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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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Point charges q1 = 14 µC and q2 = −60 µC are fixed at r1 = (5.0î − 4.0ĵ) m and r2 = (9.0î + 7.5ĵ) m. What is the force (in N) of
Lostsunrise [7]

Answer:

The force on q₁ due to q₂ is (0.00973i + 0.02798j) N

Explanation:

F₂₁ = \frac{K|q_1|q_2|}{r^2}.\frac{r_2_1}{|r_2_1|}

Where;

F₂₁ is the vector force on q₁ due to q₂

K is the coulomb's constant = 8.99 X 10⁹ Nm²/C²

r₂₁ is the unit vector

|r₂₁| is the magnitude of the unit vector

|q₁| is the absolute charge on point charge one

|q₂| is the absolute charge on point charge two

r₂₁ = [(9-5)i +(7.4-(-4))j] = (4i + 11.5j)

|r₂₁| = \sqrt{(4^2)+(11.5^2)} = \sqrt{148.25}

(|r₂₁|)² = 148.25

F_2_1=\frac{K|q_1|q_2|}{r^2}.\frac{r_2_1}{|r_2_1|} = \frac{8.99X10^9(14X10^{-6})(60X10^{-6})}{148.25}.\frac{(4i + 11.5j)}{\sqrt{148.25} }

      = 0.050938(0.19107i + 0.54933j) N

      = (0.00973i + 0.02798j) N

Therefore, the force on q₁ due to q₂ is (0.00973i + 0.02798j) N

7 0
3 years ago
Three collinear forces,F1=45N west,F2=63N east and an unknown force F3 are applied to an object.The resultant force of the three
Greeley [361]

Take east to be the positive direction. Then the resultant force from adding <em>F</em>₁ and <em>F</em>₂ is

<em>F</em>₁ + <em>F</em>₂ = (-45 N) + 63 N = 18 N

which is positive, so it's directed east.

To this we add a third force <em>F</em>₃ such that the resultant is 12 N pointing west, making it negative, so that

18 N + <em>F</em>₃ = -12 N

<em>F</em>₃ = -30 N

So <em>F</em>₃ has a magnitude of 30 N and points west.

6 0
3 years ago
What is the difference between condensation and precipitation
Alexxandr [17]

Explanation:

The water cycle is based on three parts;

1. Evaporation

2. Condensation

3. Participation

Condensation:

It is the process in which water vapor changes into liquid water or in other words, it is the transition from the gaseous state to liquid state.

Precipitation:

It is the process in which any liquid or frozen water such as snow that forms in the atmosphere and falls back to the Earth

Condensation depends on temperature and pressure whereas precipitation depends on the temperature and the concentration of the solution.

3 0
3 years ago
Read 2 more answers
An object accelerating at 16 m/s/s doubles its mass and triples its net force acting on it. What will the new acceleration be? (
nataly862011 [7]

Answer:

24 m/s²

Explanation:

The given parameters are;

The initial acceleration of the object, a = 16 m/s²

Let 'm' represent the initial mass of the object

The initial force acting on the object, F = m × a

∴ F = 16 × m = 16·m

When the mass is doubled, we have;

The new mass of the object, m₂ = 2 × m = 2·m

When the net force acting on the object triples, we have;

The new net force acting on the object, F₂ = 3 × F = 3 × 16·m = 48·m

From F = m × a, we have;

a = F/m

∴ The new acceleration of the object, a₂ = F₂/m₂

From which, by plugging in the values, we have;

a₂ = 48·m/(2·m) = 24

The new acceleration of the object, a₂ = 24 m/s².

6 0
3 years ago
When 597 J of heat are added to a gas, it expands. Its internal energy increases by 318 J. How much work does the gas do? (Unit=
damaskus [11]

The magnitude of work done by the gas is 279 J and the sign is negative so W = -279 J as work is done by the system.

<u>Explanation:</u>

According to first law of thermodynamics, the change in internal energy of the system is equal to the sum of the heat energy added or released from the system with the work done on or by the system. If the heat energy is added to the system to perform a certain work, then the heat energy is taken as positive, while it will be negative when the heat energy is released from the system.

Similarly, in this case, the heat energy of 597 J is added to the system. So the heat energy will be positive, while the gas expansion occurs means work is done by the system.

          ΔU = Q+W

Since ΔU is the change in internal energy which is given as 318 J and the heat energy added to the system is Q = 597 J.

Then the work done by the gas = ΔU - Q = 318 J - 597 J = - 279 J.

As the work is done by the system, so it will be denoted in negative sign and the magnitude of work done by the gas is 279 J.

7 0
3 years ago
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