Answer:
Explanation:
We know that the volume V for a sphere of radius r is

If we got an uncertainty
the formula for the uncertainty of V is:

We can calculate this uncertainty, first we obtain the derivative:


And using it in the formula:



The relative uncertainty is:



Using the values for the problem:

This is, a percent uncertainty of 4.77 %
Answer:
In the 1980s, astronomers turned to multiverse theory to explain the “luckily small” amount of dark energy in our universe. According to cosmologist's theories, this small amount of dark energy appeared to enable our universe to host life, while most universes in the multiverse couldn't.
Explanation:
The concept required to solve this problem is the optical relationship that exists between the apparent depth and actual or actual depth. This is mathematically expressed under the equations.

Where,
Depth of glass
Refraction index of water
Refraction index of glass
Refraction index of air
Depth of water
I enclose a diagram for a better understanding of the problem, in this way we can determine that the apparent depth in the water of the logo would be subject to



Therefore the distance below the upper surface of the water that appears to be the logo is 4.041cm
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