Answer:
Part a)

Part b)

Part c)

Explanation:
As we know that acceleration is rate of change in velocity of the object
So here we know that


Part a)
differentiate x and y two times with respect to time to find the acceleration






Now the acceleration of the object is given as

at t= 1.1 s we have

now the net force of the object is given as



now magnitude of the force will be

Part b)
Direction of the force is given as



Part c)
For velocity of the particle we have




now at t = 1.1 s

now the direction of the velocity is given as



Answer
given,
time = 10 s
ship's speed = 5 Km/h
F = m a
a is the acceleration and m is mass.
In the first case
F₁=m x a₁
where a₁ = difference in velocity / time
F₁ is constant acceleration is also a constant.
Δv₁ = 5 x 0.278
Δv₁ = 1.39 m/s

a₁ = 0.139 m/s²
F₂ =m x a₂
F₃ = F₂ + F₁
Δv₃ = 19 x 0.278
Δv₃ = 5.282 m/s
a₃=Δv₂ / t

a₃ = 0.5282 m²/s
m a₃=m a₁ + m a₂
a₃ = a₂ + a₁
0.5282 = a₂ + 0.139
a₂=0.3892 m²/s
F₂ = m x 0.3892...........(1)
F₁ = m x 0.139...............(2)
F₂/F₁
ratio = 
ratio = 2.8
Answer:
The appropriate response is "
". A further explanation is described below.
Explanation:
The torque (
) produced by the force on the dam will be:
⇒ 
On applying integration both sides, we get
⇒ 
⇒ 
⇒ ![=pgL[\frac{h^3}{2} -\frac{h^3}{3} ]](https://tex.z-dn.net/?f=%3DpgL%5B%5Cfrac%7Bh%5E3%7D%7B2%7D%20-%5Cfrac%7Bh%5E3%7D%7B3%7D%20%5D)
⇒ 