Given Information:
Initial speed = u = 3.21 yards/s
Acceleration = α = 1.71 yards/s²
Final speed = v = 7.54 yards/s
Required Information:
Distance = s = ?
Answer:
Distance = s = 13.61
Explanation:
We are given the speeds and acceleration of the runner and we want to find out how much distance he covered before being tackled.
We know from the equations of motion,
v² = u² + 2αs
Where u is the initial speed of the runner, v is the final speed of the runner, α is the acceleration of the runner and s is the distance traveled by the runner.
Re-arranging the above equation for distance yields,
2αs = v² - u²
s = (v² - u²)/2α
s = (7.54² - 3.21²)/2×1.71
s = 46.55/3.42
s = 13.61 yards
Therefore, the runner traveled a distance of 13.61 yards before being tackled.
At what angle torque is half of max
Answer:
va = 4.79 m/s
vb = 1.29 m/s
Explanation:
Momentum is conserved:
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
(3.00) (0) + (6.50) (3.50) = (3.00) v₁ + (6.50) v₂
22.75 = 3v₁ + 6.5v₂
For an elastic collision, kinetic energy is conserved.
½ m₁u₁² + ½ m₂u₂² = ½ m₁v₁² + ½ m₂v₂²
m₁u₁² + m₂u₂² = m₁v₁² + m₂v₂²
(3.00) (0)² + (6.50) (3.50)² = (3.00) v₁² + (6.50) v₂²
79.625 = 3v₁² + 6.5v₂²
Two equations, two variables. Solve with substitution:
22.75 = 3v₁ + 6.5v₂
22.75 − 3v₁ = 6.5v₂
v₂ = (22.75 − 3v₁) / 6.5
79.625 = 3v₁² + 6.5v₂²
79.625 = 3v₁² + 6.5 ((22.75 − 3v₁) / 6.5)²
79.625 = 3v₁² + (22.75 − 3v₁)² / 6.5
517.5625 = 19.5v₁² + (22.75 − 3v₁)²
517.5625 = 19.5v₁² + 517.5625 − 136.5v₁ + 9v₁²
0 = 28.5v₁² − 136.5v₁
0 = v₁ (28.5v₁ − 136.5)
v₁ = 0 or 4.79
We know v₁ isn't 0, so v₁ = 4.79 m/s.
Solving for v₂:
v₂ = (22.75 − 3v₁) / 6.5
v₂ = 1.29 m/s
The strong nuclear force holds together the quarks and gluons within protons and neutrons. It also holds together the protons and neutrons within the nucleus of the atom. The strong nuclear force holds together the quarks and gluons within protons and neutrons.
So it would be A.
