Aplane has an airspeed of 112 km km/h. it is flying on a bearing of 81 degrees° while there is a 28 km divided by h28 km/h wind out of the northeast (bearing 225degrees°). what are the ground speed and the bearing of the plane?

speed = 90.7934 km/hr

bearing = 88.55°

Explanation:

given data

plane speed v = 112 km/hr

bearing = 81°  

θ  = 90 - 81 = 9°

wind speed w = 28 km/hr

bearing (θ2)  =  225°

to find out

ground speed and the bearing of the plane

solution

we get here resultant ground speed of plane that is

speed = [ v cos(θ) + w cos(θ2)] i + [ v sin(θ) + w sin(θ2)] j

speed = [112 cos(9) + 28 cos(225)] i + [112 sin(9) + 28 sin(225)] j

speed = 90.822 i - 2.278 j

| speed | =

| speed | = 90.7934 km/hr

and

tan(θ) =  

θ = 1.4367°

bearing of the plan will be

bearing  = 90 - 1.4367

bearing = 88.55°

answer: option c. i.e. electric current flowing “downhill” from the negative electrode to the positive electrode in a battery

explanation:

the water analogy to the circuits define the pipes (water flows) to the electric current. now, we know that the electric current flows from the positive terminal to the circuit (as shown in the diagram attached).

hence, option c is the correct answer.

massxgravity x heght=weightxheight= pe

4900=490xh

h=10