Answer:
vf = 0
Explanation:
Since the initial height hi = 0, we can rewrite the energy equation as
vf^2 = vi^2 - 2ghf = (10 m/s)^2 - 2(10 m/s^2)(5 m) = 0
Therefore, his final velocity vf is
vf = 0
Answer
given,
ω₁ = 0 rev/s
ω₂ = 6 rev/s
t = 11 s
Using equation of rotational motion
The angular acceleration is
ωf - ωi = α t
11 α = 6 - 0
= 0.545 rev/s²
The angular displacement
θ₁= ωi t + (1/2) α t²
θ₁= 0 + (1/2) (0.545)(11)^2
θ₁= 33 rev
case 2
ω₁ = 6 rev/s
ω₂ = 0 rev/s
t = 14 s
Using equation of rotational motion
The angular acceleration is
ωf - ωi = α t
14 α = 0 - 6
= - 0.428 rev/s²
The angular displacement
θ₂= ωi t + (1/2) α t²
θ₂= 6 x 14 + (1/2) (-0.428)(14)^2
θ₂= 42 rev
total revolution in 25 s is equal to
θ = θ₁ + θ₂
θ = 33 + 42
θ = 75 rev
Answer:
Electrons accelerated to high velocities travel in straight lines through an empty cathode ray tube and strike the glass wall of the tube, causing excited atoms to fluoresce or glow.
Answer:
Option B. Decreases
Explanation:
Coulomb's law states that:
F = Kq₁q₂ / r²
Where:
F => is the force of attraction between two charges
K => is the electrical constant.
q₁ and q₂ => are the two charges
r => is the distance apart.
From the formula:
F = Kq₁q₂ / r²
The force of attraction (F) is inversely proportional to the square of their separating distance (r).
This implies that as the distance between them increase, the force of attraction between the two charges will decrease and as the distance between two charges decrease, the force of attraction between them will increase.
Considering the question given above and the illustration given above, the force of attraction will decrease as their distance of separation increases.
Option B gives the right answer to the question.
Because they built:different
