what happens when an electric current passes through a coil of wire instead of a single straight peice of wire

Answers

Answer 1
Answer:

Answer:

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.

Explanation:

Answer 2
Answer:

Final answer:

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.

Explanation:

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.

Learn more about Electromagnetism here:

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In general terms, the efficiency of a system can be thought of as the output per unit input. Which of the expressions is a good mathematical representation of efficiency e of any heat engine? Where, Qh: the absolute value (magnitude) of the heat absorbed from the hot reservoir during one cycle or during some time specified in the problem Qc: the absolute value (magnitude) of the heat delivered to the cold reservoir during one cycle or during some time specified in the problem W: the amount of work done by the engine during one cycle or during some time specified in the problem A) e=QhW B) e=QcQh C) e=QcW D) e=WQh E) e=WQc

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?

Answers

Answer: ∆p2 = 2* ∆p1

Explanation:

Given that all other factors remain constant. The pressure drop across the pipeline is directly proportional to the length.

i.e ∆p ~ L

Therefore,

∆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 ?

Answers

Answer:

1.34\cdot 10^(-16) C

Explanation:

The strength of the electric field produced by a charge Q is given by

E=k(Q)/(r^2)

where

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

E=3.00 \mu N/C = 3.00\cdot 10^(-6)N/C

and the fish can detect the electric field at a distance of

r=63.5 cm = 0.635 m

Substituting these numbers into the equation and solving for Q, we find the amount of charge needed:

Q=(Er^2)/(k)=((3.00\cdot 10^(-6) N/C)(0.635 m)^2)/(9\cdot 10^9 Nm^2 C^(-2))=1.34\cdot 10^(-16) C

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.

Answers

Answer:

For 25-turn electromagnet, Number of clips = 4.1

For 50-turn electromagnet number of clips = 9.6

Explanation:

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.

Hence;

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

Answers

Answer:

3.6 × 10¹² nanoseconds

Explanation:

Hour is the unit of time. Seconds is the SI unit of time.

Hour and seconds are related as:

1 hour = 60 minutes

1 minute = 60 seconds

So,

1 hour = 60 ×60 seconds = 3600 seconds

Thus,

3600  seconds are in one hour

Also,

1 sec = 10⁹ nanoseconds

Thus,

3600 sec = 3600 × 10⁹ nanoseconds = 3.6 × 10¹² nanoseconds

Thus,

3.6 × 10¹² nanoseconds are in one hour.

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

Answers

Answer:

    a = 5.53 g ,   a = -15g

Explanation:

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??

Answers

It is easiest to consider problems like this by thinking exclusively about parallel plate capacitors for which C \equiv (Q)/(V) =\kappa \epsilon_0 (A)/(d) 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 
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