Before going to answer this question first we have to understand reflection and laws of reflection.
Reflection is the optical phenomenon in which light will bounce back to the same medium from which it had originated .
Whenever a light ray will incident on a mirror or any reflecting surface, it will be reflected. The ray which falls on the reflecting surface is called incident ray and the ray which is reflected is called reflected ray.
Let us consider a normal to the point of incidence.The angle made by incident ray with the normal is called angle of incidence.Let it be denoted as[ i ]
The angle made by the reflected ray with the normal is called angle of incidence.Let it be denoted as [r]
There are two types of reflection.One is called regular and other one is called as irregular.The laws of reflection is valid for both the types of reflection.
There are two laws of reflection.
FIRST LAW -It states that the incident ray,reflected ray and the normal to the point of incidence,all lie in one plane.
SECOND LAW- It states that that the angle of incidence is equal to the angle of reflection irrespective of the type of reflection.i.e i =r
Hence the correct answer will be angle of reflection.
The two things that change are position and speed.
Answer:
a-
V= IR
9V = I ×( 12+6)
I = 9/ 18 A = 0.5 A
b
V=IR
240 = 6 A ×( 20 + R)
40 = 20 + R
R = 20 ohm
c
resultant resistance of the 2 parallel resistances= Ro
1/Ro = 1/ 5 + 1/ 20
1/Ro =( 20+5)/100
= 1/Ro = 1/4
Ro= 4 ohm
V=IR
V = 2A × ( 1+ 4 OHM)
V = 10V
d
equivalent resistance = Ro
1/Ro = 1/(2+8) + 1/(5+5)
1/Ro = 1/10 +1/10
2/10 = 1/ Ro
Ro= 10/2 = 5 ohm
V = IR
12V = I × 5Ohm
I=2.4 A
I'm going to say carrying capacity since carrying capacity is the maximum amount of organisms that can live in an ecosystem. So the reason for there being a maximum of organisms is because, like the question says, organisms usually produce more offspring than the ecosystem can support.
Answer:

Explanation:
F₁ = Force on one side of the jack
A₁ = Area of cross-section of one side of the jack
F₂ = Force on second side of the jack
A₂ = Area of cross-section of second side of the jack = 2 A₁
Using pascal's law



