Answer plz answer plzzz I am a little confused with full time

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Answer 1

Answer: I can’t read that I’m sorry make it more clear

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A 100 A current circulates around a 2.00-mm-diameter superconducting ring. A. What is the ring's magnetic dipole moment?
B. What is the on-axis magnetic field strength 4.70 cm from the ring?

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Answer:

0.000314 Am²

6.049*10^-7 T

Explanation:

From the definitions of magnetic dipole moment, we can establish that

= , where

= the magnetic dipole moment in itself

= Current, 100 A

= Area, πr² (r = diameter divided by 2). Converting to m², we have 0.000001 m²

On solving, we have

= ,

= 100 * 3.14 * 0.000001

= 0.000314 Am²

=(0)/4* 2/³, where

(0) = constant of permeability = 1.256*10^-6

z = 4.7 cm = 0.047 m

B = 1.256*10^-6 / 4*3.142 * [2 * 0.000314/0.047³]

B = 1*10^-7 * 0.000628/1.038*10^-4

B = 1*10^-7 * 6.049

B = 6.049*10^-7 T

Final answer:

The magnetic dipole moment of the superconducting ring is 0.000314 Ampere*m². The on-axis magnetic field strength 4.70 cm from the ring is 6.56 μT.

Explanation:

The magnetic dipole moment (μ) of a current (I) circulating around a loop of radius (r) is given by the formula μ = Iπr². Substituting the given values in SI units (I=100 Ampere, r=1.00 mm = 0.001 m), we get μ = 100 * π * (0.001)² = 0.000314 Ampere*m².

To find out the on-axis magnetic field strength (B) at a certain distance from the ring, we use the formula B = μ0/(4π) * (2μ/r³), where μ0 represents the permeability of free space. Plugging the values in SI units (μ0 = 4π × 10-7 T*m/A, r=4.70 cm = 0.047 m), The magnetic field is calculated to be B = (4π × 10-7 T*m/A) /(4π) * (2 * 0.000314 m² / 0.0473m³) = 6.56 × 10-6 T or 6.56 μT.

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Two 3.7 microCoulomb charges are 0.8 m apart. How much energy (in milliJoule) went into assembling these two charges? Enter a number with one digit behind the decimal point.

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Answer:

Explanation:

The potential energy of a system of two charges is given by the expression

$(K* Q_1* Q_2)/(R)$

Q₁ and Q₂ are two charges and R is distance between the charges.

Given Q₁ = Q₂ = 3.7 X 10⁻⁶ , R = .8 and K = 9 x 10⁹

Putting these values in the equation we have,

$[tex](9* 10^9* 3.7*10^(-6) 3.7* 10^(-6))/(.8)$ [/tex]

Potential energy = 154.01 x 10⁻³ J

This energy have been spent to bring these repelling two charges at this close distance . The energy spent have been stored as potential energy here which has been calculated.

Determine whether or not each of the following statement is true. If a statement is true, prove it. If the statement is false, provide a counterexample and explain how it constitutes a counterexample. Diagrams can be useful in explaining such things. If the electric potential in a certain region of space is constant, then the charge enclosed by any closed surface completely contained within that region is zero.

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Answer:

True

Explanation:

This is a representation of Gauss law.

Gauss’s law does hold for moving charges, and in this respect Gauss’s law is more general than Coulomb’s law. In words, Gauss’s law states that: The net outward normal electric flux through any closed surface is proportional to the total electric charge enclosed within that closed surface. The law can be expressed mathematically using vector calculus in integral form and differential form, both are equivalent since they are related by the divergence theorem, also called Gauss’s theorem.

Which is true about the radiation force of light shining on a surface? The force is greater if the light reflects back along its incident path than in some other direction. The force is greater if the light is absorbed instead of being reflected. The force is greater if the light is reflected in some direction other than back along the incident path.

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Answer:

The force is greater if the light is absorbed instead of being reflected

Explanation:

Light could either be reflected or absorbed. Reflection takes place where light is concentrated back to another surface whereas absorption takes place when light is incorporated into a surface thereby providing kinetic energy. The kinetic energy produced by absorption provides more force than reflection which just involves concentrating back to another surface.

A rock is pulled back in a slingshot as shown in the diagram below. The elastic on the slingshot is displaced 0.2 meters from its initial position. The rock is pulled back with a force of 10 newtons.When the rock is released, what is its kinetic energy?

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The rock takes 8.16s to return to its release point. Given that the elastic band provides a speed of 40m/s to the rock in 10 cm stretch.

What will be the speed of the rock?

Initial speed of the rock, u = 40m/s

Final position of the rock s = 0m taking the release point as reference. The rock takes 8.16s to return to its release point. Given that the elastic band provides a speed of 40m/s to the rock in 10 cm stretch.

Nuclear energy is a useful source of power but has disadvantages. The disadvantage of nuclear energy is it produces dangerous waste.

Initial speed of the rock, u = 40m/s

Final position of the rock s = 0m taking the release point as reference

From the second equation of motion:

solving above we get:

t = 0s or t = 8.16s, t =0 seconds is neglected since it represents the initial position which is the same as the final position at t = 8.16s

So, the rock takes 8.16 seconds to return to the release point.

Therefore, The rock takes 8.16s to return to its release point. Given that the elastic band provides a speed of 40m/s to the rock in 10 cm stretch.

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Answer:

Explanation:

i don't know

A Porsche sports car can accelerate at 8.8 m/s^2. Determine its acceleration in km/h^2.

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Answer:

The acceleration expressed in the new units is $114.048 Km/h^(2)$

Explanation:

To convert from $m/s^(2)$ to $Km/h^(2)$ it is necessary to remember that there are 1000 meters in 1 kilometer and 3600 seconds in 1 hour:

Then by means of a rule of three it is get:

$8.8(m)/(s^(2)).((1Km)/(1000m)).((3600s)/(1h))^(2)$

$8.8(m)/(s^(2)).((1Km)/(1000m)).((12960000s^(2))/(1h^(2)))$

Hence, the units of meters and seconds will cancel. Notice the importance of square the ratio 3600s/1h, so that way they can match with the other units:

$114.048 Km/h^2$