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Degger [83]
3 years ago
14

Two electromagnets are running with equal amounts of electric current. Electromagnet A uses a piece of iron as its metal core, w

hile electromagnet B uses a piece of silver. Which one would have a stronger magnetic field?
Physics
1 answer:
sp2606 [1]3 years ago
5 0

Answer:

Electromagnet A, which uses a piece of iron as its metal core will produce the stronger magnetic field

This is because of the ferromagnetic properties of iron

Silver is not attracted to magnets

Explanation:

An electromagnet with an iron core produces a stronger magnetic effect, because ferromagnetic materials have the effect of concentrating the magnetic flux.

An electromagnet with a gold, silver or titanium core would be perfectly functional, as would one with a core made out of wood, or a human finger, that is to say they don't produce magnetic effects. Silver isn't noticeably magnetic. In fact, only iron, nickel, cobalt, gadolinium, neodymium and samarium are magnetic in our everyday lives. An ingot of pure silver isn't attracted to a magnet.

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poizon [28]
Let us first know the given: Tennis ball has a mass of 0.003 kg, Soccer ball has a mass of 0.43 kg. Having the same velocity at 16 m/s. First the equation for momentum is P=MV P=Momentum M=Mass V=Velocity. Now let us have the solution for the momentum of tennis ball. Pt=0.003 x 16 m/s= (    kg-m/s ) I use the subscript "t" for tennis.  Momentum of Soccer ball Ps= 0.43 x 13m/s = (      km-m/s). If we going to compare the momentum of both balls, the heavier object will surely have a greater momentum because it has a larger mass, unless otherwise  the tennis ball with a lesser mass will have a greater velocity to be equal or greater than the momentum of a soccer ball.
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4 years ago
Read 2 more answers
A 50-g cube of ice, initially at 0.0°C, is dropped into 200 g of water in an 80-g aluminum container, both initially at 30°C.
MakcuM [25]

Answer:

b. 9.5°C

Explanation:

m_i = Mass of ice = 50 g

T_i = Initial temperature of water and Aluminum = 30°C

L_f = Latent heat of fusion = 3.33\times 10^5\ J/kg^{\circ}C

m_w = Mass of water = 200 g

c_w = Specific heat of water = 4186 J/kg⋅°C

m_{Al} = Mass of Aluminum = 80 g

c_{Al} = Specific heat of Aluminum = 900 J/kg⋅°C

The equation of the system's heat exchange is given by

m_i(L_f+c_wT)+m_wc_w(T-T_i)+m_{Al}c_{Al}=0\\\Rightarrow 0.05\times (3.33\times 10^5+4186\times T)+0.2\times 4186(T-30)+0.08\times 900(T-30)=0\\\Rightarrow 1118.5T-10626=0\\\Rightarrow T=\dfrac{10626}{1118.5}\\\Rightarrow T=9.50022\ ^{\circ}C

The final equilibrium temperature is 9.50022°C

4 0
3 years ago
Please i need detailed explanation​
zubka84 [21]

Answer:

2Micro Farahds

Explanation:

Its in the picture.

I Hope it helps.

4 0
3 years ago
when a circular plate of metal is heated in an oven, its radius increases at .03 cm/min, at what rate is the area increasing whe
Alinara [238K]

Answer:

Rate of change of area will be 9.796cm^2/min

Explanation:

We have given rate of change of radius \frac{dr}{dt}=0.03cm/min

Radius of the circular plate r = 52 cm

Area is given by A=\pi r^2

So \frac{dA}{dt}=2\pi r\frac{dr}{dt}

Puting the value of r and \frac{dr}{dt}

\frac{dA}{dt}=2\times 3.14\times 52\times 0.03=9.796cm^2/min

So rate of change of area will be 9.796cm^2/min

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3 years ago
Which of the following are voltage sourWhich of the following are voltage sources?
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The answer is
ohms law

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