you tend to push off with the foot still on the boat.
Newtons 3rd law of motion hold that every action has an equal and opposite reaction.
So, the amount of force you use on your back foot to push yourself onto the dock, has an equal end opposite amount of force going the other way from your sole of your foot, which pushes the boat the other way.
The same principle applies when you fire a cannon - in the pirate films, you see the body of the cannon forced back as it is fired.
Take it one step further, Henry VIII flagship, “Mary Rose fired all its cannons together one day in a broad side, the opposite force rolled the ship over and sank it.
Answer:
a. v₁ = 16.2 m/s
b. μ = 0.251
Explanation:
Given:
θ = 15 ° , r = 100 m , v₂ = 15.0 km / h
a.
To determine v₁ to take a 100 m radius curve banked at 15 °
tan θ = v₁² / r * g
v₁ = √ r * g * tan θ
v₁ = √ 100 m * 9.8 m/s² * tan 15° = 16.2 m/s
b.
To determine μ friction needed for a frightened
v₂ = 15.0 km / h * 1000 m / 1 km * 1h / 60 minute * 1 minute / 60 seg
v₂ = 4.2 m/s
fk = μ * m * g
a₁ = v₁² / r = 16.2 ² / 100 m = 2.63 m/s²
a₂ = v₂² / r = 4.2 ² / 100 m = 0.18 m/s²
F₁ = m * a₁ , F₂ = m * a₂
fk = F₁ - F₂ ⇒ μ * m * g = m * ( a₁ - a₂)
μ * g = a₁ - a₂ ⇒ μ = a₁ - a₂ / g
μ = [ 2.63 m/s² - 0.18 m/s² ] / (9.8 m/s²)
μ = 0.251
If your talking about the chemical equations like combustion, single displacement, etc. Then the equation would be double displacement:
AB + CD ---> AD + CB
Answer:
180^2 + 390^2 = force ^2 (Pythagoras) root of force^2 = 429.5N approx resultant force Acceleration = Force/Mass 429.5/270 = 1.5907 ms^-2 in a Southwesterly direction.
Explanation:
