We use them to survive, such as for oxygen, food, and shelter.
Answer:
The work done by the force is 6μJ
Explanation:
Suppose, A 30 nC charge is moved from a point where V₁ = 130 V to a point where V₂ = -70 V.
We need to calculate the work done by the force
Using formula of work done
Where, q = charge
=potential difference
Put the value into the formula
Hence, The work done by the force is 6μJ.
Interaction of Electromagnetic Radiation and Matter
It is well known that all matter is comprised of atoms. But subatomically, matter is made up of mostly empty space. For example, consider the hydrogen atom with its one proton, one neutron, and one electron. The diameter of a single proton has been measured to be about 10-15 meters. The diameter of a single hydrogen atom has been determined to be 10-10meters, therefore the ratio of the size of a hydrogen atom to the size of the proton is 100,000:1. Consider this in terms of something more easily pictured in your mind. If the nucleus of the atom could be enlarged to the size of a softball (about 10 cm), its electron would be approximately 10 kilometers away. Therefore, when electromagnetic waves pass through a material, they are primarily moving through free space, but may have a chance encounter with the nucleus or an electron of an atom.
Because the encounters of photons with atom particles are by chance, a given photon has a finite probability of passing completely through the medium it is traversing. The probability that a photon will pass completely through a medium depends on numerous factors including the photon’s energy and the medium’s composition and thickness. The more densely packed a medium’s atoms, the more likely the photon will encounter an atomic particle. <span>In other words, the more subatomic particles in a material (higher Z number), the greater the likelihood that interactions will occur </span>Similarly, the more material a photon must cross through, the more likely the chance of an encounter.
When a photon does encounter an atomic particle, it transfers energy to the particle. The energy may be reemitted back the way it came (reflected), scattered in a different direction or transmitted forward into the material. Let us first consider the interaction of visible light. Reflection and transmission of light waves occur because the light waves transfer energy to the electrons of the material and cause them to vibrate. If the material is transparent, then the vibrations of the electrons are passed on to neighboring atoms through the bulk of the material and reemitted on the opposite side of the object. If the material is opaque, then the vibrations of the electrons are not passed from atom to atom through the bulk of the material, but rather the electrons vibrate for short periods of time and then reemit the energy as a reflected light wave. The light may be reemitted from the surface of the material at a different wavelength, thus changing its color.
<span>X-Rays and Gamma Rays
</span>X-rays and gamma rays also transfer their energy to matter though chance encounters with electrons and atomic nuclei. However, X-rays and gamma rays have enough energy to do more than just make the electrons vibrate. When these high energy rays encounter an atom, the result is an ejection of energetic electrons from the atom or the excitation of electrons. The term "excitation" is used to describe an interaction where electrons acquire energy from a passing charged particle but are not removed completely from their atom. Excited electrons may subsequently emit energy in the form of x-rays during the process of returning to a lower energy state.
1. distance plus direction
displacement
2. 30 m/s northwest, for example
velocity
3. total distance divided by total time
average speed
4. what an object going around a curve at a constant speed is
undergoing
acceleration
5. 30 m/s, for example
speed
6. any non-directional quantity
scalar
During a rodeo, a clown runs 7.7 m north, turns 49.9 degrees east of north, and runs 6.4 m. Then after waiting for the bull to come near, the clown turns due east and runs 19.8 m to exit the arena. The magnitude of the clown’s displacement is 27 m.
<u>Explanation:
</u>
As the clown is running in the north direction for about 7.7 m and then he turns 49.9 degrees east of north. In the east of north, he covers a distance of 6.4 m and then turns east to exit the arena after covering a distance of 19.8 m. Let’s have a simple diagram to easily understand the problem.
In first step, the clown runs 7.7 m in north direction, so the image will be as in fig 1. Then he takes a direction of north east and covers a distance of 6.4 m, so the image will be modified as in fig 2. Then after the bull comes, he turns east and runs 19.8 m to exit the arena, so the image will be as in figure 3.
So, the extension of North line and the East line at a point shown as the dotted line in the above image, forms the total displacement as the hypotenuse of a right angled triangle. The extended dotted lines is nothing but the horizontal and vertical components of the angle 49.9 degree.
By using Pythagoras theorem, the total displacement can be found as
Similarly, the adjacent side of this imaginary triangle is the distance covered by the clown in the North direction.
Thus, the total displacement covered by the clown is
Thus, the total displacement by the clown is 27 m.