The net speed due west is = distance traveled in west / time taken = 120/0.5 = 240 km/h.
so airspeed due west is = net speed - speed of plane = 240-220= 20 km/h.
airspeed due south is = distance traveled in west / time taken= 20/0.5= 40 km/h.
the magnitude of the wind velocity = √[(airspeed due south )² + (airspeed due west)²] = √ ( 40^2 + 20^2 ) = 44.72 km/h
the angle of airspeed south of west is tan⁻¹ ( airspeed due south / airspeed due west )= tan⁻¹(40/20)=63.43 degrees.
if wind velocity is 40 km/h due south, her velocity should have 20 km/h component in north.
so component west = sqrt ( 220^2 - 40^2 ) = 216.33 km/h.
the angle north of west is arctan( 40/216.33 ) = 10.47 degrees.
no its nuclear fusion requires very high temp
Explanation:
j took the test trust me :)
Answer:
the constant force applied by Kitty is 8 N.
Explanation:
Given;
work done by Kitty, W = 32 J
distance moved by Kitty's trolley, d = 4 m
Let the constant force applied by Kitty = F
The work done by Kitty is calculated as follows;
W = F x d
F = W / d
F = 32 / 4
F = 8 N.
Therefore, the constant force applied by Kitty is 8 N.
(2nd variant, 7kg is 3 times faster than 21kg)