B
add them all by direction
13 East
10 West
subtract difference
3 E
Answer:
The ocean is 6485.6m deep when measured from the vessel
Explanation:
v=1474m/s
t=8.88s
let d represent distance from the vessel to the ocean bottom.
an echo travels a distance equivalent to 2d, that is to and fro after it reflects from the obstacle.
d= 6485.6m
Firstly they have a acceleration downwards due the force downwards due they gravitational field acting on it's mass.
as it falls it gains speed, and as it gains speed the air Resistance which is a upward force actin on the drop increases, eventually the rain drop's upward and downward forces are balanced and hence there is no RESULTANT force therefore no acceleration, so the drops falls in constant speed (terminal verlocity is a better term)
Are you wondering that why is the raindrop still moving given that the forces are balanced? If so according to Newton's 1st law an object will keep moving or Remain at rest until a RESULTANT force acts on it.
Answer:
Explanation:
Given that,
Mass of block
M = 2kg
Spring constant k = 300N/m
Velocity v = 12m/s
At t = 0, the spring is neither stretched nor compressed. Then, it amplitude is zero at t=0
xo = 0
It velocity is 12m/s at t=0
Then, it initial velocity is
Vo = 12m/s
Then, amplitude is given as
A = √[xo + (Vo²/ω²)]
Where
xo is the initial amplitude =0
Vo is the initial velocity =12m/s
ω is the angular frequency and it can be determine using
ω = √(k/m)
Where
k is spring constant = 300N/m
m is the mass of object = 2kg
Then,
ω = √300/2 = √150
ω = 12.25 rad/s²
Then,
A = √[xo + (Vo²/ω²)]
A = √[0 + (12²/12.5²)]
A = √[0 + 0.96]
A = √0.96
A = 0.98m
