what happens when an electric current passes through a coil of wire instead of a single straight peice of wire
An electric current passing through a coil of wire gives a strong form of magnetism called electromagnetism. When the electric current passes through a single straight piece of wire the electromagnetism is weak.
Passing an electric current through a coil of wire generates a magnetic field. The strength of this field can be modified by changing the amount of current or the number of turns in the coil.
When an electric current passes through a coil of wire, as opposed to a straight piece, it creates a magnetic field around the coil. This is the principle behind electromagnets and many electrical appliances we use on a daily basis. The strength of the magnetic field depends on the amount of current and the number of turns in the coil. For example, the more turns the wire has, the stronger the magnetic field.
For a very rough pipe wall the friction factor is constant at high Reynolds numbers. For a length L1 the pressure drop over the length is p1. If the length of the pipe is then doubled, what is the relation of the new pressure drop p2 to the original pressure drop p1 at the original mass flow rate?
Answer: ∆p2 = 2* ∆p1
Given that all other factors remain constant. The pressure drop across the pipeline is directly proportional to the length.
i.e ∆p ~ L
∆p2/L2 = ∆p1/L1
Since L2 = 2 * L1
∆p2/2*L1 = ∆p1/L1
Eliminating L1 we have,
∆p2/2 = ∆p1
Multiplying both sides by 2
∆p2 = 2 * ∆p1
The 9-inch-long elephant nose fish in the Congo River generates a weak electric field around its body using an organ in its tail. When small prey, or even potential mates, swim within a few feet of the fish, they perturb the electric field. The change in the field is picked up by electric sensor cells in the skin of the elephant nose. These remarkable fish can detect changes in the electric field as small as 3.00 μN/C. How much charge, modeled as a point charge, in the fish would be needed to produce such a change in the electric field at a distance of 63.5 cm ?
The strength of the electric field produced by a charge Q is given by
Q is the charge
r is the distance from the charge
k is the Coulomb's constant
In this problem, the electric field that can be detected by the fish is
and the fish can detect the electric field at a distance of
Substituting these numbers into the equation and solving for Q, we find the amount of charge needed:
Calculate the slope of the 25-coil line and the 50-coil line to determine the average number of paper clips that a 1 V battery would pick up.
For 25-turn electromagnet, Number of clips = 4.1
For 50-turn electromagnet number of clips = 9.6
To calculate the slope of the 25-coil line and the 50-coil line to determine the average number of paper clips that a 1 V battery would pick up.
Using the equations gotten from the graph in the previous question and 1.0 V as the value for x, we get
For 25-turn electromagnet y = 3.663x * 0.5
(rounded to one decimal place) Number of clips = 4.1
For 50-turn electromagnet y = 7.133x 2.5
(rounded to one decimal place) Number of clips = 9.6
How many nanoseconds are in one hour? How do you write the following in scientific notation?2,560,000m
3.6 × 10¹² nanoseconds
Hour is the unit of time. Seconds is the SI unit of time.
Dr. John Paul Stapp was a U.S. Air Force officer who studied the effects of extreme acceleration on the human body. On December 10, 1954, Stapp rode a rocket sled, accelerating from rest to a top speed of 282 m/s (1015 km/h) in 5.2 s and was brought jarringly back to rest in only 1 s. Calculate his (a) magnitude of acceleration in his direction of motion and (b) magnitude of acceleration opposite to his direction of motion. Express each in multiples of g (9.80 m/s2) by taking its ratio to the acceleration of gravity. g g
a = 5.53 g , a = -15g
This is an exercise in kinematics.
a) Let's look for the acceleration
as part of rest v₀ = 0
v = v₀ + a t
a = v / t
a = 282 / 5.2
a = 54.23 m / s²
in relation to the acceleration of gravity
a / g = 54.23 / 9.8
a = 5.53 g
b) let's look at the acceleration to stop
va = 0
0 = v₀ -2 a y
a = vi / y
a = 282/2 1
a = 141 m /s²
a / G = 141 / 9.8
a = -15g
why does the value of capacitance of a capacitor increases in parallel combination but not in series??
It is easiest to consider problems like this by thinking exclusively about parallel plate capacitors for which where Q is the charge separated (+Q on one plate, -Q on the other), V is the voltage difference between the plates, A is the area of each plate, and d is the separation between the plates.
When capacitors are connected in parallel, the voltage across each capacitor is the same. But with two capacitors, it will require more charge to reach the voltage V than it would with just one capacitor. In fact, if capacitor 1 requires charge