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Sonbull [250]
3 years ago
5

1. Consider the point exactly halfway between the two wires. Can you adjust the current so that, with current passing through ea

ch of the wires, the net field is zero here? 2. Note the distance between the field lines from one of the wires. Are they equally space? Explain. 3. With currents of different magnitudes passing through the two wires, how do the forces on the two wires compare? What is the reason for this?

Physics
1 answer:
chubhunter [2.5K]3 years ago
8 0

Answer:

hello your question is incomplete attached below is missing part of the question

answer:

1 )  Magnetic field due to long current carrying wire : B = \frac{U_{0} I}{2\pi d}

Therefore the net magnetic field due the both wires ; B = B_{1} + B_{2} . when we adjust the current I_{1} = I_{2} then the Netfield (B ) = zero

2) The distance between the field lines are not equally spaced and this is because the separation between field lines increases with the increase in the distance between the wires

3) Increase in current through the wire will lead to increase in force and this can be explained via this equation

F = \frac{U_{0}I_{1}I_{2}   }{2\pi d }

Explanation:

1 )  Magnetic field due to long current carrying wire : B = \frac{U_{0} I}{2\pi d}

Therefore the net magnetic field due the both wires ; B = B_{1} + B_{2} . when we adjust the current I_{1} = I_{2} then the Netfield (B ) = zero

2) The distance between the field lines are not equally spaced and this is because the separation between field lines increases with the increase in the distance between the wires

3) Increase in current through the wire will lead to increase in force and this can be explained via this equation

F = \frac{U_{0}I_{1}I_{2}   }{2\pi d }

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The work done by the frictional force is 600J.

Explanation:

The work W done by the frictional force is

W= Fd.

Now, F = 60N and d =10m; therefore,

W= (60N)(10m)

\boxed{W = 600J.}

Hence, the work done by friction is 660J.

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3 years ago
In a chemical reaction blank are the substances left over
Zepler [3.9K]

Answer: In a chemical reaction blank are the substances left over

Explanation: The substance left over after a reaction takes place

3 0
3 years ago
The density of ice is 0.93 g/cm3. what is the volume, in cm3, of a block of ice whose mass is 5.00 kg? remember to select an ans
serious [3.7K]

The Volume of the ice block is 5376.344 cm^3.

The density of a material is define as the mass per unit volume.

Here, the density of ice given is 0.93 g/cm^3

Mass of the ice block  given is 5 kg or 5000 g

Now calculate the volume of the ice block

density=mass/volume

0.93=5000/Volume

Volume =5376.344 cm^3

Therefore the volume of  ice block is 5376.344 cm^3

7 0
3 years ago
A confined aquifer with a porosity of 0.15 is 30 m thick. The potentiometric surface elevation at two observation wells 1000 m a
AlekseyPX

Answer:

Part (a) The flow rate per unit width of the aquifer is 1.0875 m³/day

Part (b) The specific discharge of the flow is 0.0363 m/day

Part (c) The average linear velocity of the flow is 0.242 m/day

Part (d) The time taken for a tracer to travel the distance between the observation wells is 4132.23 days = 99173.52 hours

Explanation:

Part (a) the flow rate per unit width of the aquifer

From Darcy's law;

q = -Kb\frac{dh}{dl}

where;

q is the flow rate

K is the permeability or conductivity of the aquifer = 25  m/day

b is the aquifer thickness

dh is the change in th vertical hight = 50.9m - 52.35m = -1.45 m

dl is the change in the horizontal hight = 1000 m

q = -(25*30)*(-1.45/1000)

q = 1.0875 m³/day

Part (b) the specific discharge of the flow

V = \frac{Q}{A} = \frac{q}{b} = -K\frac{dh}{dl}\\\\V = -(25 m/d).(\frac{-1.45 m}{1000 m}) = 0.0363 m/day

V = 0.0363 m/day

Part (c) the average linear velocity of the flow assuming steady unidirectional flow

Va = V/Φ

Φ is the porosity = 0.15

Va = 0.0363 / 0.15

Va = 0.242 m/day

Part (d) the time taken for a tracer to travel the distance between the observation wells

The distance between the two wells = 1000 m

average linear velocity = 0.242 m/day

Time = distance / speed

Time = (1000 m) / (0.242 m/day)

Time = 4132.23 days

        = 4132.23 days *\frac{24 .hrs}{1.day} = 99173.52, hours

4 0
3 years ago
You toss a ball straight up with an initial speed of 30m/s. How high does it go, and how long is it in the air (neglecting air r
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Explanation:

Given that,

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Here, a = -g

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So, the time for upward motion is 3.06 seconds. It means that it will in air for 3.06×2 = 6.12 seconds

Let d is the maximum distance covered by it.

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Hence, it will go to a height of 45.91 m and it will in the air for 6.12 seconds.

8 0
3 years ago
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