Answer: Brightness consistency
There are three types of perceptual consistency
Types of Perceptual Constancy: Shape, Size, and Brightness Size constancy
Since the moon and sun affect light, brightness consistency is occurring.
Brightness constancy is our visual ability to perceive objects as having the same level of brightness even though the level of lighting changes.
Answer:
Longest wavelength, lowest intensity
Explanation:
Answer:
The answer is letter b. All of these should be considered when deciding on a report format.
Explanation:
A Professional Report is a type of formal document about a topic or information that is intended for a specific audience or purpose. The report's style of writing needs a lot of knowledge from the writer. Oftentimes, it involves the following important elements: <em>Title, Summary, Body, Discussion, Conclusion and Recommendation. </em>
The writer should write according to his target audience and purpose. He also needs to consider the length of his report, as well as the suitable words and sentences that he should use.
Thus, all of the choices are important in writing a professional report. So, the answer is letter b.
Answer:
Explanation:
Given that,
Weight of jet
W = 2.25 × 10^6 N
It is at rest on the run way.
Two rear wheels are 16m behind the front wheel
Center of gravity of plane 10.6m behind the front wheel
A. Normal force entered on the ground by front wheel.
Taking moment about the the about the real wheel.
Check attachment for better understanding
So,
Clock wise moment = anti-clockwise moment
W × 5.4 = N × 16
2.25 × 10^6 × 5.4 = 16•N
N = 2.25 × 10^6 × 5.4 / 16
N = 7.594 × 10^5 N
B. Normal force on each of the rear two wheels.
Using the second principle of equilibrium body.
Let the rear wheel normal be Nr and note, the are two real wheels, then, there will be two normal forces
ΣFy = 0
Nr + Nr + N — W = 0
2•Nr = W—N
2•Nr = 2.25 × 10^6 — 7.594 × 10^5
2•Nr = 1.491 × 10^6
Nr = 1.491 × 10^6 / 2
Nr = 7.453 × 10^5 N
Answer:
The electric flux is 
Explanation:
Given:
- Radius of the disc R=0.50 m
- Angle made by disk with the horizontal

- Magnitude of the electric Field

The flux of the Electric Field E due to the are dA in space can be found out by using Gauss Law which is as follows

where
is the total Electric Flux- E is the Electric Field
- dA is the Area through which the electric flux is to be calculated.
Now according to question we have

Hence the electric flux is calculated.