Please help easy 8th grade question

A long, thin superconducting wire carrying a 17 A current passes through the center of a thin, 3.0-cm-diameter ring. A uniform e

nalin [4]

Answer:

a

$\frac{dE}{dt} =- 2.72 *10^{15} \ N/C \cdot s$

b

The direction of the electric field is opposite that of the current

Explanation:

From the question we are told that

The current is $I = 17\ A$

The diameter of the ring is $d = 3.0 \ cm = 0.03 \ m$

Generally the radius is mathematically represented as

$r = \frac{d}{2}$

$r = \frac{0.03}{2}$

$r = 0.015 \ m$

The cross-sectional area is mathematically represented as

$A = \pi r^2$

=> $A = 3.142 * (0.015^2)$

=> $A = 7.07 *10^{-4 } \ m^ 2$

Generally according to ampere -Maxwell equation we have that

$\oint \= B \cdot \= ds = \mu_o I + \epsilon_o \mu _o\frac{ d \phi }{dt }$

Now given that $\= B = 0$ it implies that

$\oint \= B \cdot \= ds = 0$

So

$\mu_o I + \epsilon_o \mu _o\frac{ d \phi }{dt } = 0$

Where $\epsilon _o$ is the permittivity of free space with value $\epsilon_o = 8.85*10^{-12 } \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2$

$\mu_o$ is the permeability of free space with value

$\mu_o = 4\pi * 10^{-7} N/A^2$

$\phi$ is magnetic flux which is mathematically represented as

$\phi = E * A$

Where E is the electric field strength

So

$\mu_o I + \epsilon_o \mu _o \frac{ d [EA] }{dt } = 0$

=> $\frac{dE}{dt} =- \frac{I}{\epsilon_o * A }$

=> $\frac{dE}{dt} =- \frac{17}{8.85*10^{-12} * 7.07*10^{-4} }$

=> $\frac{dE}{dt} =- 2.72 *10^{15} \ N/C \cdot s$

The negative sign shows that the direction of the electric field is opposite that of the current

8 0

3 years ago

A parachute on a racing dragster opens and changes the speed of the car from 95 m/s to 35 m/s in a period of 6.5 seconds. What i

tatiyna

Answer:

30 miles

Explanation:

6 0

3 years ago

A single crystal of zinc is oriented for a tensile test such that its slip plane normal makes an angle of 65∘ with the tensile a

rosijanka [135]

Answer: (a) 30° (b)0.9MPa

Explanation:

Ф = 65°

(a) First, you find the cosines for the possible λ values given

λ ∈ ( 30°, 65°, 78°)

30° ; cos 30 = 0.87

65° ; cos 65 = 0.42

78° ; cos 78 = 0.21

(make sure when using your calculator to calculate for cos, input the values as degrees. e.g cos 30 in deg form is 0.87, but in rad form is 0.15).

among these calculated cosine values, slip will occur along the direction for which (cosФ cos λ) is maximum

The maximum value of Schmid factor is 0.87. Thus, the most favored slip direction is 30° with the tensile axis.

(b)The plastic deformation begins at a tensile stress of 2.5MPa. Also, the value of the angle between the slip plane normal and the tensile axis is mentioned as 65°

From the expression for Schmid’s law:

τ = σ*cos(Φ)*cos(λ)

Substituting 2.5MPa for σ, 30° for λ and 65° for Φ

τ = 2.5*cos(65°)*cos(30°)

= 2.5 × 0.42 × 0.87 = 0.91

The critical resolved shear stress for zinc, τ = 0.9 MPa

6 0

3 years ago

Oxygen-18 is a naturally-occuring, stable isotope and is commonly used is scientific studies as a tracer. Using the periodic tab

Molodets [167]

Answer:

1. Proton number = 8

2. Electron number = 8

3. Neutron = 10

Explanation:

Mass number = 18

Proton = 8(isotope still retain their atomic number)

Electron = proton = 8( since it is a neutral atom)

Neutron number =?

Neutron = Mass number - proton

Neutron number = 18 - 8 = 10

4 0

4 years ago

A circular swimming pool has a diameter of 14 m, the sides are 4 m high, and the depth of the water is 3 m. How much work (in Jo

vredina [299]

The work required is Wa = 2954112 J

Given:

swimming pool diameter = 14 m

length of sides = 4 m

height of water = 3 m

To Find:

work required to pump water

Solution: The radius of the swimming pool is

r = 14/2 = 7 m

The work is mathematically given as

W = Force x distance

Now force is mathematically given as

F = density x area x height of pool = p*(πr²)dx

Now the work done to pump all of the water over the side

W = ∫p*(πr²)(H-x)dx = ∫1000*9.81*(π*7^2)(4-x)dx

W = 64000*9.8π∫(4-x) dx = 64000*9.8π{4(3) - 3/2}

W = 2954112 J

So, work required is Wa = 2954112 J

Learn more about Work here:

brainly.com/question/8119756

#SPJ4

7 0

2 years ago