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Novosadov [1.4K]
3 years ago
9

When you drop a strong magnet through the center of a copper pipe, what happens to the copper piper and the magnet?

Physics
1 answer:
Sedbober [7]3 years ago
8 0
Copper is; unlike iron and steel; not ferromagnetic, but diamagnetic. This means that induced magnetic fields in copper will counter the applied force.

When you drop a strong magnet through a copper pipe, the moving magnetic field will induce currents (Lenz’ Law). These currents will now induce their own magnetic field. This magnetic field counters the falling magnetic.

Result: the magnet will fall way slower than if it was falling through a plastic pipe.
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Which scenario will drain the battery in the circuit diagram fastest: three 20 W bulbs in positions one, two, and three or 80 W
kakasveta [241]

Answer:

See the explanation below.

Explanation:

If you connect the three bulbs 20 W each will have a total power of 60W.

Now we need to understand to assign the meaning of the word gap, that is, if the circuit is open at that point or if there is no bulb connected at that point.

If the circuit is open at Point 2, there will be no current in the circuit, so the battery will drain faster with the three bulbs 20W.

In the second event, where gap means that there are no bulbs connected at that point, it means that you have two bulbs connected in series of 80W each.

In this case the bulbs will consume 160W thus drain the battery faster than the three 20W bulbs connected in series.

3 0
3 years ago
A centrifuge in a medical laboratory rotates at an angular speed of 3,500 rev/min, clockwise (when viewed from above). When swit
Ratling [72]

Answer:

The magnitude of angular acceleration is 232.38\ rad/s^2.

Explanation:

Given that,

Initial angular velocity, \omega_i=3500\ rev/min=366.5\ rad/s

When it switched off, it comes o rest, \omega_f=0

Number of revolution, \theta=46=289.02\ rad

We need to find the magnitude of angular acceleration. It can be calculated using third equation of rotational kinematics as :

\omega_f^2-\omega_i^2=2\alpha \theta\\\\\alpha =\dfrac{-\omega_i^2}{2\theta}\\\\\alpha =\dfrac{-(366.51)^2}{2\times 289.02}\\\\\alpha =-232.38\ rad/s^2  

So, the magnitude of angular acceleration is 232.38\ rad/s^2. Hence, this is the required solution.

6 0
3 years ago
A 4.79 g bullet moving at 642.3 m/s penetrates a tree trunk to a depth of 4.35 cm. Use work and energy considerations to find th
vladimir1956 [14]

Answer:

Force, F=2.27\times 10^4\ N

Explanation:

Given that,

Mass of the bullet, m = 4.79 g = 0.00479 kg

Initial speed of the bullet, u = 642.3 m/s

Distance, d = 4.35 cm = 0.0435 m

To find,

The magnitude of force required to stop the bullet.

Solution,

The work energy theorem states that the work done is equal to the change in its kinetic energy. Its expression is given by :

F.d=\dfrac{1}{2}m(v^2-u^2)

Finally, it stops, v = 0

F.d=-\dfrac{1}{2}m(u^2)

F=\dfrac{-mu^2}{2d}

F=\dfrac{-0.00479\times (642.3)^2}{2\times 0.0435}

F = -22713.92 N

F=2.27\times 10^4\ N

So, the magnitude of the force that stops the bullet is 2.27\times 10^4\ N

7 0
3 years ago
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