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3 years ago

Which is an example of scientific evidence that can be used to critique a new scientific argument?

1 answer:

4 0

An example of scientific evidence that can be used to critique a new scientific argument would be a new study that came and has shown that meat causes cancer. Such a study can be used when one would like to talk/argument about the issues of meat consumption.

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A long, straight, cylindrical wire of radius R carries a current uniformly distributed over its cross section.

Mice21 [21]

Answer:

Explanation:

We shall solve this question with the help of Ampere's circuital law.

Ampere's ,law

∫ B dl = μ₀ I , B is magnetic field at distance x from the axis within wire

we shall find magnetic field at distance x . current enclosed in the area of circle of radius x

= I x π x² / π R²

= I x² / R²

B x 2π x = μ₀ x current enclosed

B x 2π x = μ₀ x I x² / R²

B = μ₀ I x / 2π R²

Maximum magnetic B₀ field will be when x = R

B₀ = μ₀I / 2π R

Given

B = B₀ / 3

μ₀ I x / 2π R² = μ₀I / 2π R x 3

x = R / 3

b ) The largest value of magnetic field is on the surface of wire

B₀ = μ₀I / 2π R

At distance x outside , let magnetic field be B

Applying Ampere's circuital law

∫ B dl = μ₀ I

B x 2π x = μ₀ I

B = μ₀ I / 2π x

Given B = B₀ / 3

μ₀ I / 2π x = μ₀I / 2π R x 3

x = 3R .

3 0

3 years ago

Olivia wants to find out whether a substance will fluoresce. She says she should put it in a microwave oven. Do you agree with h

Sophie [7]

I disagree.
Many substances show fluorescence under ULTRAVIOLET light, NOT microwaves. :)

5 0

3 years ago

Read 2 more answers

Two 10-cm-diameter charged rings face each other, 25.0cm apart. Both rings are charged to +20.0nC. What is the electric field st

Black_prince [1.1K]

Answer:

Part A:

$E_{midpoint}=0$

Part B:

$E_{center}=2711.7558 N/C$

Explanation:

Part A:

Formula of Electric Field Strength:

$E=\frac{1}{4\pi\epsilon}\frac{xQ}{(x^2+R^2)^{3/2}}$

Where:

x is the distance from the ring

R is the radius of the ring

$\epsilon$ is constant permittivity of free space=8.854*10^-12 farads/meter

Q is the charge

For right Ring E at the midpoint can be calculated as:

x for right plate=25/2=12.5 cm=0.125 m

Radius=R=10/2=5 cm=0.05 m

$E_{right}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.125)*(20*10^{-19})}{((0.125)^2+(0.05)^2)^{3/2}}\\E_{right}=9208.1758 N/C$

For Left Ring E at the midpoint can be calculated as:

Since charge on both plates is +ve and same in magnitude, the electric field will be same for both plates.

$E_{left}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.125)*(20*10^{-19})}{((0.125)^2+(0.05)^2)^{3/2}}\\E_{left}=9208.1758 N/C$

Electric Field at midpoint:

Both rings have same magnitude but the direction of fields will be opposite as they have same charge on them.

$E_{midpoint}=E_{left}-E_{right}\\E_{midpoint}=9208.1758-9208.1758\\E_{midpoint}=0$

Part B:

At center of left ring:

Due to left ring Electric field at center is zero because x=0.

$E_{left}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0)*(20*10^{-19})}{((0)^2+(0.05)^2)^{3/2}}\\E_{left}=0 N/C$

Due to right ring Electric field at center of left ring:

Now: x=25 cm= o.25 m (To the center of left ring)

$E_{right}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.25)*(20*10^{-19})}{((0.25)^2+(0.05)^2)^{3/2}}\\E_{right}=2711.7558 N/C$

Electric Field Strength at center of left ring is same as that of right ring.

$E_{center}=2711.7558 N/C$

5 0

3 years ago

1. A plane travels a distance of 500

Yanka [14]

Answer:

Option C, 139 m/s

Explanation:

s = d/t

s = 500*10^3/3600

s = 138.8 = 139 m/s

5 0

3 years ago

To do your homework correctly you should (5 points)

pychu [463]

Answer:

C according to me

Explanation:

ITS HARD

4 0

3 years ago

Read 2 more answers