Answer:
Approximately
, assuming that the volume of these two charged objects is negligible.
Explanation:
Assume that the dimensions of these two charged objects is much smaller than the distance between them. Hence, Coulomb's Law would give a good estimate of the electrostatic force between these two objects regardless of their exact shapes.
Let
and
denote the magnitude of two point charges (where the volume of both charged object is negligible.) In this question,
and
.
Let
denote the distance between these two point charges. In this question,
.
Let
denote the Coulomb constant. In standard units,
.
By Coulomb's Law, the magnitude of electrostatic force (electric force) between these two point charges would be:
.
Substitute in the values and evaluate:
.
The cornea is responsible of refraction light 1/3 in eye.
<h3>What is the function of the cornea?</h3>
In addition to the protective function, it plays a fundamental role in the formation of vision. Transparent, it works like a lens over the iris, focusing light from the pupil towards the retina.
Normally, the cornea and lens deflect (refract) incoming light rays, focusing them on the retina. The shape of the cornea is fixed, but the lens changes shape to focus on objects at different distances from the eye.
See more about cornea at brainly.com/question/2297282
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Answer:
(a) the speed of the block after the bullet embeds itself in the block is 3.226 m/s
(b) the kinetic energy of the bullet plus the block before the collision is 500J
(c) the kinetic energy of the bullet plus the block after the collision is 16.13J
Explanation:
Given;
mass of bullet, m₁ = 0.1 kg
initial speed of bullet, u₁ = 100 m/s
mass of block, m₂ = 3 kg
initial speed of block, u₂ = 0
Part (A)
Applying the principle of conservation linear momentum, for inelastic collision;
m₁u₁ + m₂u₂ = v(m₁ + m₂)
where;
v is the speed of the block after the bullet embeds itself in the block
(0.1 x 100) + (3 x 0) = v (0.1 + 3)
10 = 3.1v
v = 10/3.1
v = 3.226 m/s
Part (B)
Initial Kinetic energy
Ki = ¹/₂m₁u₁² + ¹/₂m₂u₂²
Ki = ¹/₂(0.1 x 100²) + ¹/₂(3 x 0²)
Ki = 500 + 0
Ki = 500 J
Part (C)
Final kinetic energy
Kf = ¹/₂m₁v² + ¹/₂m₂v²
Kf = ¹/₂v²(m₁ + m₂)
Kf = ¹/₂ x 3.226²(0.1 + 3)
Kf = ¹/₂ x 3.226²(3.1)
Kf = 16.13 J