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3 years ago

Do you think it is possible to control the magnetic properties of a magnet? Can a magnet be turned on and off?

Physics

1 answer:

Sunny_sXe [5.5K]3 years ago

6 0

Answer:

Yes it is possible to control to some extent.

Explanation:

In general there are two types of magnets : permanent and temporary (electromagnets).

Electromagnets can be controlled since it basically depends on electricity. By switching on and off the electric supply the magnets also can be switched on and off respectively. We can also control the intensity of magnetic power.

On the other hand permanent magnet cannot be switched on and off but the magnetic properties can be altered event to an extent when it loses all its magnetic properties. It can be caused by high temperature, physical impact and also exposure to other magnetic fields. For every element there is a point of temperature called curie temperature above which the permanent magnet loses its magnetic properties. This can be brought back again by induced magnetism. The only issue is that induced magnetism work in most cases but not in all.

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An 100 kg object traveling at 50 m/s collides (perfectly inelastic) with a 50 kg object initially at rest.

qaws [65]

Answer:

Option C. 5,000 kg m/s

Explanation:

Linear Momentum on a System of Particles

Is defined as the sum of the momenta of each particles in a determined moment. The individual momentum is the product of the mass of the particle by its speed

P=mv

The question refers to an 100 kg object traveling at 50 m/s who collides with another object of 50 kg object initially at rest. We compute the moments of each object

$m_1=(100\ kg)(50\ m/s)=5,000\ kg\ m/s$

$m_2=(50\ kg)(0\ m/s) = 0$

The sum of the momenta of both objects prior to the collision is

$P=5,000\ kg\ m/s+0\ kg\ m/s$

$\boxed{ P=5,000\ kg\ m/s}$

7 0

3 years ago

How much work is required to compress 5.05 mol of air at 19.5°C and 1.00 atm to one-eleventh of the original volume by an isothe

Rus_ich [418]

Explanation:

(a) For an isothermal process, work done is represented as follows.

W = $-nRT ln(\frac{V_{2}}{V_{1}})$

Putting the given values into the above formula as follows.

W = $-nRT ln(\frac{V_{2}}{V_{1}})$

= $- 5.05 mol \times 8.314 J/mol K \times (19.5 + 273) K \times ln (\frac{\frac{V_{1}}{11}}{V_{1}})$

= $-12280.82 \times ln (0.09)$

= $-12280.82 \times -2.41$

= 29596.78 J

or, = 29.596 kJ (as 1 kJ = 1000 J)

Therefore, the required work is 29.596 kJ.

(b) For an adiabatic process, work done is as follows.

W = $\frac{P_{1}V^{\gamma}_{1}(V^{1-\gamma}_{2} - V(1-\gamma)_{1})}{(1 - \gamma)}$

= $\frac{-nRT_{1}(11^{\gamma - 1} - 1)}{1 - \gamma}$

= $\frac{-5.05 \times 8.314 J/mol K \times 292.5 (11^{1.4 - 1} - 1)}{1 - 1.4}$

= 49.41 kJ

Therefore, work required to produce the same compression in an adiabatic process is 49.41 kJ.

(c) We know that for an isothermal process,

$P_{1}V_{1} = P_{2}V_{2}$

or, $P_{2} = \frac{P_{1}V_{1}}{V_{2}}$

= $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})$

= 11 atm

Hence, the required pressure is 11 atm.

(d) For adiabatic process,

$P_{1}V^{\gamma}_{1} = P_{2}V^{\gamma}_{2}$

or, $P_{2} = P_{1} (\frac{V_{1}}{V_{2}})^{1.4}$

= $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})^{1.4}$

= 28.7 atm

Therefore, required pressure is 28.7 atm.

6 0

3 years ago

What is the magnitude (size) and direction of the cumulative force acting on the car shown in the picture above?

mrs_skeptik [129]

Answer:

5070

Explanation:

add them up and then you get your answers 

5 0

3 years ago

Read 2 more answers

A 1750-kilogram cars travels at a constant speed of 15.0 meters per second around a horizontal, circular track with a radius of

bulgar [2K]

m = mass of the car moving in horizontal circle = 1750 kg

v = Constant speed of the car moving in the horizontal circle = 15 m/s

r = radius of the horizontal circular track traced by the car = 45.0 m

F = magnitude of the centripetal force acting on the car

To move in a circle . centripetal force is required which is given as

F = m v²/r

inserting the above values in the formula

F = (1750) (15)²/(45)

F = (1750) (225)/(45)

F = 1750 x 5

F = 8750 N

6 0

3 years ago

A 5.0 kg object moving at 10 m/s on a frictionless surfaces collides with but does not stick to a 2.0 kg object that is initiall

lilavasa [31]

Answer:

I= 20 i {N.s}

Explanation:

In order to obtain the impulse on the 2 kg ball, you have to apply the equation of Impulse:

I=FΔt

Where I is the impulse vector, F is the net force and Δt is the interval of time when the force is applied.

In this case:

Δt=0.01 s

F= 2000 i N

where i is the unit vector in the x direction.

Replacing the values in the formula:

I=(2000)(0.01)i

Therefore:

I= 20 i {N.s}

3 0

3 years ago