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mrs_skeptik [129]
3 years ago
14

Technician A says that magnetic lines of force can be seen by placing iron filings on a piece of paper and then holding them ove

r a magnet. Technician B says that the effects of magnetic lines of force can be seen using a compass. Which technician is correct?
Physics
2 answers:
aev [14]3 years ago
5 0

Answer:

Technician A is correct

Magnetic lines of force can be seen by placing iron filings on a piece of paper and then holding them over a magnet.

Explanation:

As we know that the magnetic field lines are always originated out from the North pole of magnet and then terminate at south pole of the magnet

Here we know that when magnetic material aligns itself in the direction of external magnetic field

So here when we put the iron filings on the cardboard and then place a magnet near it then all the iron filing will align itself in the direction of the magnetic field

Now if we place a compass near the magnet then it will aligns itself in the tangential direction of the magnetic field at the same position

So here we can say that Technician A is correct

Eva8 [605]3 years ago
4 0

Answer:

Both A and B are correct

Explanation:

  • Magnetic line of force can be seen by placing iron fillings on a piece of paper and then hold a magnet over them, the iron fillings align themselves in the magnetic field of the magnetic along the line of forces thus showing the loops of the magnetic line of forces.
  • Also, we know that unlike poles attract where as like poles repel each other. This fact can be proved when we hang a bar magnet it aligns itself in the North-South direction because of the the magnetic field due to the alignment of the geographical North-South poles and the poles of the bar magnet and the compass works the same way.

Thus both the technicians are correct.

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mote1985 [20]

Answer:

OC, OD, OA, OB

Explanation:

7 0
3 years ago
An aluminum cup of mass 150 g contains 800 g of water in thermal equilibrium at 80.0°C. The combination of cup and water is cool
gladu [14]

Answer:

Heat Flow Rate : ( About ) 87 W

Explanation:

The heat flowing out of the system each minute, will be represented by the following equation,

Q( cup ) + Q( water ) = m( cup ) * c( al ) * ΔT + m( w ) * c( w ) * ΔT

So as you can see, the mass of the aluminum cup is 150 grams. For convenience, let us convert that into kilograms,

150 grams = .15 kilograms - respectively let us convert the mass of water to kilograms,

800 grams = .8 kilograms

Now remember that the specific heat of aluminum is 900 J / kg * K, and the specific heat of water = 4186 J / kg * K. Therefore let us solve for " the heat flowing out of the system per minute, "

Q( cup ) + Q( water ) = .15 * ( 900 J / kg * K )  * 1.5 + .8 * ( 4186 J / kg * K ) * 1.5,

Q( cup ) + Q( water ) = 5225.7 Joules

And the heat flow rate should be Joules per minute,

5225.7 Joules / 60 seconds = ( About ) 87 W

5 0
3 years ago
What is physical quantity ? Give examples.​
Dahasolnce [82]

Explanation:

physical quantity is any physical property that can be qualified that,is, be measured using numbers e.g mass, amount of substance,time and length

7 0
3 years ago
Dalton proposed the first atomic model. Which of these statements accurately reflected his thinking at that time?. . . . A.. Ato
QveST [7]
A.. Atoms are alike...
4 0
3 years ago
Read 2 more answers
A parallel-plate vacuum capacitor has 8.38 J of energy stored in it. The separation between the plates is 2.30 mm. If the separa
Elanso [62]

Answer:

Explanation:

plate separation = 2.3 x 10⁻³ m

capacity C₁ = ε A / d

= ε A / 2.3 x 10⁻³

C₂ = ε A / 1.15 x 10⁻³

\frac{C_2}{C_1} = \frac{2.3}{1.15}

a ) when charge remains constant

energy = \frac{q^2}{2C}

q is charge and C is capacity

energy stored initially E₁= \frac{q^2}{2C_1}

energy stored finally E₂ = \frac{q^2}{2C_2}

\frac{E_1}{E_2} = \frac{C_2}{C_1} = \frac{2.3}{1.15}

E_2 = \frac{1.15}{2.3 } \times E_1

= \frac{1.15}{2.3 } \times 8.38

= 4.19 J

b )

In this case potential diff remains constant

energy of capacitor = 1/2 C V²

energy is proportional to capacity as V is constant .

\frac{E_2}{E_1} = \frac{C_2}{C_1}

\frac{E_2}{8.38} = \frac{2.3}{1.15}

E_2 = 16.76 .

8 0
3 years ago
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