Explanation:
|F|= B/2
B/2 = √(A²+B²+2ABcosθ) ------(1)
Since the resultant of A and B is perpendicular to Vector A
tan90°= BSinθ/(A+BCosθ)
(A+BCosθ)=0
Cosθ=-A/B ----(2)
Using equation (1)
B/2 = √(A²+B²+2ABcosθ)
B/2 = √(A²+B²+2AB×-A/B)
B/2=√(A²+B²-2A²)
B/2=√(B²-A²)
B²/4=B²-A²
A²=B²-B²/4
A²=3B²/4
A=√3B/2
Using equation (2)
Cosθ=-A/B
Cosθ=-[√3B/2]/B
Cosθ=-√3/2
θ= cos^-1 (-√3/2)
θ= 150°
The body of a porter carrying load on head on a level road uses more energy than is required when the porter is resting completely. In this way the porter is doing some work.
However, when we think of theoretical definition of work, there is no force resisting the horizontal movement of the load when carried on a level road, and therefore there is no work done when a porter carrying a load on his head walk along a level road.
When the porter is going down the stairs the gravitational energy acting on the load is assisting the downward movement of the load. In this way energy is released by downward movement of the load. Thus the porter is doing negative work on the load.
Answer:
Both transmit data or information.
Both get weaker when traveling long distances.
Answer:
if this is true or false, its true i think (not 100% sure)
Explanation:
