<span>Here I think you have to find the velocity in x and y components where x is east and y is north
So as air speed indicator shows the negative speed in y component and adding it in
air speed while multiplying with the direction component we will get the velocity as velocity is a vector quantity so direction is also required
v=-28 m/s y + 18 m/s (- x/sqrt(2) - y/sqrt(2))
solving
v= -12.7 m/s x-40.7 m/s y
if magnitude of velocity or speed is required then
speed= sqrt(12.7^2 + 40.7^2)
speed= 42.63 m/s
if angle is asked
angle = arctan (40.7/12.7)
angle = 72.67 degrees south of west</span>
Neap tide = tide where there is the least difference between high and low water levels
Spring tide = tide where there is the greatest difference between high and low water levels
Equator = an imaginary line drawn around earth dividing it into northern and southern hemispheres
Seasons = the divisions of the year marked by specific weather patterns and daylight hours.
Hope this helps!
Answer:
The velocity of the particle from T = 0 s to T = 4 s is;
0.5 m/s
Explanation:
The given parameters from the graph are;
The initial displacement (covered) at time, t₁ = 0 s is x₁ = 1 m
The displacement covered at time, t₂ = 4 s is x₂ = 3 m
The graph of distance to time, from time t = 0 to time t = 4 is a straight line graph, with the velocity given by the rate of change of the displacement to the time which is dx/dt which is also the slope of the graph given as follows;
The velocity of the particle from t = 0 s to t = 4 s = 1/2 m/s = 0.5 m/s.
it is c. safest when passing a large truck
hope this helps* :)
