1). From the frame of reference of a passenger on the airplane looking out of his window, the tree appears to be moving, at roughly 300 miles per hour toward the left of the picture.
2). The SI unit best suited to measuring the height of a building is the meter.
3). 'Displacement' is the straight-line distance and direction from the start-point to the end-point, regardless of the path that was followed to get there.
The ball started out in the child's hand, and it ended up 2 meters away from her in the direction of the wall. So the displacement of the ball from the beginning to the end of the story is: 2 meters toward the wall.
Answer: Gravity
Explanation: Gravity is pulling down on the ball, making it stay on the floor
Answer:

Explanation:
As in any sample you will have 75.8% of Cl-35 iosotopes and 24.3% of Cl-37 iosotopes you can get the average atomic mass as:

Answer:
Ionic bond
Explanation:
Also called electrovalent bond, type of linkage formed from the electrostatic attraction between oppositely charged ions in a chemical compound.
Hope this helps! brainliest welcomed! :)
Answer:
<h3>a.</h3>
- After it has traveled through 1 cm :

- After it has traveled through 2 cm :

<h3>b.</h3>
- After it has traveled through 1 cm :

- After it has traveled through 2 cm :

Explanation:
<h2>
a.</h2>
For this problem, we can use the Beer-Lambert law. For constant attenuation coefficient
the formula is:

where I is the intensity of the beam,
is the incident intensity and x is the length of the material traveled.
For our problem, after travelling 1 cm:




After travelling 2 cm:




<h2>b</h2>
The optical density od is given by:
.
So, after travelling 1 cm:




After travelling 2 cm:



