Answer:
1.) Micrometres screw gauge
2.) Tape rule.
Explanation:
Given that the diameter and the length of a thin wire, approximately 1m in length, are measured as accurately as possible.
what are the best instruments to use ?
To measure the diameter of a thin wire, the best instrument to use is known as micrometres screw gauge.
And to measure the length of a thin wire up to 1 m, the measuring device can be tape rule or long metre rule.
The answer is B: 6; 4. Carbon has four electrons in its outermost shell, which are its valence electrons.
The three properties of electromagnetic waves are; they travel at the speed of light, they include ultraviolet waves, and they can transfer energy through empty space.
<h2>Further Explanation</h2><h3>A wave</h3>
- A wave is a transmission of a disturbance. It involves transmission of energy from one point which is the source to another point.
- Waves may be classified depending on the need for a transmission medium or based on the vibration of particles relative to the direction of wave motion.
- Waves may be either transverse or longitudinal based on the direction of wave motion relative to the vibration of particles
- Additionally waves may be classified as either electromagnetic wave or mechanical based on the need for a transmission medium.
<h3>Electromagnetic waves </h3>
- Electromagnetic waves are types of waves that do not require a material medium for transmission.
- All waves of the electromagnetic spectrum are electromagnetic transverse waves that do not require a material medium for transmission.
- They include; radio waves, microwaves, infrared, visible light, ultra-violet, x-rays, and gamma rays.
- All waves of the electromagnetic spectrum travel with a speed of light, 3.0 x10^8 m/s.
- Additionally, electromagnetic waves possess energy that is given by; E = hf; where h is the plank's constant and f is the frequency.
keywords: Wave, electromagnetic wave, electromagnetic spectrum
<h2>Learn more about: </h2>
Level: High school
Subject: Physics
Topic: Electromagnetic spectrum
Sub-topic: Properties of an electromagnetic waves
For the purpose of the exercise, we can assume that the Earth is at perihelion (closest point to the Sun) on December 21st (in reality, it happens around January 4th) and that the Earth is at aphelion (farthest point from the Sun) on June 21st (in reality, this happens around July 4th). The distance Earth-Sun is the following:
- Perihelion: 147.1 milion km
- Aphelion: 152.1 milion km
- Average distance: 149.6 milion km
So, we can calculate the percentage change with respect to the average distance as:
The axial field is the integration of the field from each element of charge around the ring. Because of symmetry, the field is only in the direction of the axis. The field from an element ds in the ring is
<span>dE = (qs*ds)cos(T)/(4*pi*e0)*(x^2 + R^2) </span>
<span>where x is the distance along the axis from the plane of the ring, R is the radius of the ring, qs is the linear charge density, T is the angle of the field from the x-axis. </span>
<span>However, cos(T) = x/sqrt(x^2 + R^2) </span>
<span>so the equation becomes </span>
<span>dE = (qs*ds)*[x/sqrt(x^2 + R^2)]/(4*pi*e0)*(x^2 + R^2) </span>
<span>dE =[qs*ds/(4*pi*e0)]*x/(x^2 + R^2)^1.5 </span>
<span>Integrating around the ring you get </span>
<span>E = (2*pi*R/4*pi*e0)*x/(x^2 + R^2)^1.5 </span>
<span>E = (R/2*e0)*x*(x^2 + R^2)^-1.5 </span>
<span>we differentiate wrt x, the term R/2*e0 is a constant K, and the derivative is </span>
<span>dE/dx = K*{(x^2 + R^2)^-1.5 +x*[(-1.5)*(x^2 + R^2)^-2.5]*2x} </span>
<span>dE/dx = K*{(x^2 + R^2)^-1.5 - 3*x^2*(x^2 + R^2)^-2.5} </span>
<span>to find the maxima set this = 0, giving </span>
<span>(x^2 + R^2)^-1.5 - 3*x^2*(x^2 + R^2)^-2.5 = 0 </span>
<span>mult both side by (x^2 + R^2)^2.5 to get </span>
<span>(x^2 + R^2) - 3*x^2 = 0 </span>
<span>-2*x^2 + R^2 = 0 </span>
<span>-2*x^2 = -R^2 </span>
<span>x = (+/-)R/sqrt(2) </span>
