They are called mechanical waves
Define an x-y coordinate system such that
The positive x-axis = the eastern direction, with unit vector

.
The positive y-axis = the northern direction, with unit vector

.
The airplane flies at 340 km/h at 12° east of north. Its velocity vector is

The wind blows at 40 km/h in the direction 34° south of east. Its velocity vector is
![\vec{v}_{2} =40(cos(34^{o})\hat{i} - sin(24^{o})]\hat{j}) = 33.1615\hat{i} -22.3677\hat{j})](https://tex.z-dn.net/?f=%5Cvec%7Bv%7D_%7B2%7D%20%3D40%28cos%2834%5E%7Bo%7D%29%5Chat%7Bi%7D%20-%20sin%2824%5E%7Bo%7D%29%5D%5Chat%7Bj%7D%29%20%3D%2033.1615%5Chat%7Bi%7D%20-22.3677%5Chat%7Bj%7D%29)
The plane's actual velocity is the vector sum of the two velocities. It is

The magnitude of the actual velocity is
v = √(121.1615² + 306.0473²) = 329.158 km/h
The angle that the velocity makes north of east is
tan⁻¹ (306.04733/121.1615) = 21.6°
Answer:
The actual velocity is 329.2 km/h at 21.6° north of east.
The answer you are looking for is:
D.) It facilitates the moment of the current though a wire.
Hope that helps!!
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Answer:
Explanation:
potential energy of compressed spring
= 1/2 k d²
= 1/2 x 730 d²
= 365 d²
This energy will be given to block of mass of 1.2 kg in the form of kinetic energy .
Kinetic energy after crossing the rough patch
= 1/2 x 1.2 x 2.3²
= 3.174 J
Loss of energy
= 365 d² - 3.174
This loss is due to negative work done by frictional force
work done by friction = friction force x width of patch
= μmg d , μ = coefficient of friction , m is mass of block , d is width of patch
= .44 x 1.2 x 9.8 x .05
= .2587 J
365 d² - 3.174 = .2587
365 d² = 3.4327
d² = 3.4327 / 365
= .0094
d = .097 m
= 9.7 cm
If friction increases , loss of energy increases . so to achieve same kinetic energy , d will have to be increased so that initial energy increases so compensate increased loss .