D) waves are used to transmit the rail signal though the air. these waves are encoded at different frequencies for different stations
Answer: V = 504m/a
F = 4N
Explanation: please find the attached file for the solution
Answer:
The velocity of the truck after this elastic collision is 15.7 m/s
Explanation:
It is given that,
Mass of the car, 
Mass of the truck, 
Initial velocity of the car, 
Initial velocity of the truck, u₂ = 0
After the collision the velocity of the car is, v₁ = -11 m/s
Let v₂ is the velocity of the truck after this elastic collision. Using the conservation of momentum as :
So, the velocity of the truck after this elastic collision is 15.7 m/s. Hence, the correct option is (c).
A)<span>
dQ = ρ(r) * A * dr = ρ0(1 - r/R) (4πr²)dr = 4π * ρ0(r² -
r³/R) dr
which when integrated from 0 to r is
total charge = 4π * ρ0 (r³/3 + r^4/(4R))
and when r = R our total charge is
total charge = 4π*ρ0(R³/3 + R³/4) = 4π*ρ0*R³/12 = π*ρ0*R³ / 3
and after substituting ρ0 = 3Q / πR³ we have
total charge = Q ◄
B) E = kQ/d²
since the distribution is symmetric spherically
C) dE = k*dq/r² = k*4π*ρ0(r² - r³/R)dr / r² = k*4π*ρ0(1 -
r/R)dr
so
E(r) = k*4π*ρ0*(r - r²/(2R)) from zero to r is
and after substituting for ρ0 is
E(r) = k*4π*3Q(r - r²/(2R)) / πR³ = 12kQ(r/R³ - r²/(2R^4))
which could be expressed other ways.
D) dE/dr = 0 = 12kQ(1/R³ - r/R^4) means that
r = R for a min/max (and we know it's a max since r = 0 is a
min).
<span>E) E = 12kQ(R/R³ - R²/(2R^4)) = 12kQ / 2R² = 6kQ / R² </span></span>
They are similar to the earth. Not relative to the outer planets
