Answer:
Bow Line
Explanation:
If the wind or current is pushing your boat away from the dock, bow line should be secured first.
1- We should cast off the bow and stern lines.
2-With the help of an oar or boat hook, keep the boat clear of the dock.
3-Leave the boat on its own for sometime and let the wind or current carry the boat away from the dock.
4 - As you see there is sufficient clearance, shift into forward gear and slowly leave the area.
Answer:
the Arrow X shows the direction of amplitude
Explanation:
As the amplitude is the maximum displacement of a wave from the mean position
Answer:
0.137m²
Explanation:
Pressure = Force/Area
Given
Force = 41,500N
Pressure = 3.00atm
since 1atm = 101325.00 N/m²
3atm = 3(101325.00)
3atm = 303,975N/m²
Pressure = 303,975N/m²
Get the area
Area = Force/Pressure
Area = 41500/303,975
Area = 0.137m²'
Hence the surface area of the inside of the tire is 0.137m²
Answer:
5.95 m
Explanation:
Given that the biggest loop is 40.0 m high. Suppose the speed at the top is 10.8 m/s and the corresponding centripetal acceleration is 2g
For the car to stick to the loop without falling down, at the top of the ride, the centripetal force must be equal to the weight of the car. That is,
(MV^2) / r = mg
V^2/ r = centripetal acceleration which is equal to 2g
2 × 9.8 = 10.8^2 / r
r = 116.64 /19.6
r = 5.95 m
Answer:
73.5 m/s
Explanation:
The position of the first ball is:
y = y₀ + v₀ t + ½ at²
y = h + (0)(18) + ½ (-9.8)(18)²
y = h − 1587.6
The position of the second ball is:
y = y₀ + v₀ t + ½ at²
y = h + (-v) (18−6) + ½ (-9.8)(18−6)²
y = h − 12v − 705.6
Setting the positions equal:
h − 1587.6 = h − 12v − 705.6
-1587.6 = -12v − 705.6
1587.6 = 12v + 705.6
882 = 12v
v = 73.5
The second ball is thrown downwards with a speed of 73.5 m/s
