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:
0.66c
Explanation:
Use length contraction equation:
L = L₀ √(1 − (v²/c²))
where L is the contracted length,
L₀ is the length at 0 velocity,
v is the velocity,
and c is the speed of light.
900 = 1200 √(1 − (v²/c²))
3/4 = √(1 − (v²/c²))
9/16 = 1 − (v²/c²)
v²/c² = 7/16
v = ¼√7 c
v ≈ 0.66 c
Answer:
Explanation:
The formula to determine the size of a capillary tube is
h = 2•T•Cos θ / r•ρ•g
Where
h = height of liquid level
T = surface tension
r = radius of capillary tube
ρ = density of liquid
θ = angle of contact = 0°
g =acceleration due to gravity=9.81m/s²
The liquid is water then,
ρ = 1000 kg / m³
Given that,
T = 0.0735 N/m
h = 0.25mm = 0.25 × 10^-3m
Then,
r = 2•T•Cos θ / h•ρ•g
r = 2 × 0.0735 × Cos0 / 2.5 × 10^-3 × 1000 × 9.81
r = 5.99 × 10^-3m
Then, r ≈ 6mm
The radius of the capillary tube is 6mm
So, the minimum size is
Volume = πr²h
Volume = π × 6² × 0.25
V = 2.83 mm³
The minimum size of the capillary tube is 2.83mm³
Good luck with solving this
An object need to move in a straight line in the same direction in equal intervals of time in order for total distance traveled and displacement to be equal.