Sully uses a battery and a coil of wire to create an electromagnet. Using the same materials, if he wants to increase the streng
1 answer:
An electromagnet is a type of magnet in which the magnetic field is produced using the current. The simplest form of an electromagnet is a wire wrapped around in a coil.
The strength of magnetic field of such magnet is given with this equation:
Where N is the number of loops in the coil, I is the strength of the current flowing through the coil, L is the length of the coil, and
is <span>permeability of the electromagnet core material.
From this equation, we can see that increasing both the current and number of loops will increase the strength of the magnet.
Both BLANKS should be Increase. When you use the additional battery you will have more voltage and more voltage means more electricity.</span>
The strength of magnetic field of such magnet is given with this equation:
Where N is the number of loops in the coil, I is the strength of the current flowing through the coil, L is the length of the coil, and
From this equation, we can see that increasing both the current and number of loops will increase the strength of the magnet.
Both BLANKS should be Increase. When you use the additional battery you will have more voltage and more voltage means more electricity.</span>
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Answer:
Explanation:
Given:
temperature at the hotter end,
temperature at the cooler end,
length of rod through which the heat travels,
cross-sectional area of rod,
mass of ice melted at zero degree Celsius,
time taken for the melting of ice,
thermal conductivity k=?
By Fourier's Law of conduction we have:
......................................(1)
where:
=rate of heat transfer
dT= temperature difference across the length dx
Now, we need the total heat transfer according to the condition:
we know the latent heat of fusion of ice,
Now the heat rate:
Now using eq,(1)
Answer:
<em>1.01 W/m</em>
Explanation:
diameter of the pipe d = 30 mm = 0.03 m
radius of the pipe r = d/2 = 0.015 m
external air temperature Ta = 20 °C
temperature of pipe wall Tw = 150 °C
convection coefficient at outer tube surface h = 11 W/m^2-K
From the above,<em> we assumed that the pipe wall and the oil are in thermal equilibrium</em>.
area of the pipe per unit length A = =
m^2/m
convectional heat loss Q = Ah(Tw - Ta)
Q = 7.069 x 10^-4 x 11 x (150 - 20)
Q = 7.069 x 10^-4 x 11 x 130 = <em>1.01 W/m</em>