Answer:
is a reflection.
The image is real light rays actually focus at the image location). As the object moves towards the mirror the image location moves further away from the mirror and the image size grows (but the image is still inverted). When the object is that the focal point, the image is at infinity.
Explanation:
Answer:
D
Explanation:
I’m pretty sure it’s correct but I don’t really know. Just trying to pass science
Answer:
A. 
B. 
C. 
Explanation:
Given:
- spring constant,

- mass attached,

A)
for a spring-mass system the frequency is given as:



B)
frequency is given as:



C)
Time period of a simple harmonic motion is given as:


Answer:
The order of magnitude of the distance from the sun to Earth is 10⁸ km.
Explanation:
The order of magnitude of the distance from the sun to Earth can be calculated as follows:

Where:
c: is the speed of light = 3x10⁸ m/s
t: is the time = 8 min
Hence, the distance is:

Therefore, the order of magnitude of the distance from the sun to Earth is 10⁸ km.
I hope it helps you!
Answer:
if we measure the change in height of the gas within the had and obtain a straight line in relation to the depth we can conclude that the air complies with Boye's law.
Explanation:
The air in the tube can be considered an ideal gas,
P V = nR T
In that case we have the tube in the air where the pressure is P1 = P_atm, then we introduce the tube to the water to a depth H
For pressure the open end of the tube is
P₂ = P_atm + ρ g H
Let's write the gas equation for the colon
P₁ V₁ = P₂ V₂
P_atm V₁ = (P_atm + ρ g H) V₂
V₂ = V₁ P_atm / (P_atm + ρ g h)
If the air obeys Boyle's law e; volume within the had must decrease due to the increase in pressure, if we measure the change in height of the gas within the had and obtain a straight line in relation to the depth we can conclude that the air complies with Boye's law.
The main assumption is that the temperature during the experiment does not change