Answer: D
Explanation:
Because the energy from the first ball immediately impacts the other balls.
Answer:
See below
Explanation:
KE = 1/2 m v^2 multiply both sides by 2
2 (KE) = mv^2 divide both sides by m
2(KE) / m = v^2 sqrt both sides
√ [(2KE)/m ] = v
Answer:
electric field E = (1 /3 e₀) ρ r
Explanation:
For the application of the law of Gauss we must build a surface with a simple symmetry, in this case we build a spherical surface within the charged sphere and analyze the amount of charge by this surface.
The charge within our surface is
ρ = Q / V
Q ’= ρ V
'
The volume of the sphere is V = 4/3 π r³
Q ’= ρ 4/3 π r³
The symmetry of the sphere gives us which field is perpendicular to the surface, so the integral is reduced to the value of the electric field by the area
I E da = Q ’/ ε₀
E A = E 4 πi r² = Q ’/ ε₀
E = (1/4 π ε₀) Q ’/ r²
Now you relate the fraction of load Q ’with the total load, for this we use that the density is constant
R = Q ’/ V’ = Q / V
How you want the solution depending on the density (ρ) and the inner radius (r)
Q ’= R V’
Q ’= ρ 4/3 π r³
E = (1 /4π ε₀) (1 /r²) ρ 4/3 π r³
E = (1 /3 e₀) ρ r
Answer:
S = 11.025 m
Explanation:
Given,
The time taken by the pebble to hit the water surface is, t = 1.5 s
Acceleration due to gravity, g = 9.8 m/s²
Using the II equations of motion
S = ut + 1/2 gt²
Here u is the initial velocity of the pebble. Since it is free-fall, the initial velocity
u = 0
Therefore, the equation becomes
S = 1/2 gt²
Substituting the given values in the above equation
S = 0.5 x 9.8 x 1.5²
= 11.025 m
Hence, the distance from the edge of the well to the water's surface is, S = 11.025 m
To solve this problem we will apply the principle of buoyancy of Archimedes and the relationship given between density, mass and volume.
By balancing forces, the force of the weight must be counteracted by the buoyancy force, therefore
Here,
m = mass
g =Gravitational energy
The buoyancy force corresponds to that exerted by water, while the mass given there is that of the object, therefore
Remember the expression for which you can determine the relationship between mass, volume and density, in which
In this case the density would be that of the object, replacing
Since the displaced volume of water is 0.429 we will have to
The density of water under normal conditions is 
, so
The density of the object is 
