Answer:
Speed is same as that before it entered glass.
Explanation:
Given:
A light ray enters and passes through the glass as shown in the diagram.
We have to analyze its speed.
Speed of light in air is 
and speed of light in glass is 
Whenever a light ray enters a glass block or slab there is bending of light at the interface of the two media.
So speed of light will decrease in glass medium and again it passes to the air.
So
Speed of light in air will again increase or will be equivalent to the earlier speed when it was entering the glass block.
Finally
Speed is same as that before it entered glass as it in the same medium (air).
Answer:
Our eyes are most sensitive to the wavelengths corresponding to the yellow and green colors of the spectrum. Flashy signs and some fire engines are painted in a yellowish-green color to attract our attention.
The answer is; pressure
The sound is a longitudinal wave meaning the particles vibrate parallel to the direction of the wave. Sound waves, therefore, produce compression (akin to the crest in a transverse wave) and rarefaction regions (akin to a trough in a transverse wave) as its energy is propagated in the medium.
Given parameters:
Mass of object = 6.7kg
Velocity = 8m/s
Unknown parameter:
Kinetic energy = ?
Energy is defined as the ability to do work. There are two forms of energy;
Kinetic and potential energy.
Kinetic energy is the energy due to the motion of a body. Whereas, potential energy is the energy due to the position of a body usually at rest.
Kinetic energy is mathematically expressed as;
Kinetic energy = 
where m is the mass of the body
v is the velocity of the body
Since we have been given both mass and velocity, input the parameter to solve for the unknown;
Kinetic energy = 
x 6.7 x 8² = 214.4J
So the kinetic energy of the body is 214.4J
Answer:
v_f = 24.3 m / s
Explanation:
A) In this exercise there is no friction so energy is conserved.
Starting point. On the roof of the building
Em₀ = K + U = ½ m v₀² + m g y₀
Final point. On the floor
Em_f = K = ½ m v_f²
Emo = Em_g
½ m v₀² + m g y₀ = ½ m v_f²
v_f² = v₀² + 2 g y₀
let's calculate
v_f = √(10² + 2 9.8 25)
v_f = 24.3 m / s
