Two opposite forces generate a force whose magnitude is the difference of the two magnitudes, and whose direction is the same of the strongest one.
We have 3320 newtons forward, and 550 backwards.
So, the net force is forward, and has magnitude 
If it were possible to move a star towards the earth then its apparent magnitude number would decrease while its absolute magnitude number would stay the same.
Definition of apparent magnitude:
The luminosity of a celestial body (such as a star) as observed from the earth compare absolute magnitude.
So for example, the apparent magnitude of the Sun is -26.7 and is the brightest celestial object we can see from Earth. However, if the Sun were 10 parsecs away, its apparent magnitude would be +4.7, only about as bright as Ganymede appears to us on Earth.
Definition of absolute magnitude:
Absolute magnitude is a measure of the luminosity of a celestial object on an inverse logarithmic astronomical magnitude scale.
To learn more about apparent magnitude here
brainly.com/question/2949443
#SPJ4
Answer:
12m
Explanation:
To obtain the answer to the question given, we must observe the characteristics of image formed by a plane mirror.
The image formed by a plane mirror have the following characteristics:
1. Laterally inverted.
2. Same distance as the object from the mirror.
3. Same height as the object.
4. Virtual.
With the above information, we can calculate the distance between the boy and his image as follow:
Initially:
Object distance (u) = 4m
Image distance (v) = 4m
The boy moved 2m away, therefore:
Object distance (u) = 2 + 4 = 6m
Image distanc(v) = 2 + 4 = 6m
The distance between the boy and his image will be the sum of his distance (u) and image distance (v) i.e (u + v)
The distance between the boy and his image = 6 + 6 = 12m
Therefore, the distance between the boy and his image is 12m.
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:
it's C
Explanation:
have a nice day............
