnet when we bring a magnet near the iron particles. We will observe the crowdness of particles at the ends of magnet. This indicates the presence of two poles in a magnet. Hence poles are present in a magnet in pair. If a magnet is divided into two parts, each part also possesses a pair of poles.

Two properties of a magnet are: A magnet always has two poles: north pole and south pole. Like magnetic poles repel each other and unlike magnetic poles attract each other.

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Deductive reasoning?
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i hope this helps :)
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How long does a player sit out of a game of handball for committing a second or third foul?
yulyashka [42]

Answer:

Walking’ - If a handball player takes more than three steps without dribbling (bouncing the ball) or holds the ball for more than 3 seconds without bouncing it, shooting or passing, then that is deemed ‘walking' and possession is lost.

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Askmeanything2♡

5 0
2 years ago
An airplane of mass 1.60 ✕ 104 kg is moving at 66.0 m/s. The pilot then increases the engine's thrust to 7.70 ✕ 104 N. The resis
Ivan

(a) No, because the mechanical energy is not conserved

Explanation:

The work-energy theorem states that the work done by the engine on the airplane is equal to the gain in kinetic energy of the plane:

W=\Delta K (1)

However, this theorem is only valid if there are no non-conservative forces acting on the plane. However, in this case there is air resistance acting on the plane: this means that the work-energy theorem is no longer valid, because the mechanical energy is not conserved.

Therefore, eq. (1) can be rewritten as

W=\Delta K + E_{lost}

which means that the work done by the engine (W) is used partially to increase the kinetic energy of the airplane (\Delta K) and part is lost because of the air resistance (E_{lost}).

(b) 77.8 m/s

First of all, we need to calculate the net force acting on the plane, which is equal to the difference between the thrust force and the air resistance:

F=7.70\cdot 10^4 N - 5.00 \cdot 10^4 N=2.70\cdot 10^4 N

Now we can calculate the acceleration of the plane, by using Newton's second law:

a=\frac{F}{m}=\frac{2.70\cdot 10^4 N}{1.60\cdot 10^4 kg}=1.69 m/s^2

where m is the mass of the plane.

Finally, we can calculate the final speed of the plane by using the equation:

v^2- u^2 = 2aS

where

v=? is the final velocity

u=66.0 m/s is the initial velocity

a=1.69 m/s^2 is the acceleration

S=5.00 \cdot 10^2 m is the distance travelled

Solving for v, we find

v=\sqrt{u^2+2aS}=\sqrt{(66.0 m/s)^2+2(1.69 m/s^2)(5.00\cdot 10^2 m)}=77.8 m/s

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