The best way to pull the boat in the launch lane is when the towing vehicle has reached the boat ramp as well as the trailer. It is best to retrieve the boat in this manner for it is more appropriate, convenient and it is a way of showing courtesy towards others because it is a proper manner.
1)
Answer:
Part 1)
H = 30.6 m
Part 2)
t = 2.5 s
Part 3)
t = 2.5 s
Part 4)

Explanation:
Part 1)
initial speed of the ball upwards

so maximum height of the ball is given by



Part 2)
As we know that final speed will be zero at maximum height
so we will have



Part 3)
Since the time of ascent of ball is same as time of decent of the ball
so here ball will same time to hit the ground back
so here it is given as
t = 2.5 s
Part 4)
since the acceleration due to earth will be same during its return path as well as the time of the motion is also same
so here its final speed will be same as that of initial speed
so we have

2)
Answer:
a = 9.76 m/s/s
Explanation:
As we know that the object is released from rest
so the displacement of the object in vertical direction is given as



3)
Answer:
v = 29.7 m/s
Explanation:
acceleration of the rocket is given as

time taken by the rocket
t = 0.33 min
final speed of the rocket is given as



4)
Answer:
Part 1)
y = 25.95 m
Part 2)
d = 6.72 m
Explanation:
Part 1)
As it took t = 2.3 s to hit the water surface
so here we will have



Part 2)
Distance traveled by it in horizontal direction is given as



Answer:corrosion (i believe)
Explanation:
Answer:
v = 10 V and E = 2 10³ N/C
Explanation:
The electrical potentials and the electric field at one point are related by the expression
ΔV = - ∫ E. dS
Where the bold indicates vector quantities, E is the electric field and S is the line of displacement of the load, in general displacement is perpendicular to the equipotential lines, which reduces the product scales to the ordinary product.
If the potential difference is the most usual that is V = 10 V, the electric field is
s = 0.5 cm = 0.5 10⁻² m
E = ΔV / S
E = 10/0.5 10⁻²
E = 2 10³ N / C
Answer:
0.368 cm
Explanation:
x = distance by which the mercury rise
d = depth of the water = 10 cm = 0.10 m
ρ = density of water = 1000 kgm⁻³
ρ' = density of mercury = 13600 kgm⁻³
P₀ = atmospheric pressure
Using equilibrium of pressure on both side
P₀ + ρ g d = P₀ + ρ' g (2x)
(1000) (0.10) = (13600) (2x)
x = 0.00368 m
x = 0.368 cm