<span>The distance between the sources of a two point surface
interference determines the number of lines that can be created for the nodal
and anti-nodal lines. When the sources are moved closer together, the number of
lines becomes minimal. Thus, the spaces between the lines become bigger. The
answer to this item is letter A. The spacing increases. </span>
Answer:
341 m/s
Explanation:
Use Bernoulli equation:
P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂
Assuming no elevation change, h₁ = h₂.
P₁ + ½ ρ v₁² = P₂ + ½ ρ v₂²
The velocity of the air at the nose is 0 m/s, so:
P₁ = P₂ + ½ ρ v₂²
ΔP = ½ ρ v₂²
Plugging in values:
75000 Pa = ½ (1.29 kg/m³) v²
v = 341 m/s
The kinetic energy of the proton is 3.4 kev
1 kev = 1.602 × 10^-16 joules
therefore 3.4 kev is equivalent to;
3.4 × (1.602 ×10^-16)= 5.4468 × 10^-16 J
Kinetic energy is calculated by the formula 1/2mv² where m is the mass and v is the velocity.
Therefore V = √((2 × ( 5.4468×10^-16))/ (1.67 ×10^-27))
= 8.077 × 10^5 m/s
To solve the problem it is necessary to use the concepts of Orbital Speed considering its density, and its angular displacement.
In general terms the Orbital speed is described as,

PART A) If the orbital speed of a star in this galaxy is constant at any radius, then,




PART B) This time we have
, where
is the angular velocity (constant at this case)




PART C) If the total mass interior to any radius r is a constant,




20% would be probably right cuz 100% is total and 80+ what gives u 100 which is 20