Answer:
The human eye and brain together translate light into color. Light receptors within the eye transmit messages to the brain, which produces the familiar sensations of color. ... Rather, the surface of an object reflects some colors and absorbs all the others. We perceive only the reflected colors.
Explanation:
Answer:
1.48×10⁻⁷ Newtons
Explanation:
From the question,
According to newton's law of universal gravitation.
F = Gmm'/r²........................ Equation 1
F = gravitational force, G = gravitational constant, m = mass of the first ball, m' = mass of the second ball, r = distance between the balls.
Given: m = m' = 8 kg, r = 17 cm = 0.17 m,
Constant : G = 6.67×10⁻¹¹ Nm²/kg²
Substitute these values into equation 1
F = (6.67×10⁻¹¹×8×8)/(0.17²)
F = 1.48×10⁻⁷ N
Answer:
W = 8.01 × 10^(-17) [J]
Explanation:
To solve this problem we need to know the electron is a subatomic particle with a negative elementary electrical charge (-1,602 × 10-19 C), The expression to calculate the work is given by:
W = q*V
where:
q = charge = 1,602 × 10^(-19) [C]
V = voltage = 500 [V]
W = work [J]
W = 1,602 × 10^(-19) * 500
W = 8.01 × 10^(-17) [J]
Answer:
1) The greatest height attained by the ball equals 20.387 meters.
2) The time it takes for the ball to reach 15 meters approximately equals 1 second.
Explanation:
The greatest height will be attained when the ball stop's in the air and starts falling back to the earth.
thus using third equation of kinematics we obtain the height attained as

where
'v' is the final speed of the ball
'u' is the initial speed of the ball
'a' is the acceleration that the ball is under which in this case equals 9.81 
's' is the distance it covers
Thus for maximum height applying the values in the equation we get

Using the same equation we can find the speed of the ball when it reaches 15 meters of height as

the time it takes to reduce the velocity to this value can be found by first equation of kinematics as
