Answer:
4-mid ocean ridge at divergent plate boundaries.
Answer:
The first interval is walked slowly, this is a straight line with a small slope
Second interval stops, which gives a horizontal line, indicating the same position
Third interval, walk back, straight downhill
Explanation:
In this problem we have a uniform movement, this means that the acceleration in each intervals
x = v t
The first interval is walked slowly, this is a straight line with a small slope
Second interval stops, which gives a horizontal line, indicating the same position
Third interval, walk back, straight downhill
Answer:
Explanation:
Let the time period of A and B be Ta and Tb respectively. Similarly , amplitude of A and B be Aa and Ab respectively.
Given ,
Tb = 2 Ta
Ab = 2 Aa
maximum speed of a particle in SHM = ω A , where ω is angular velocity and A is amplitude .
ω = 2π / T
If maximum speed of A be Va
Va = ω A
= 2π( Aa / Ta )
If maximum speed of B be Vb
Vb = ω A
= 2π (Ab / Tb)
Va / Vb = (2π Aa / Ta) x (Tb / 2π Ab)
= (Aa / Ab) x (Tb / T a )
= .5 x 2 = 1
Va = Vb
Their maximum velocities will be equal.