When light crosses the interface between a medium with higher refractive index and a medium with lower refractive index, there is a maximum value of the angle of incidence after which there is no refraction, but all the light is reflected, and this maximum value is called critical angle.
The critical angle is given by
where n1 is the refractive index of the first medium while n2 is the refractive index of the second medium. In our problem, n1=1.33 and n2=1.58, so the critical angle is
Answer:
Evaporation
Explanation:
Evaporation is a form of mass tranfer phenomena where by water are moved from the earth surface into the atmosphere as vapours,it is path of the water cycle a decription of the path moved by land water until it turns into rain, humidity,air and temperature are factors that influence evaporation though evaporation can happen at all temperature
Answer:
a = 1.055 x 10¹⁷ m/s²
Explanation:
First, we will find the force on electron:
where,
F = Force = ?
E = Electric Field = 6 x 10⁵ N/C
q = charge on electron = 1.6 x 10⁻¹⁹ C
Therefore,
F = 9.6 x 10⁻¹⁴ N
Now, we will calculate the acceleration using Newton's Second Law:
where,
a = acceleration = ?
m = mass of electron = 9.1 x 10⁻³¹ kg
therefore,
<u>a = 1.055 x 10¹⁷ m/s²</u>
Answer:
d) An object that is speeding up always has a positive acceleration, regardless of the direction it travels.
Explanation:
a ) a) An object that is slowing down while traveling in the negative x-direction always has a positive acceleration.
It has negative acceleration in the negative x-direction.
b) An object that is speeding up while traveling in the negative x-direction always has a positive acceleration.
It has a positive acceleration in the negative x-direction'
c) An object that is slowing down always has a negative acceleration, regardless of the direction it travels.
It has a positive acceleration in opposite direction.
e ) An object that is slowing down always has a positive acceleration, regardless of the direction it travels.
It has a positive acceleration only in opposite direction .
Answer:
Range of projectile, R
For projection above ground surface, the range of the angle of projection with respect to horizontal direction, θ, is 0° ≤ θ ≤ 90° and the corresponding range of 2θ is 0° ≤ 2θ ≤ 180°.S
hope you're looking for this.
