Answer:
Speed of the boat, v = 4.31 m/s
Explanation:
Given that,
Height of the bridge, h = 32 m
The model boat is 11 m from the point of impact when the key was released, d = 11 m
Firstly, we will find the time needed for the boat to get in this position using second equation of motion as :

Here, u = 0 and a = g


t = 2.55 seconds
Let v is the speed of the boat. It can be calculated as :


v = 4.31 m/s
So, the speed of the boat is 4.31 m/s. Hence, this is the required solution.
If the planet earth has no land masses, the idealized zonal
precipitation pattern would likely have regions that are wet in the equator and
there will be more of mid-latitudes if the earth has no land masses at all and
it does not exist.
Answer:
Explanation:
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Answer:
24 m/s
Explanation:
Using v = u + at where u = initial velocity of the motorboat = 0 m/s (since the boat starts from rest), a = acceleration = 4 m/s², t = time = 6 s and v = velocity of the motorboat after 6.0 s.
Substituting the values of the variables into the equation, we have
v = u + at
= 0 m/s + 4 m/s² × 6.0 s
= 0 m/s + 24 m/s
= 24 m/s
Answer:
L=31.9 mm
δ = 0.22 mm
Explanation:
Given that
v= 14 m/s
ρ=997 kg/m³
μ= 0.891 × 10⁻3 kg/m·s
As we know that when Reynolds number grater than 5 x 10⁵ then flow will become turbulent.



L=0.0319 m
L=31.9 mm
The thickness of the boundary layer at that location L given as


δ = 0.00022 m
δ = 0.22 mm