Absolute pressure in tank is P1 = 260 kPa and local ambient absolute pressure is P2 =100 kPa. If liquid density in pipe is 13600
1 answer:
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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>
Answer:
The electric potential is approximately 5.8 V
The resulting direction of the electric field will lie on the line that joins the charges but since it is calculated in the midpoint and the charges are the same we can directly say that its magnitude is zero
Explanation:
The two protons can be considered as point charges. Therefore, the electric potential is given by the point charge potential:
(1)
where is the charge of the particle,
the electric permittivity of the vacuum (I assuming the two protons are in a vacuum) and
is the distance from the point charge to the point where the potential is being measured. Because the electric potential is an scalar, we can simply add the contribution of the two potentials in the midpoint between the protons. Thus:
Substituting the values ,
and
we obtain:
The resulting direction of the electric field will lie on the line that joins the charges but since it is calculated in the midpoint and the charges are the same we can directly say that its magnitude is zero.